problem solving with trigonometric ratios

Solve Problems Using Trigonometric Ratios

Problems on Trigonometric Ratios

Some trigonometric solutions based problems on trigonometric ratios are shown here with the step-by-step explanation.

1. If sin θ = 8/17, find other trigonometric ratios of <θ.

Let us draw a ∆ OMP in which ∠M = 90°.

Then sin θ = MP/OP = 8/17.

Let MP = 8k and OP = 17k, where k is positive.

By Pythagoras’ theorem, we get

Therefore, sin θ = MP/OP = 8k/17k = 8/17

cos θ = OM/OP = 15k/17k = 15/17

tan θ = Sin θ/Cos θ = (8/17 × 17/15) = 8/15

csc θ = 1/sin θ = 17/8

sec θ = 1/cos θ = 17/15 and

cot θ = 1/tan θ = 15/8.

2. If Cos A = 9/41, find other trigonometric ratios of ∠A.

Let us draw a ∆ ABC in which ∠B = 90°.

Then cos θ = AB/AC = 9/41.

Let AB = 9k and AC = 41k, where k is positive.

Therefore, sin A = BC/AC = 40k/41k = 40/41

cos A = AB/AC = = 9k/41k = 9/41

tan A = Sin A/Cos A = (40/41 × 41/9) = 40/9

csc A = 1/sin A = 41/40

sec A = 1/cos A = 41/9 and

cot A = 1/tan A = 9/40.

3. Show that the value of sin θ and cos θ cannot be more than 1.

We know, in a right angle triangle the hypotenuse is the longest side.

sin θ = perpendicular/hypotenuse = MP/OP < 1 since perpendicular cannot be greater than hypotenuse; sin θ cannot be more than 1.

Similarly, cos θ = base/hypotenuse = OM/OP < 1 since base cannot be greater than hypotenuse; cos θ cannot be more than 1.

4. Is that possible when A and B be acute angles, sin A = 0.3 and cos B = 0.7?

Since A and B are acute angles, 0 ≤ sin A ≤ 1 and 0 ≤ cos B ≤ 1, that means the value of sin A and cos B lies between 0 to 1. So, it is possible that sin A = 0.3 and cos B = 0.7

5. If 0° ≤ A ≤ 90° can sin A = 0.4 and cos A = 0.5 be possible?

Let us draw a ∆ ABC in which ∠B = 90° and ∠BAC = θ.

Then sin θ = BC/AC = 1/2.

Let BC = k and AC = 2k, where k is positive.

= 3√3/2 - 4 × 3√3/8

= 3√3/2 - 3√3/2

Hence, (3cos θ - 4 cos³ θ) = 0.

7. Show that sin α + cos α > 1 when 0 ° ≤ α ≤ 90°

From the right triangle MOP,

Sin α = perpendicular/ hypotenuse

Cos α = base/ hypotenuse

Now, Sin α + Cos α

= perpendicular/ hypotenuse + base/ hypotenuse

= (perpendicular + base)/hypotenuse, which is > 1, Since we know that the sum of two sides of a triangle is always greater than the third side.

8. If cos θ = 3/5, find the value of (5csc θ - 4 tan θ)/(sec θ + cot θ)

Let ∠A = θ°

Then cos θ = AB/AC = 3/5.

Let AB = 3k and AC = 5k, where k is positive.

Therefore, sec θ = 1/cos θ = 5/3

tan θ = BC/AB =4k/3k = 4/3

cot θ = 1/tan θ = 3/4 and

csc θ = AC/BC = 5k/4k = 5/4

Now (5csc θ -4 tan θ)/(sec θ + cot θ)

= (5 × 5/4 - 4 × 4/3)/(5/3 + 3/4)

= (25/4 -16/3)/(5/3 +3/4)

= 11/12 × 12/29

9. Express 1 + 2 sin A cos A as a perfect square.

1 + 2 sin A cos A

10. If sin A + cos A = 7/5 and sin A cos A =12/25, find the values of sin A and cos A.

sin A + cos A = 7/5

⇒ cos A = 7/5 - sin θ

Now from sin θ/cos θ = 12/25

We get, sin θ(7/5 - sin θ) = 12/25

or, 5 sin θ(5 sin θ - 4) - 3(5 sin θ - 4) = 0

or, (5 sin θ - 3) (5 sin θ - 4) = 0

⇒ (5 sin θ - 3) = 0 or, (5 sin θ - 4) = 0

⇒ sin θ = 3/5 or, sin θ = 4/5

When sin θ = 3/5, cos θ = 12/25 × 5/3 = 4/5

Again, when sin θ = 4/5, cos θ = 12/25 × 5/4 = 3/5

Therefore, sin θ =3/5, cos θ = 4/5

or, sin θ =4/5, cos θ = 3/5.

11. If 3 tan θ = 4, evaluate (3sin θ + 2 cos θ)/(3sin θ - 2cos θ).

Solution: Given,

3 tan θ = 4

⇒ tan θ = 4/3

(3sin θ + 2 cos θ)/(3sin θ - 2cos θ)

= (3 tan θ + 2)/(3 tan θ - 2), [dividing both numerator and denominator by cos θ]

= (3 × 4/3 + 2)/(3 × 4/3 -2), putting the value of tan θ = 4/3

12. If (sec θ + tan θ)/(sec θ - tan θ) = 209/79, find the value of θ.

Solution: (sec θ + tan θ)/(sec θ - tan θ) = 209/79

⇒ [(sec θ + tan θ) - (sec θ - tan θ)]/[(sec θ + tan θ) + (sec θ - tan θ)] = [209 – 79]/[209 + 79], (Applying componendo and dividendo)

⇒ 2 tan θ/2 sec θ =130/288

⇒ sin θ/cos θ × cos θ = 65/144

⇒ sin θ = 65/144.

13. If 5 cot θ = 3, find the value of (5 sin θ - 3 cos θ)/(4 sin θ + 3 cos θ).

Given 5 cot θ = 3

⇒ cot θ = 3/5

Now (5 sin θ - 3 cos θ)/(4 sin θ + 3 cos θ)

= (5 - 3 cot θ)/(4 sin θ + 3 cot θ), [dividing both numerator and denominator by sin θ]

= (5 - 3 × 3/5)/(4 + 3 × 3/5)

= (5 - 9/5)/(4 + 9/5)

= (16/5 × 5/29)

⇒ sin θ(sin θ - 2) - 1(sin θ - 2) = 0

⇒ (sin θ - 2)(sin θ - 1) = 0

⇒ (sin θ - 2) = 0 or, (sin θ - 1) = 0

⇒ sin θ = 2 or, sin θ = 1

So, value of sin θ can’t be greater than 1,

Therefore sin θ = 1

Basic Trigonometric Ratios  

Relations Between the Trigonometric Ratios

Reciprocal Relations of Trigonometric Ratios

Trigonometrical Identity

Problems on Trigonometric Identities

Elimination of Trigonometric Ratios  

Eliminate Theta between the equations

Problems on Eliminate Theta  

Trig Ratio Problems

Proving Trigonometric Ratios

Trig Ratios Proving Problems

Verify Trigonometric Identities  

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8.2: The Trigonometric Ratios

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• Richard W. Beveridge
• Clatsop Community College

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There are six common trigonometric ratios that relate the sides of a right triangle to the angles within the triangle. The three standard ratios are the sine, cosine and tangent. These are often abbreviated sin, cos and tan. The other three (cosecant, secant and cotangent) are the reciprocals of the sine, cosine and tangent and are often abbreviated csc, sec, and cot.

Given an angle situated in a right triangle, the sine function is defined as the ratio of the side opposite the angle to the hypotenuse, the cosine is defined as the ratio of the side adjacent to the angle to the hypotenuse and the tangent is defined as the ratio of the side opposite the angle to the side adjacent to the angle. \[ \begin{aligned} \sin \theta &=\frac{o p p}{h y p} \\ \cos \theta &=\frac{a d j}{h y p} \\ \tan \theta &=\frac{o p p}{a d j} \end{aligned} \] A common mneumonic device to help remember these relationships is -SOHCAHTOA-which identifies the sin as Opp over Hyp Cos as Adj over Hyp and the Tan as Opp over Adj.

An acute angle placed in the other position of a right triangle would have different oppposite and adjacent sides although the hypotenuse would remain the same.

since we know the legs of the triangle, we can substitute these values for \(a\) and \(b\) in the Pythagorean Theorem: \[ \begin{array}{c} 3^{2}+5^{2}=c^{2} \\ 9+25=c^{2} \\ 34=c^{2} \\ \sqrt{34}=c \end{array} \] Now that we know the hypotenuse \((\sqrt{34}),\) we can determine the sin, cos and tan for the angle \(\theta\) \[ \begin{aligned} \sin \theta &=\frac{3}{\sqrt{34}} \\ \cos \theta &=\frac{5}{\sqrt{34}} \\ \tan \theta &=\frac{3}{5} \end{aligned} \] Find \(\sin \theta, \cos \theta\) and \(\tan \theta\) for the given angle \(\theta\)

Again, in order to find the sin, cos and tan of the angle \(\theta,\) we must find the missing side of the triangle by using the Pythagorean Theorem. since, in this case, we know the hypotenuse and one of the legs, the value of the hypotenuse must be substituted for \(c\) and the length of the leg we're given can be substituted for either \(a\) or \(b\)

\[ \begin{array}{c} 4^{2}+b^{2}=9^{2} \\ 16+b^{2}=81 \\ b^{2}=65 \\ b=\sqrt{65} \end{array} \] Now that we know the length of the other leg of the triangle \((\sqrt{65}),\) we can determine the sin, cos and tan for the angle \(\theta\) \[ \begin{aligned} \sin \theta &=\frac{\sqrt{65}}{9} \\ \cos \theta &=\frac{4}{9} \\ \tan \theta &=\frac{\sqrt{65}}{4} \end{aligned} \] In addition to the examples above, if we are given the value of one of the trigonometric ratios, we can find the value of the other two. Example Given that \(\cos \theta=\frac{1}{3},\) find \(\sin \theta\) and \(\tan \theta\) Given the information about the cosine of the angle \(\theta,\) we can create a triangle that will allow us to find \(\sin \theta\) and \(\tan \theta\)

Using the Pythagorean Theorem, we can find the missing side of the triangle: \[ \begin{array}{c} a^{2}+1^{2}=3^{2} \\ a^{2}+1=9 \end{array} \] \(a^{2}=8\) \(a=\sqrt{8}=2 \sqrt{2}\) Then \(\sin \theta=\frac{\sqrt{8}}{3}\) and \(\tan \theta=\frac{\sqrt{8}}{1}=\sqrt{8}\) You might say to yourself, "Wait a minute, just because the cosine of the angle \(\theta\) is \(\frac{1}{3},\) that doesn't necessarily mean that the sides of the triangle are 1 and \(3,\) they could be 2 and \(6,\) or 3 and 9 or any values \(n\) and \(3 n . "\)

This is true, and if the sides are expressed as \(n\) and \(3 n,\) then the missing side would be \(n \sqrt{8},\) so that whenever we find a trigonometric ratio, the \(n^{\prime}\) s will cancel out, so we just leave them out to begin with and call the sides 1 and 3 Example Given that \(\tan \theta=\frac{\sqrt{5}}{7},\) find \(\sin \theta\) and \(\cos \theta\) First we'll take the infomation about the tangent and use this to draw a triangle.

Then use the Pythagorean Theorem to find the missing side of the triangle: \[ \begin{array}{c} \sqrt{5}^{2}+7^{2}=c^{2} \\ 5+49=c^{2} \\ 54=c^{2} \\ \sqrt{54}=3 \sqrt{6}=c \end{array} \] So then: \[ \sin \theta=\frac{\sqrt{5}}{\sqrt{54}}=\sqrt{\frac{5}{54}} \] \[ \cos \theta=\frac{7}{\sqrt{54}}=\frac{7}{3 \sqrt{6}} \]

Exercises 1.2 Find \(\sin \theta, \cos \theta\) and \(\tan \theta\) for the given triangles.

Use the information given to find the other two trigonometric ratios. 11. \(\quad \tan \theta=\frac{1}{2}\) 12. \(\quad \sin \theta=\frac{3}{4}\) 13. \(\quad \cos \theta=\frac{3}{\sqrt{20}}\) 14. \(\quad \tan \theta=2\) 15. \(\sin \theta=\frac{5}{\sqrt{40}}\) 16. \(\sin \theta=\frac{7}{10}\) 17. \(\cos \theta=\frac{9}{40}\) 18. \(\quad \tan \theta=\sqrt{3}\) 19. \(\cos \theta=\frac{1}{2}\) 20. \(\cos \theta=\frac{3}{7}\) 21. \(\sin \theta=\frac{\sqrt{5}}{7}\) 22. \(\quad \tan \theta=1.5\)

Trigonometric Ratios

A series of free, online High School Geometry Video Lessons and solutions. In these lessons, we will learn

• how to find the sine of an angle in a right triangle,
• how to find the cosine of an angle in a right triangle,
• how to find the tangent of an angle in a right triangle,
• how to use inverse trigonometric functions to find an angle with a given trigonometric value.

Related Pages Using SOH-CAH-TOA Trigonometry Word Problems Inverse trigonometry Lessons On Trigonometry Trigonometry Worksheets

The following diagram shows the trigonometric ratios using SOHCAHTOA. Scroll down the page if you need more examples and solutions on how to use the trigonometric ratios.

Trigonometric Ratios: Sine

Right triangles have ratios to represent the angles formed by the hypotenuse and its legs. Sine ratios, along with cosine and tangent ratios, are ratios of the lengths of two sides of the triangle. Sine ratios in particular are the ratios of the length of the side opposite the angle they represent over the hypotenuse. Sine ratios are useful in trigonometry when dealing with triangles and circles.

How to define the sine ratio and identify the sine of an angle in a right triangle?

Identify the hypotenuse of a right triangle. sin θ = opposite/hypotenuse

A word problem involving the trigonometric ratio of sine to calculate the height of a pole

Example: A 55 ft wire connects a point on the ground to the top of a pole. The cable makes an angle of 60° with the ground. Find the height of the pole to the nearest foot.

Trigonometric Ratios: Cosine

Right triangles have ratios that are used to represent their base angles. Cosine ratios, along with sine and tangent ratios, are ratios of two different sides of a right triangle.Cosine ratios are specifically the ratio of the side adjacent to the represented base angle over the hypotenuse. In order to find the measure of the angle, we must understand inverse trigonometric functions.

How to use the Cosine formula (the CAH Formula) to find missing sides or angles?

Cosine θ = adjacent/hypotenuse

Example: Find the missing side or angle a) cos 30° = x/2 b) cos 37° = 4.2/x c) cos θ = 63/80

How to apply the Sine and Cosine Ratios?

Sine, cosine are trigonometric ratios for the acute angles and involve the length of a leg and the hypotenuse of a right triangle.

Angle of elevation - When looking up at an object, the angle your line of sight makes with a horizontal line is called the angle of elevation.

Angle of depression - When looking down at an object, the angle your line of sight makes with a horizontal line is called the angle of depression.

• Find sin U and sin W. Write each answer as a fraction and as a decimal rounded to four places.
• Find cos S and cos R. Write each answer as a fraction and as a decimal rounded to four places.
• You walk from one corner of a basketball court to the opposite corner. write and solve a proportion using a trigonometric ratio to approximate the distance of the walk.
• You are at the top of a roller coaster 100 feet above the ground. The angle of depression is 44°. About how far do you ride down the hill?
• A railroad crossing arm that is 20 feet long is stuck with an angle of elevation of 35°. Find the lengths of x and y.
• Use a special right triangle to find the sine and cosine of a 30° angle.

Trigonometric Ratios: Tangent

Right triangles have ratios that are used to represent their base angles. Tangent ratios, along with cosine and sine ratios, are ratios of two different sides of a right triangle. Tangent ratios are the ratio of the side opposite to the side adjacent the angle they represent. In order to find the measure of the angle itself, one must understand inverse trigonometric functions.

How to use the Tangent formula (the TOA Formula)?

Tangent θ = opposite/adjacent

Example: Find the missing side or angle a) tan 28° = x/40 b) tan 41° = 1.9/x c) tan θ = 11/8

Applications of Trigonometric Ratios (Word Problems Involving Tangent, Sine and Cosine)

• Find the area of the parallelogram.
• A 70 foot ramp rises from the first floor to the second floor of a parking garage. The ramp makes an angle with the ground. How high above the first floor is the second floor?
• You see Mr. Wandera flying a kite in the park. The kite string is 65 meters long. What angle does the string need to form with the ground so that the kite is 30 feet off of the ground?
• From the top of a 100-foot lookout tower, a forest ranger spotted a fire at a 25° angle of depression. How far was the fire from the base of the lookout tower?
• An 8 foot ladder is leaning against a wall. The ladder makes a 53° angle with the wall. How high does the ladder reach?

Inverse Trigonometric Functions

Once we understand the trigonometric functions sine, cosine, and tangent, we are ready to learn how to use inverse trigonometric functions to find the measure of the angle the function represents. Inverse trigonometric functions, found on any standard scientific or graphing calculator, are a vital part of trigonometry and will be encountered often in Calculus.

How to use inverse trigonometric functions to find an angle with a given trigonometric value and how to use inverse trigonometric functions to solve a right triangle?

Example: Use the calculator to find an angle θ in the interval [0, 90] that satisfies the equation.

• sin θ = 0.7523
• tan θ = 3.54
• Solve the given right triangle if a = 44.3 cm and b = 55.9 cm.
• Find each angle in a 3,4,5 triangle

How to use inverse trig to find a missing angle?

Inverse trig functions are used to find missing angles rather than missing sides.

Find missing Angles - Using Inverse Sine, Cosine, Tangent

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

PROBLEMS ON TRIGONOMETRIC RATIOS

Problem 1 :

For the measures in the figure shown below, compute sine, cosine and tangent ratios of the angle θ.

In the given right angled triangle, note that for the given angle θ, PR is the ‘opposite’ side and PQ is the ‘adjacent’ side.

Then, 

sin θ  =  opposite side/hypotenuse  =  PR/QR  =  35/37

cos θ  =  adjacent side/hypotenuse = PQ/QR  =  12/37

tan θ  =  opposite side / adjacent side  =  PR/PQ  =  35/12

Problem 2 :

Find the six trigonometric ratios of the angle θ using the diagram shown below. 

In the given right angled triangle, note that for the given angle θ, AC is the ‘opposite’ side and AB is the ‘adjacent’ side.

And also, the length of the adjacent side 'AB' is not given. 

Find the length of AB.

By Pythagorean Theorem,

BC 2   =  AB 2  + AC 2

25 2   =  AB 2  + 7 2

625  =  AB 2 + 49

Subtract 49 from each side. 

576  =  AB 2

24 2   =  AB 2

24  =  AB  

sin θ  =  opposite side/hypotenuse  =   AC/ BC  =  7/25

cos θ  =  adjacent side/hypotenuse  =  AB/BC  =  24/25

tan θ  =  opposite side/adjacent side  =  AC/AB  =  7/24

csc θ  =  1/s in θ  =  25/7

sec θ  =  1/cos θ  =   25/24

cot θ  =  1/tan θ  =   24/7

Problem 3 :

If tanA = 2/3, then find all the other trigonometric ratios. 

tanA  =  opposite side/adjacent side  =  2/3

AC 2   =  AB 2  + BC 2

AC 2   =  3 2  + 2 2

AC 2   =  9 + 4

AC 2   =  13

AC  =  √13

sinA   =  opposite side/hypotenuse  =   BC/ AC  =  2/ √ 13

cosA   =  adjacent side/hypotenuse  =  AB/AC  =  3/ √13

cscA   =  1/s inA   =   √13 /2

secA   =  1/cosA   =   √13 /3

cotA   =  1/tanA   =  3/2

Problem 4 :

If sec θ   =  2/3, then find the value of

(2sinθ - 3cosθ)/(4sinθ - 9cosθ)  

sec θ   =  hypotenuse/adjacent side  =  13/5

13 2   =  5 2  + AC 2

169  =  25 + AC 2

Subtract 25 from each side. 

144  =  AC 2

12 2   =  AC 2

12  =   AC

sin θ   =  opposite side/hypotenuse  =   AC/BC  =  12/1 3

cos θ    =  adjacent side/hypotenuse  =  AB/BC  =  5/ 13

(2sinθ - 3cosθ)/(4sinθ - 9cosθ) : 

=  (2  ⋅  12/13 - 3  ⋅  5/13)/(4  ⋅ 12/13 - 9  ⋅ 5/13)

=  (24 /13 - 1 5/13)/(48 /13 - 45/13)

=  [(24 - 15)/13]/[(48 - 45)/13]

=  (9/13)/(3/13)

=  (9/13)  ⋅  (13/3)

=  9/3

(2sinθ - 3cosθ)/(4sinθ - 9cosθ)   =  3

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Trigonometric Ratios

Trigonometric ratios are the ratios of the length of sides of a triangle. These ratios in trigonometry relate the ratio of sides of a right triangle to the respective angle. The basic trigonometric ratios are sin, cos, and tan, namely sine, cosine, and tangent ratios. The other important trig ratios, cosec, sec, and cot, can be derived using the sin, cos, and tan respectively.

The word "Trigonometry" originated from the words, "Trigonon" which means "triangle" and "Metron" which means "to measure". It is a branch of mathematics that deals with the relation between the angles and sides of a right-angled triangle. In fact, trigonometry is one of the most ancient subjects which is studied by scholars all over the world. Let us understand the trigonometric ratios in detail in the following sections.

What are Trigonometric Ratios?

In trigonometry, there are six trigonometric ratios, namely, sine , cosine , tangent , secant, cosecant, and cotangent. These ratios are written as sin, cos, tan, sec, cosec(or csc), and cot in short. Let us have a look at the right-angled triangle shown below. Trigonometric ratios can be used to determine the ratios of any two sides out of a total of three sides of a right-angled triangle in terms of the respective angles.

The values of these trigonometric ratios can be calculated using the measure of an acute angle , θ in the right-angled triangle given below. This implies that the value of the ratio of any two sides of the triangle here depends on angle C. We can alternatively find the values of these trig ratios for angle A. Also, only the base and perpendicular will interchange for the given right triangle in that case.

These six trigonometric ratios can be defined as,

Sine: The sine ratio for any given angle is defined as the ratio of the perpendicular to the hypotenuse. In the given triangle, sine of angle θ can be given as, sin θ = AB/AC.

Cosine: The cosine ratio for any given angle is defined as the ratio of the base to the hypotenuse. In the given triangle, cosine of angle θ can be given as, cos θ = BC/AC.

Tangent: The tangent ratio for any given angle is defined as the ratio of the perpendicular to the base. In the given triangle, the tangent of angle θ can be given as, tan θ = AB/BC.

Cosecant: The cosecant ratio for any given angle is defined as the ratio of the hypotenuse to the perpendicular. In the given triangle, cosecant of angle θ can be given as, cosec θ = AC/AB.

Secant: The secant ratio for any given angle is defined as the ratio of the hypotenuse to the base. In the given triangle, secant of angle θ can be given as, sec θ = AC/BC.

Cotangent: The cotangent ratio for any given angle is defined as the ratio of the base to the perpendicular. In the given triangle, cotangent of angle θ can be given as, cot θ = BC/AB.

Let us understand these and more trigonometric ratio formulas in detail in the next section.

Trigonometric Ratios Formulas

Trigonometric ratios can be calculated by taking the ratio of any two sides of the right-angled triangle. We can evaluate the third side using the Pythagoras theorem , given the measure of the other two sides. We can use the abbreviated form of trigonometric ratios to compare the length of any two sides with the angle in the base. The angle θ is an acute angle (θ < 90º) and in general is measured with reference to the positive x-axis, in the anticlockwise direction. The basic trigonometric ratios formulas are given below,

• sin θ = Perpendicular/Hypotenuse
• cos θ = Base/Hypotenuse
• tan θ = Perpendicular/Base
• sec θ = Hypotenuse/Base
• cosec θ = Hypotenuse/Perpendicular
• cot θ = Base/Perpendicular

Now, let us observe the reciprocal trigonometric ratio formulas of the above-mentioned trigonometric ratios. As we observe, we notice that sin θ is a reciprocal of cosec θ, cos θ is a reciprocal of sec θ, tan θ is a reciprocal of cot θ, and vice-versa. So, the new set of formulas for trigonometric ratios is:

• sin θ = 1/cosec θ
• cos θ = 1/sec θ
• tan θ = 1/cot θ
• cosec θ = 1/sin θ
• sec θ = 1/cos θ
• cot θ = 1/tan θ

Trigonometric Ratios Table

In the trigonometric ratios table, we use the values of trigonometric ratios for standard angles 0°, 30°, 45°, 60°, and 90º. It is easy to predict the values of the table and to use the table as a reference to calculate values of trigonometric ratios for various other angles, using the trigonometric ratio formulas for existing patterns within trigonometric ratios and even between angles. Now, we will summarize the value of trigonometric ratios for specific angles in the table below:

Trigonometric Ratios Identities

There are many trigonometric ratios identities that we use to make our calculations easier and simpler. These include identities of complementary angles, supplementary angles, Pythagorean identities, and sum, difference, product identities.

Trigonometric Ratios of Complementary Angles Identities

The complementary angles are a pair of two angles such that their sum is equal to 90°. The complement of an angle θ is (90° - θ). The trigonometric ratios of complementary angles are:

• sin (90°- θ) = cos θ
• cos (90°- θ) = sin θ
• cosec (90°- θ) = sec θ
• sec (90°- θ) = cosec θ
• tan (90°- θ) = cot θ
• cot (90°- θ) = tan θ

Pythagorean Triogometric Ratios Identities

The Pythagorean trigonometric ratios identities in trigonometry are derived from the Pythagoras theorem. Applying Pythagoras theorem to the right-angled triangle below, we get:

Opposite 2 + Adjacent 2 = Hypotenuse 2

Dividing both sides by Hypotenuse 2

Opposite 2 /Hypotenuse 2 + Adjacent 2 /Hypotenuse 2 = Hypotenuse 2 /Hypotenuse 2

• sin 2 θ + cos 2 θ = 1

This is one of the important Pythagorean identities. In the same way, we can derive two other Pythagorean trigonometric ratios identities:

• 1 + tan 2 θ = sec 2 θ
• 1 + cot 2 θ = cosec 2 θ

Sum, Difference, Product Trigonometric Ratios Identities

The sum, difference, and product trigonometric ratios identities include the formulas of sin(A+B), sin(A-B), cos(A+B), cos(A-B), etc.

• sin (A + B) = sin A cos B + cos A sin B
• sin (A - B) = sin A cos B - cos A sin B
• cos (A + B) = cos A cos B - sin A sin B
• cos (A - B) = cos A cos B + sin A sin B
• tan (A + B) = (tan A + tan B)/ (1 - tan A tan B)
• tan (A - B) = (tan A - tan B)/ (1 + tan A tan B)
• cot (A + B) = (cot A cot B - 1)/(cot B - cot A)
• cot (A - B) = (cot A cot B + 1)/(cot B - cot A)
• 2 sin A⋅cos B = sin(A + B) + sin(A - B)
• 2 cos A⋅cos B = cos(A + B) + cos(A - B)
• 2 sin A⋅sin B = cos(A - B) - cos(A + B)

Half, Double, and Triple-Angles Trigonometric Ratios Identities

Double Angle Trigonometric Ratios Identities

The double angle trigonometric identities can be obtained by using the sum and difference formulas.

For example, from the above formula sin (A+B) = sin A cos B + cos A sin B

Substitute A = B = θ on both sides here, we get:

sin (θ + θ) = sinθ cosθ + cosθ sinθ sin 2θ = 2 sinθ cosθ

In the same way, we can derive the other double angle identities.

• sin 2θ = 2 sinθ cosθ
• cos 2θ = cos 2 θ - sin 2 θ cos 2θ = 2 cos 2 θ - 1 cos 2θ = 1 - 2 sin 2 θ cos 2θ = (1 - tan 2 θ)/(1 + tan 2 θ)
• tan 2θ = (2 tanθ)/ (1 - tan 2 θ)
• sec 2θ = sec 2 θ/(2-sec 2 θ)
• cosec 2θ = (sec θ. cosec θ)/2
• cot 2θ = (cot θ - tan θ)/2

Half Angle Trigonometric Ratios Identities

Using one of the above double angle formulas, cos 2θ = 1 - 2 sin 2 θ 2 sin 2 θ = 1- cos 2θ sin 2 θ = (1 - cos 2θ)/(2) sin θ = ±√[(1 - cos 2θ)/2]

Replacing θ by θ/2 on both sides,

sin (θ/2) = ±√[(1 - cos θ)/2]

This is the half-angle formula of sine.

In the same way, we can derive the other half-angle formulas.

• sin (θ/2) =±√[(1 - cos θ)/2]
• cos (θ/2) = ±√[(1 + cos θ)/2]
• tan (θ/2) = ±√[(1 - cos θ)(1 + cosθ)]

Triple Angle Trigonometric Ratios Identities

• sin 3θ = 3sin θ - 4sin 3 θ
• cos 3θ = 4cos 3 θ - 3cos θ
• tan 3θ = (3tanθ - tan 3 θ)/(1 - 3tan 2 θ)

Related Topics:

• Trigonometry
• Trigonometric Identities
• Trigonometric Table
• Trigonometric Formulas

Important Notes on Trigonometric Ratios:

• The values of trigonometric ratios do not change with the change in the side lengths of the triangle if the angle remains the same.
• All trigonometric functions are periodic in nature.
• Trigonometric ratios are used to find the missing sides or angles in a triangle.

Examples on Trigonometric Ratios

Example 1: In a right-angled triangle ABC, right-angled at B, hypotenuse AC = 10 units, base BC = 8 units and perpendicular AB = 6 units and if ∠ACB = θ, then find the trigonometric ratios tan θ, sin θ, and cos θ.

We know, sin θ = perpendicular/hypotenuse

cos θ = base/hypotenuse

tan θ = perpendicular/base

⇒ sin θ = 6/10 = 3/5

⇒ cos θ = 8/10 = 4/5

⇒ tan θ = 6/8 = 3/4

Answer: sin θ, cos θ, and tan θ for the given triangle are 3/5, 4/5, and 3/4 respectively.

Example 2: A building is at a distance of 210 feet from point A on the ground. Find the height of the building if tan θ = 4/3?

Solution: The triangle formed is a right-angled triangle. Now apply the trigonometric ratio of tan⁡θ to calculate the height of the building.

4/3 = Height/210 ft

Height = (4 × 210/3) = 280 ft

Answer: The height of the building is 280 ft.

Example 3: The triangle is right-angled at C with AB = 29 units and AC = 20 units. Can you verify the trigonometric ratios identity cos 2 θ + sin 2 θ = 1 using these values?

Solution: We will find BC using the Pythagorean theorem,

BC = √(AB 2 - AC 2 )

= √(29 2 - 20 2 )

= √841 2 - 400 2 )

Now let's determine the values of sinθ and cosθ

sinθ = AC/AB = 20/29

cosθ = BC/AB = 21/29

Now let's verify the identity.

cos 2 θ + sin 2 θ = (21/29) 2 + (20/29) 2 = (400 + 441)/841 = 1

Answer: Hence, the trigonometric ratios identity is verified.

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Practice Questions on Trigonometric Ratios

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FAQs on Trigonometric Ratios

How to find trigonometric ratios.

Trigonometric Ratios can be calculated either by using the given acute angle or determining the ratios of the sides of the right-angled triangle. The trigonometric ratios formulas to be used are:

What are the Six Trigonometric Ratios?

The main six trigonometric ratios are sine, cosine, tangent, cosecant, secant, and cotangent. The trigonometric ratios can be calculated using the formulas given in the article.

What are the Applications of Trigonometric Ratios?

There are various applications of trigonometric ratios such as:

• Sine and cosine are used to represent sound waves.
• Some trigonometric ratios are used in the field of architecture, constructing buildings, etc.
• They are used in creating maps.

What are Trigonometric Ratios of Specific Angles?

Trigonometric ratios can be determined for different angles. But for convenience of calculations, we memorize the trigonometric ratios of some specific angles like 0°, 30°, 45°, 60°, 90°. Refer to the trigonometric ratios table for the values of the ratios at these angles.

What are Trigonometric Ratios of Complementary Angles?

Complementary angles are two angles whose sum is 90°. Formulas for trigonometric ratios of complementary angles are:

What is the Relationship Between Sin, Cos, and Tan Trigonometric Ratios?

We know, that sin can be given as perpendicular/hypotenuse, cos as base/hypotenuse, and tan as perpendicular/base. The relationship between the trigonometric ratios sin, cos, and tan can therefore be given as, tan θ = sin θ/cos θ.

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Trigonometry : Solving Word Problems with Trigonometry

Study concepts, example questions & explanations for trigonometry, all trigonometry resources, example questions, example question #1 : solving word problems with trigonometry.

You can draw the following right triangle using the information given by the question:

Since you want to find the height of the platform, you will need to use tangent.

You can draw the following right triangle from the information given by the question.

In order to find the height of the flagpole, you will need to use tangent.

You can draw the following right triangle from the information given in the question:

In order to find out how far up the ladder goes, you will need to use sine.

In right triangle ABC, where angle A measures 90 degrees, side AB measures 15 and side AC measures 36, what is the length of side BC?

This triangle cannot exist.

Example Question #5 : Solving Word Problems With Trigonometry

A support wire is anchored 10 meters up from the base of a flagpole, and the wire makes a 25 o angle with the ground. How long is the wire, w? Round your answer to two decimal places.

23.81 meters

28.31 meters

21.83 meters

To make sense of the problem, start by drawing a diagram. Label the angle of elevation as 25 o , the height between the ground and where the wire hits the flagpole as 10 meters, and our unknown, the length of the wire, as w. 

Now, we just need to solve for w using the information given in the diagram. We need to ask ourselves which parts of a triangle 10 and w are relative to our known angle of 25 o . 10 is opposite this angle, and w is the hypotenuse. Now, ask yourself which trig function(s) relate opposite and hypotenuse. There are two correct options: sine and cosecant. Using sine is probably the most common, but both options are detailed below.

We know that sine of a given angle is equal to the opposite divided by the hypotenuse, and cosecant of an angle is equal to the hypotenuse divided by the opposite (just the reciprocal of the sine function). Therefore:

To solve this problem instead using the cosecant function, we would get:

The reason that we got 23.7 here and 23.81 above is due to differences in rounding in the middle of the problem. 

Example Question #6 : Solving Word Problems With Trigonometry

When the sun is 22 o above the horizon, how long is the shadow cast by a building that is 60 meters high?

To solve this problem, first set up a diagram that shows all of the info given in the problem. 

Next, we need to interpret which side length corresponds to the shadow of the building, which is what the problem is asking us to find. Is it the hypotenuse, or the base of the triangle? Think about when you look at a shadow. When you see a shadow, you are seeing it on something else, like the ground, the sidewalk, or another object. We see the shadow on the ground, which corresponds to the base of our triangle, so that is what we'll be solving for. We'll call this base b.

Therefore the shadow cast by the building is 150 meters long.

If you got one of the incorrect answers, you may have used sine or cosine instead of tangent, or you may have used the tangent function but inverted the fraction (adjacent over opposite instead of opposite over adjacent.)

Example Question #7 : Solving Word Problems With Trigonometry

From the top of a lighthouse that sits 105 meters above the sea, the angle of depression of a boat is 19 o . How far from the boat is the top of the lighthouse?

423.18 meters

318.18 meters

36.15 meters

110.53 meters

To solve this problem, we need to create a diagram, but in order to create that diagram, we need to understand the vocabulary that is being used in this question. The following diagram clarifies the difference between an angle of depression (an angle that looks downward; relevant to our problem) and the angle of elevation (an angle that looks upward; relevant to other problems, but not this specific one.) Imagine that the top of the blue altitude line is the top of the lighthouse, the green line labelled GroundHorizon is sea level, and point B is where the boat is.

Merging together the given info and this diagram, we know that the angle of depression is 19 o  and and the altitude (blue line) is 105 meters. While the blue line is drawn on the left hand side in the diagram, we can assume is it is the same as the right hand side. Next, we need to think of the trig function that relates the given angle, the given side, and the side we want to solve for. The altitude or blue line is opposite the known angle, and we want to find the distance between the boat (point B) and the top of the lighthouse. That means that we want to determine the length of the hypotenuse, or red line labelled SlantRange. The sine function relates opposite and hypotenuse, so we'll use that here. We get:

Example Question #8 : Solving Word Problems With Trigonometry

Angelina just got a new car, and she wants to ride it to the top of a mountain and visit a lookout point. If she drives 4000 meters along a road that is inclined 22 o to the horizontal, how high above her starting point is she when she arrives at the lookout?

9.37 meters

1480 meters

3708.74 meters

10677.87 meters

1616.1 meters

As with other trig problems, begin with a sketch of a diagram of the given and sought after information.

Angelina and her car start at the bottom left of the diagram. The road she is driving on is the hypotenuse of our triangle, and the angle of the road relative to flat ground is 22 o . Because we want to find the change in height (also called elevation), we want to determine the difference between her ending and starting heights, which is labelled x in the diagram. Next, consider which trig function relates together an angle and the sides opposite and hypotenuse relative to it; the correct one is sine. Then, set up:

Therefore the change in height between Angelina's starting and ending points is 1480 meters. 

Example Question #9 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 50 feet apart. The shorter building is 40 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 48 o . How high is the taller building?

To solve this problem, let's start by drawing a diagram of the two buildings, the distance in between them, and the angle between the tops of the two buildings. Then, label in the given lengths and angle. 

Example Question #10 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 80 feet apart. The shorter building is 55 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 32 o . How high is the taller building?

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Solving Real-Life Problems Using Trigonometry

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Introduction to Trigonometry

Trigonometry (from Greek trigonon "triangle" + metron "measure")

Want to learn Trigonometry? Here is a quick summary. Follow the links for more, or go to Trigonometry Index

Trigonometry helps us find angles and distances, and is used a lot in science, engineering, video games, and more!

Right-Angled Triangle

The triangle of most interest is the right-angled triangle . The right angle is shown by the little box in the corner:

Another angle is often labeled θ , and the three sides are then called:

• Adjacent : adjacent (next to) the angle θ
• Opposite : opposite the angle θ
• and the longest side is the Hypotenuse

Why a Right-Angled Triangle?

Why is this triangle so important?

Imagine we can measure along and up but want to know the direct distance and angle:

Trigonometry can find that missing angle and distance.

Or maybe we have a distance and angle and need to "plot the dot" along and up:

Questions like these are common in engineering, computer animation and more.

And trigonometry gives the answers!

Sine, Cosine and Tangent

The main functions in trigonometry are Sine, Cosine and Tangent

They are simply one side of a right-angled triangle divided by another.

For any angle " θ ":

(Sine, Cosine and Tangent are often abbreviated to sin, cos and tan .)

Example: What is the sine of 35°?

Using this triangle (lengths are only to one decimal place):

sin(35°) = Opposite Hypotenuse = 2.8 4.9 = 0.57...

The triangle could be larger, smaller or turned around, but that angle will always have that ratio .

Calculators have sin, cos and tan to help us, so let's see how to use them:

Example: How Tall is The Tree?

We can't reach the top of the tree, so we walk away and measure an angle (using a protractor) and distance (using a laser):

• We know the Hypotenuse
• And we want to know the Opposite

Sine is the ratio of Opposite / Hypotenuse :

sin(45°) = Opposite Hypotenuse

Get a calculator, type in "45", then the "sin" key:

sin(45°) = 0.7071...

What does the 0.7071... mean? It is the ratio of the side lengths, so the Opposite is about 0.7071 times as long as the Hypotenuse.

We can now put 0.7071... in place of sin(45°):

0.7071... = Opposite Hypotenuse

And we also know the hypotenuse is 20 :

0.7071... = Opposite 20

To solve, first multiply both sides by 20:

20 × 0.7071... = Opposite

Opposite = 14.14m (to 2 decimals)

The tree is 14.14m tall

Try Sin Cos and Tan

Play with this for a while (move the mouse around) and get familiar with values of sine, cosine and tangent for different angles, such as 0°, 30°, 45°, 60° and 90°.

Also try 120°, 135°, 180°, 240°, 270° etc, and notice that positions can be positive or negative by the rules of Cartesian coordinates , so the sine, cosine and tangent change between positive and negative also.

So trigonometry is also about circles !

Unit Circle

What you just played with is the Unit Circle .

It is a circle with a radius of 1 with its center at 0.

Because the radius is 1, we can directly measure sine, cosine and tangent.

Here we see the sine function being made by the unit circle:

Note: you can see the nice graphs made by sine, cosine and tangent .

Degrees and Radians

Angles can be in Degrees or Radians . Here are some examples:

Repeating Pattern

Because the angle is rotating around and around the circle the Sine, Cosine and Tangent functions repeat once every full rotation (see Amplitude, Period, Phase Shift and Frequency ).

When we want to calculate the function for an angle larger than a full rotation of 360° (2 π radians) we subtract as many full rotations as needed to bring it back below 360° (2 π radians):

Example: what is the cosine of 370°?

370° is greater than 360° so let us subtract 360°

370° − 360° = 10°

cos(370°) = cos(10°) = 0.985 (to 3 decimal places)

And when the angle is less than zero, just add full rotations.

Example: what is the sine of −3 radians?

−3 is less than 0 so let us add 2 π radians

−3 + 2 π = −3 + 6.283... = 3.283... rad ians

sin(−3) = sin(3.283...) = −0.141 (to 3 decimal places)

Solving Triangles

Trigonometry is also useful for general triangles, not just right-angled ones .

It helps us in Solving Triangles . "Solving" means finding missing sides and angles.

Example: Find the Missing Angle "C"

Angle C can be found using angles of a triangle add to 180° :

So C = 180° − 76° − 34° = 70°

We can also find missing side lengths. The general rule is:

When we know any 3 of the sides or angles we can find the other 3 (except for the three angles case)

See Solving Triangles for more details.

Other Functions (Cotangent, Secant, Cosecant)

Similar to Sine, Cosine and Tangent, there are three other trigonometric functions which are made by dividing one side by another:

Trigonometric and Triangle Identities

And as you get better at Trigonometry you can learn these:

Enjoy becoming a triangle (and circle) expert!

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Trigonometry Questions

Trigonometry questions given here involve finding the missing sides of a triangle with the help of trigonometric ratios and proving trigonometry identities. We know that trigonometry is one of the most important chapters of Class 10 Maths. Hence, solving these questions will help you to improve your problem-solving skills.

What is Trigonometry?

The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). Trigonometry is the study of relationships between the sides and angles of a triangle.

The basic trigonometric ratios are defined as follows.

sine of ∠A = sin A = Side opposite to ∠A/ Hypotenuse

cosine of ∠A = cos A = Side adjacent to ∠A/ Hypotenuse

tangent of ∠A = tan A = (Side opposite to ∠A)/ (Side adjacent to ∠A)

cosecant of ∠A = cosec A = 1/sin A = Hypotenuse/ Side opposite to ∠A

secant of ∠A = sec A = 1/cos A = Hypotenuse/ Side adjacent to ∠A

cotangent of ∠A = cot A = 1/tan A = (Side adjacent to ∠A)/ (Side opposite to ∠A)

Also, tan A = sin A/cos A

cot A = cos A/sin A

Also, read: Trigonometry

Trigonometry Questions and Answers

1. From the given figure, find tan P – cot R.

From the given,

In the right triangle PQR, Q is right angle.

By Pythagoras theorem,

PR 2 = PQ 2 + QR 2

QR 2 = (13) 2 – (12) 2

= 169 – 144

tan P = QR/PQ = 5/12

cot R = QR/PQ = 5/12

So, tan P – cot R = (5/12) – (5/12) = 0

2. Prove that (sin 4 θ – cos 4 θ +1) cosec 2 θ = 2

L.H.S. = (sin 4 θ – cos 4 θ +1) cosec 2 θ

= [(sin 2 θ – cos 2 θ) (sin 2 θ + cos 2 θ) + 1] cosec 2 θ

Using the identity sin 2 A + cos 2 A = 1,

= (sin 2 θ – cos 2 θ + 1) cosec 2 θ

= [sin 2 θ – (1 – sin 2 θ) + 1] cosec 2 θ

= 2 sin 2 θ cosec 2 θ

= 2 sin 2 θ (1/sin 2 θ)

3. Prove that (√3 + 1) (3 – cot 30°) = tan 3 60° – 2 sin 60°.

LHS = (√3 + 1)(3 – cot 30°)

= (√3 + 1)(3 – √3)

= 3√3 – √3.√3 + 3 – √3

= 2√3 – 3 + 3

RHS = tan 3 60° – 2 sin 60°

= (√3) 3 – 2(√3/2)

= 3√3 – √3

Therefore, (√3 + 1) (3 – cot 30°) = tan 3 60° – 2 sin 60°.

Hence proved.

4. If tan(A + B) = √3 and tan(A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B.

tan(A + B) = √3

tan(A + B) = tan 60°

A + B = 60°….(i)

tan(A – B) = 1/√3

tan(A – B) = tan 30°

A – B = 30°….(ii)

Adding (i) and (ii),

A + B + A – B = 60° + 30°

Substituting A = 45° in (i),

45° + B = 60°

B = 60° – 45° = 15°

Therefore, A = 45° and B = 15°.

5. If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

sin 3A = cos(A – 26°); 3A is an acute angle

cos(90° – 3A) = cos(A – 26°) {since cos(90° – A) = sin A}

⇒ 90° – 3A = A – 26

⇒ 3A + A = 90° + 26°

⇒ 4A = 116°

⇒ A = 116°/4

6. If A, B and C are interior angles of a triangle ABC, show that sin (B + C/2) = cos A/2.

We know that, for a given triangle, the sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

B + C = 180° – A

Dividing both sides of this equation by 2, we get;

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Take sin on both sides,

sin (B + C)/2 = sin (90° – A/2)

⇒ sin (B + C)/2 = cos A/2 {since sin(90° – x) = cos x}

7. If tan θ + sec θ = l, prove that sec θ = (l 2 + 1)/2l.

tan θ + sec θ = l….(i)

We know that,

sec 2 θ – tan 2 θ = 1

(sec θ – tan θ)(sec θ + tan θ) = 1

(sec θ – tan θ) l = 1 {from (i)}

sec θ – tan θ = 1/l….(ii)

tan θ + sec θ + sec θ – tan θ = l + (1/l)

2 sec θ = (l 2 + 1)l

sec θ = (l 2 + 1)/2l

8. Prove that (cos A – sin A + 1)/ (cos A + sin A – 1) = cosec A + cot A, using the identity cosec 2 A = 1 + cot 2 A.

LHS = (cos A – sin A + 1)/ (cos A + sin A – 1)

Dividing the numerator and denominator by sin A, we get;

= (cot A – 1 + cosec A)/(cot A + 1 – cosec A)

Using the identity cosec 2 A = 1 + cot 2 A ⇒ cosec 2 A – cot 2 A = 1,

= [cot A – (cosec 2 A – cot 2 A) + cosec A]/ (cot A + 1 – cosec A)

= [(cosec A + cot A) – (cosec A – cot A)(cosec A + cot A)] / (cot A + 1 – cosec A)

= cosec A + cot A

9. Prove that: (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

[Hint: Simplify LHS and RHS separately]

LHS = (cosec A – sin A)(sec A – cos A)

= (cos 2 A/sin A) (sin 2 A/cos A)

= cos A sin A….(i)

RHS = 1/(tan A + cot A)

= (sin A cos A)/ (sin 2 A + cos 2 A)

= (sin A cos A)/1

= sin A cos A….(ii)

From (i) and (ii),

i.e. (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

10. If a sin θ + b cos θ = c, prove that a cosθ – b sinθ = √(a 2 + b 2 – c 2 ).

a sin θ + b cos θ = c

Squaring on both sides,

(a sin θ + b cos θ) 2 = c 2

a 2 sin 2 θ + b 2 cos 2 θ + 2ab sin θ cos θ = c 2

a 2 (1 – cos 2 θ) + b 2 (1 – sin 2 θ) + 2ab sin θ cos θ = c 2

a 2 – a 2 cos 2 θ + b 2 – b 2 sin 2 θ + 2ab sin θ cos θ = c 2

a 2 + b 2 – c 2 = a 2 cos 2 θ + b 2 sin 2 θ – 2ab sin θ cos θ

a 2 + b 2 – c 2 = (a cos θ – b sin θ ) 2

⇒ a cos θ – b sin θ = √(a 2 + b 2 – c 2 )

Video Lesson on Trigonometry

Practice Questions on Trigonometry

Solve the following trigonometry problems.

• Prove that (sin α + cos α) (tan α + cot α) = sec α + cosec α.
• If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
• If sin θ + cos θ = √3, prove that tan θ + cot θ = 1.
• Evaluate: 2 tan 2 45° + cos 2 30° – sin 2 60°
• Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

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Trig Identities: A Crash Course in Complex Math Concepts

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Fundamental trigonometric identities, aka trig identities or trigo identities, are equations involving trigonometric functions that hold true for any value you substitute into their variables.

These identities are essential tools if you want to solve trigonometric equations and perform complex calculations in mathematics, physics or engineering . Understanding all the trigonometric identities can help you simplify seemingly complicated problems, especially in geometry and calculus.

The Foundation of Trigonometry

Fundamental trigonometric identities, double angle trigonometric identities, triple angle trigonometric identities, half angle identities, sum and difference identities.

Trigonometry is a branch of mathematics . At the heart of trigonometry lie the trigonometric functions, which relate the angles of a triangle to the ratios of its sides.

The most basic trigonometric functions are sine, cosine and tangent, which instructors often teach using the mnemonic SOH-CAH-TOA in right-angled triangles.

From these basic trig functions, we derive other crucial functions, such as secant, cosecant and cotangent, all of which play vital roles in further developing trigonometric theory.

You might hear people refer to sine, cosine, tangent, secant, cosecant and cotangent as the six trigonometric ratios , or trig ratios.

Trigonometric identities form a cornerstone of higher mathematics. They encapsulate all the trigonometric ratios and relationships in a framework that enhances the solving of equations and understanding of geometric and algebraic concepts.

Trigonometric identities encompass a wide range of formulas, but people generally group them into categories based on their specific applications and forms.

There are three main categories comprising eight fundamental trigonometric identities. These categories include reciprocal identities, Pythagorean identities and quotient identities.

Reciprocal Identities

These identities express the basic trigonometric functions in terms of their reciprocal functions:

• Sine and cosecant : csc( θ ) = 1/sin( θ )
• Cosine and secant : sec( θ ) = 1/cos( θ )
• Tangent and cotangent : cot( θ ) = 1/tan( θ )

Pythagorean Identities

The Pythagorean trigonometric identities stem from the Pythagorean theorem , also known as the Pythagoras theorem, after the Greek scholar who came up with the mathematical statement.

The trig identities based on the Pythagorean theorem are fundamental to connecting the squares of the primary trigonometric functions:

• Basic Pythagorean identity : sin 2 ( θ ) + cos 2 ( θ ) = 1
• Derived for tangent : 1 + tan 2 ( θ ) = sec 2 ( θ )
• Derived for cotangent : cot 2 ( θ ) + 1 = csc 2 ( θ )

Quotient Identities

These identities relate the functions through division:

• Tangent as a quotient : tan( θ ) = sin( θ )/cos( θ )
• Cotangent as a quotient : cot( θ ) = cos( θ )/sin( θ )

Of course, there are many more trigonometric identities beyond just these core identities that have applications in specific scenarios, such as double angle, triple angle, half angle and sum and difference identities.

The double angle formulas are trigonometric identities that express trigonometric functions of double angles — that is, angles of the form 2 θ — in terms of trigonometric functions of single angles ( θ ).

These formulas are crucial in various mathematical computations and transformations, particularly in calculus, geometry and solving trigonometric equations.

The primary double angle formulas include those for sine, cosine and tangent.

Cosine Double Angle Formula

The cosine double angle formula is:

cos(2 θ ) = cos 2 ( θ ) – sin 2 ( θ )

You can also represent this in two alternative forms using the Pythagorean identity sin 2 ( θ ) + cos 2 ( θ ) = 1 :

Sine Double Angle Formula

The sine double angle formula is:

This formula is derived from the sum identities and is useful for solving problems involving products of sine and cosine.

Tangent Double Angle Formula

The tangent double angle formula is:

This expression arises from dividing the sine double angle formula by the cosine double angle formula and simplifying using the definition of tangent.

Triple angle formulas, while less commonly used, offer shortcuts in specific scenarios, such as in certain integrals and polynomial equations. These are identities that allow the calculation of the sine, cosine and tangent of three times a given angle (3θ) using the trigonometric functions of the angle itself (θ).

For example, the sine triple angle formula is:

This formula is derived by using the sine double angle formula and the angle sum identity.

Triple angle formulas can be derived from double angle and sum identities and are useful in specific mathematical and engineering contexts, such as simplifying complex trigonometric expressions or solving higher-degree trigonometric equations.

Half angle identities are trigonometric formulas that allow you to prove trigonometric identities for the sine, cosine and tangent of half of a given angle.

Half angle formulas are particularly useful in solving trigonometric equations, integrating trigonometric functions, and simplifying expressions when the angle involved is halved. Half angle formulas are derived from the double angle identities and other fundamental trigonometric identities.

The half angle identities for sine, cosine and tangent use the following half angle formulas:

• Sine half angle identity : sin⁡( θ /2) = ±√((1 – cos θ )/2)
• Cosine half angle identity : cos⁡( θ /2) = ±√((1 + cos θ ​​)/2)
• Tangent half angle identity : tan( θ /2) = sin( θ )/(1 + cos( θ )) = 1 – (cos( θ )/sin( θ ))

In the case of the sine and cosine half angle formulas, the sign depends on the quadrant in which θ /2​ resides. The tangent half angle formula you can also express in terms of sine and cosine directly.

These identities are derived by manipulating the double angle identities. For example, the cosine double angle identity cos(2 θ ) = 2cos 2 ( θ ) can be rearranged to express cos 2 ( θ ) in terms of cos(2 θ ) , and then taking the square root (and adjusting for sign based on the angle's quadrant) gives the half angle formula for cosine.

Half angle identities are crucial for simplifying the integration of trigonometric functions, particularly when integral limits involve pi (π) or when integrating periodic functions. They also play a vital role in various fields of science and engineering where wave functions and oscillations are analyzed.

Sum identities in trigonometry are essential formulas that allow for the calculation of the sine, cosine and tangent of the sum of two angles. Conversely, difference formulas allow you to calculate the sine, cosine and tangent of the difference between two angles.

These identities are incredibly useful for simplifying expressions, solving trigonometric equations and performing complex calculations.

We created this article in conjunction with AI technology, then made sure it was fact-checked and edited by a HowStuffWorks editor.

Please copy/paste the following text to properly cite this HowStuffWorks.com article:

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5. Trigonometry

5.2 Solve Applications: Sine, Cosine and Tangent Ratios.

Learning Objectives

By the end of this section it is expected that you will be able to

• Find missing side of a right triangle using sine, cosine, or tangent ratios
• Find missing angle of a right triangle using sine, cosine, or tangent ratios
• Solve applications using right angle trigonometry

Sine, Cosine, and Tangent Ratios

We know that any right triangle has three sides and a right angle. The side opposite to the right angle is called the hypotenuse. The other two angles in a right triangle are acute angles (with a measure less than 90 degrees). One of those angles we call reference angle and we use θ (theta) to represent it.

The hypotenuse is always the longest side of a right triangle. The other two sides are called opposite side and adjacent side. The names of those sides depends on which of the two acute angles is being used as a reference angle.

In the right triangle each side is labeled with a lowercase letter to match the uppercase letter of the opposite vertex.

Label the sides of the triangle and find the hypotenuse, opposite, and adjacent.

We labeled  the sides with a lowercase letter to match the uppercase letter of the opposite vertex.

c is hypotenuse

a is opposite

b is adjacent

Label the sides of the triangle and find the hypotenuse, opposite and adjacent.

y is hypotenuse

z is opposite

x is adjacent

Trigonometric Ratios

Trigonometric ratios are the ratios of the sides in the right triangle. For any right triangle we can define three basic trigonometric ratios: sine, cosine, and tangent.

Let us refer to Figure 1 and define the three basic trigonometric ratios as:

Three Basic Trigonometric Ratios

Where θ is the measure of a reference angle measured in degrees.

Very often we use the abbreviations for sine, cosine, and tangent ratios.

Some people remember the definition of the trigonometric ratios as SOH CAH TOA.

For the given triangle find the sine, cosine and tangent ratio.

For the given triangle find the sine cosine and tangent ratio.

When calculating we will usually round the ratios to four decimal places and at the end our final answer to one decimal place unless stated otherwise.

For the given triangle find the sine, cosine and tangent ratios. If necessary round to four decimal places.

We have two possible reference angles: R an S.

Using the definitions, the trigonometric ratios for angle R are:

Using the definitions, the trigonometric ratios for angle S:

For the given triangle find the sine, cosine, and tangent ratios. If necessary round to four decimal places.

Now, let us use a scientific calculator to find the trigonometric ratios.  Can you find the sin, cos, and tan buttons on your calculator? To find the trigonometric ratios make sure your calculator is in Degree Mode.

Using a calculator find the trigonometric ratios. If necessary, round to 4 decimal places.

Make sure your calculator is in Degree Mode.

a)  Using a calculator find that sin 30° = 0.5

b)  Using a calculator find that cos 45° = 0.7071 Rounded to 4 decimal places.

c)  Using a calculator find that tan 60° = 1.7321 Rounded to 4 decimal places.

Find the trigonometric ratios. If necessary, round to 4 decimal places.

a)  sin 60°

b)  cos 30°

c)  tan 45°

a)   sin 60° = 0.8660

b)   cos 30° = 0.8660

c)   tan 45° = 1

Finding Missing Sides of a Right Triangle

In this section you will be using trigonometric ratios to solve right triangle problems. We will adapt our problem solving strategy for trigonometry applications. In addition, since those problems will involve the right triangle, it is helpful to draw it (if the drawing is not given) and label it with the given information.We will include this in the first step of the problem solving strategy for trigonometry applications.

HOW TO: Solve Trigonometry Applications

• Read the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts.
• Identify what we are looking for.
• Label what we are looking for by choosing a variable to represent it.
• Find the required trigonometric ratio.
• Solve the ratio using good algebra techniques.
• Check the answer by substituting it back into the ratio in step 4 and by making sure it makes sense in the context of the problem.
• Answer the question with a complete sentence.

In the next few examples, having given the measure of one acute angle and the length of one side of the right triangle, we will solve the right triangle for the missing sides.

 Find the missing sides. Round  your final answer to two decimal places

 Find the missing sides. Round  your final answer to one decimal place.

Find the hypotenuse. Round your final answer to one decimal place.

Finding Missing Angles of a Right Triangle

Sometimes we have a right triangle with only the sides given. How can we find the missing angles? To find the missing angles, we use the inverse of the trigonometric ratios. The inverse buttons sin -1 , cos -1 , and tan -1 are on your scientific calculator.

Find the angles. Round your final answer to one decimal place.

a)  sin A = 0.5

b)  cos B = 0.9735

c)  tan C = 2.89358

Use your calculator and press the 2nd FUNCTION key and then press the SIN, COS, or TAN key

a)  A = sin -1 0.5

b)  B = cos -1 0.9735

c) C = tan -1 2.89358

a)  sin X = 1

b)  cos Y = 0.375

c)  tan Z = 1.676767

In the example below we have a right triangle with two sides given. Our acute angles are missing. Let us see what the steps are to find the missing angles.

Find the missing angle X. Round your final answer to one decimal place.

Find the missing angle A. Round your final answer to one decimal place.

Find the missing angle C. Round your final answer to one decimal place.

Solving a Right Triangle

From the section before we know that any triangle has three sides and three interior angles. In a right triangle, when all six parts of the triangle are known, we say that the right triangle is solved.

Solve the right triangle. Round your final answer to one decimal place.

Since the sum of angles in any triangle is 180°, the measure of angle B can be easy calculated.

We solved the right triangle

Solve the right triangle. Round to two decimal places.

The missing angle F = 180° – 90° – 26.39° = 63.64°

Solve the right triangle. Round to one decimal place.

Solve Applications Using Trigonometric Ratios

In the previous examples we were able to find missing sides and missing angles of a right triangle. Now, let’s use the trigonometric ratios to solve real-life  problems.

Many applications of trigonometric ratios involve understanding of an angle of elevation or angle of depression.

The angle of elevation is an angle between the horizontal line (ground) and the observer’s line of sight.

The angle of depression is the angle between horizontal line (that is parallel to the ground) and the observer’s line of sight.

James is standing 31 metres away from the base of the Harbour Centre in Vancouver. He looks up to the top of the building at a 78° angle. How tall is the Harbour Centre?

Marta is standing 23 metres away from the base of the tallest apartment building in Prince George and looks at the top of the building at a 62° angle. How tall is the building?

43.3 metres

Thomas is standing at the top of the building that is 45 metres high and looks at her friend that is standing on the ground, 22 metres from the base of the building. What is the angle of depression?

Hemanth is standing on the top of a cliff 250 feet above the ground and looks at his friend that is standing on the ground, 40 feet from the base of the cliff. What is the angle of depression?

Key Concepts

• Read the problem and make sure all the words and ideas are understood. Draw the right triangle and label  the given parts.
• Check the answer by substituting it back into the ratio solved in step 5 and by making sure it makes sense in the context of the problem.

5.2 Exercise Set

Label the sides of the triangle.

• If the reference angle in Question 1 is B, Find the adjacent ?
• If the reference angle in Question 2 is Z, find the opposite?

Use your calculator to find the given ratios. Round to four decimal places if necessary:

For the given triangles, find the sine, cosine and tangent of the θ.

For the given triangles, find the missing side. Round it to one decimal place.

For the given triangles, find the missing sides. Round it to one decimal place.

Solve the triangles. Round to one decimal place.

• Kim stands 75 metres from the bottom of a tree and looks up at the top of the tree at a 48° angle. How tall is the tree?
• A tree makes a shadow that is 6 metres long when the angle of elevation to the sun is 52°. How tall is the tree?
• A ladder that is 15 feet is leaning against a house and makes a 45° angle with the ground. How far is the base of the ladder from the house?
• Roxanne is flying a kite and has let out 100 feet of string. The angle of elevation with the ground is 38°. How high is her kite above the ground?
• Marta is flying a kite and has let out 28 metres of string. If the kite is 10 metres above the ground, what is the angle of elevation?
• An airplane takes off from the ground at the angle of 25°. If the airplane traveled 200 kilometres, how high above the ground is it?

• y = 19.3, z = 8.2

Attribution:

This chapter has been adapted from “Solve Applications: Sine, Cosine and Tangent Ratios” in Introductory Algebra  by Izabela Mazur, which is under a CC BY 4.0 Licence . See the Adaptation Statement for more information.

Business/Technical Mathematics Copyright © 2021 by Izabela Mazur and Kim Moshenko is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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High school geometry

Course: high school geometry   >   unit 5.

• Triangle similarity & the trigonometric ratios

Trigonometric ratios in right triangles

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