# TRIGONOMETRIC RATIOS OF COMPOUND ANGLES

## About "Trigonometric ratios of compound angles"

Trigonometric ratios of compound angles :

An angle made up of the algebraic sum of two or more angles is called a compound angle.

Here, we are going see the formulas for trigonometric ratios of compound angles.

sin (A + B)  =  sinA cosB + cosA sinB

sin (A - B)  =  sinA cosB - cosA sinB

cos (A + B)  =  cosA cosB - sinA cosB

cos (A - B)  =  cosA cosB + sinA cosB

tan (A + B)  =  [tanA + tanB] / [1 - tanA tanB]

tan (A - B)  =  [tanA - tanB] / [1 + tanA tanB]

## Trigonometric ratio table From the above table, we can get the values of trigonometric ratios for standard angles such as 0°, 30°, 45°, 60°, 90°

Now, let us look at some practice problems on "Trigonometric ratios of compound angles".

## Trigonometric ratios of compound angles - Practice problems

Example 1 :

Find the value of cos15°

Solution :

First, we have to write the given angle 15° in terms of  sum or difference of two standard angles.

So, we have 15°  =  45° - 30°

cos15°  =  cos (45° - 30°)

cos15°  =  cos45° cos30° + sin45° sin30°

Using the above trigonometric ratio table, we have

cos15°  =  (√2/2) x (√3/2) + (√2/2) x (1/2)

cos15°  =  (√6 / 4)  +  (√2/4)

cos15°  =  (√6 + √2) / 4

Hence, the value of  cos15° is equal to  (√6 + √2) / 4

Let us look at the next problem on "Trigonometric ratios of compound angles"

Example 2 :

Find the value of cos105°

Solution :

First, we have to write the given angle 105° in terms of  sum or difference of two standard angles.

So, we have 105°  =  60° + 45°

cos105°  =  cos (60° + 45°)

cos105°  =  cos60° cos45° - sin60° sin45°

Using the above trigonometric ratio table, we have

cos105°  =  (1/2) x (√2/2) - (√3/2) x (√2/2)

cos105°  =  (√2 / 4)  -  (√6/4)

cos15°  =  (√2 - √6) / 4

Hence, the value of  cos15° is equal to (√2 - √6) / 4

Let us look at the next problem on "Trigonometric ratios of compound angles"

Example 3 :

Find the value of sin75°

Solution :

First, we have to write the given angle 75° in terms of  sum or difference of two standard angles.

So, we have 75°  =  45° + 30°

sin75°  =  sin (45° + 30°)

sin75°  =  sin45° cos30° + cos45° sin30°

Using the above trigonometric ratio table, we have

sin75°  =  (√2/2) x (√3/2) + (√2/2) x (1/2)

sin75°  =  (√6 / 4) + (√2/4)

sin75°  =  (√6 + √2) / 4

Hence, the value of sin75° is equal to (√6 + √2) / 4

Let us look at the next problem on "Trigonometric ratios of compound angles"

Example 4 :

Find the value of tan15°

Solution :

First, we have to write the given angle 15° in terms of  sum or difference of two standard angles.

So, we have 15°  =  45° - 30°

tan15°  =  tan (45° - 30°)

tan15°  =  [tan45° -  tan30°] / [1 + tan45° tan30°]

Using the above trigonometric ratio table, we have

tan15°  =  [1 - 1/√3] / [1 + 1x1/√3]

tan15°  =  [(√3 - 1)/√3]  /  [(√3 + 1)/√3]

tan15°  =  [(√3 - 1)/√3]  x  [(√3/(√3 + 1)]

tan15°  =  (√3 - 1) / (√3 + 1)

By rationalizing the denominator, we get

tan15°  =  2 - √3

Hence, the value of tan15° is equal to 2 - √3

Let us look at the next problem on "Trigonometric ratios of compound angles"

Example 5 :

If A and B are acute angles, sinA = 3/5, cosB = 12/13,

find cos (A+B)

Solution :

cos (A + B)  =  cosAcosB - sinAsinB

From the formula of cos (A + B), the value of sinA and cosA are given in the question.

We need the value of sinB and cosA.

We know that,

sin ² B  =  1 - cos ² B

sin ² B  =  1 - (12/13)²

sin ² B  =  1 - 144/169

sin ² B  =  (169 - 144)/169

sin ² B  =  25/169

sin ² B  =  (5/13)² --------> sinB  =  5/13

cos ² A  =  1 - sin ² A

cos ² A  =  1 - (3/5)²

cos ² A  =  1 - 9/25

cos ² A  =  (25 - 9)/25

cos ² A  =  16 / 25

cos ² A  =  (4/5)² --------> cosA  =  4 / 5

Plugging the values of cosA, cosB, sinA and sinB in the formula of cos (A+B), we get,

cos (A + B)  =  (4/5) x (12/13)  -  (3/5) x (5/13)

cos (A + B)  =  (4/5) x (12/13)  -  (3/5) x (5/13)

cos (A + B)  =  (48 / 65)  -  (15 / 65)

cos (A + B)  =  (48 - 15) / 65

cos (A + B)  =  33 / 65

Hence, the value of cos(A+B) is 33/65.

## Compound angles without standard angles

As we have seen in the above examples, some angles can not be written in terms of sum or difference of two standard angles.

For example,

Let us consider sin 225°

Here, 225° can not be written in terms of sum or difference of two standard angles.

All that we can do is, 225° can be written in terms sum or difference of two angles where one of the angles will be quadrantal angles such as 0°, 90°, 180°, 270°

So, sin 225° can be written as

sin (180° + 45°)  or  sin (270° - 45°)

To evaluate sin (180° + 45°)  or  sin (270° - 45°), we have to know ASTC formula

ASTC formla has been explained clearly in the figure given below. More clearly In the first quadrant (0° to 90°), all trigonometric ratios are positive.

In the second quadrant (90° to 180°), sin and csc are positive and other trigonometric ratios are negative.

In the third quadrant (180° to 270°), tan and cot are positive and other trigonometric ratios are negative.

In the fourth quadrant (270° to 360°), cos and sec are positive and other trigonometric ratios are negative.

## Important conversions

When we have the angles 90° and 270° in the trigonometric ratios in the form of

(90° + θ)

(90° - θ)

(270° + θ)

(270° - θ)

We have to do the following conversions,

sin θ <------> cos θ

tan θ <------> cot θ

csc θ <------> sec θ

For example,

sin (270° + θ)  =  - cos θ

cos (90° - θ)  =  sin θ

For the angles 0° or 360° and  180°, we should not make the above conversions.

## Evaluation of trigonometric ratios using ASTC formula

Now, let us evaluate sin (180° + 45°)

To evaluate sin (90° + θ), we have to consider the following important points.

(i)  (180° + 45°) will fall in the III rd quadrant.

(ii)  When we have 180°, "sin" will not be changed.

(iii)  In the III rd quadrant, the sign of "sin" is negative.

Considering the above points, we have

sin 225°  =  sin (180° + 45°)

sin 225°  =  - sin 45°

sin 225°  =  - √2 / 2

When we are not able to write the given angle in terms of sum or difference of two standard angles, we have to proceed the problem in this way.

After having gone through the stuff given above, we hope that the students would have understood "Trigonometric ratios of compound angles"

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