# TRIGONOMETRIC RATIOS OF COMPOUND ANGLES

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

Formulas for trigonometric ratios of compound angles :

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 cos B

cos (A - B)  =  cos A cos B + sin A cos 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]

## 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°.

## Practice Problems

Problem 1 :

Find the value of cos 15°.

Solution :

Write the given angle 15° in terms of sum or difference of two standard angles.

Then,

15°  =  45° - 30°

cos15°  =  cos (45° - 30°)

cos 15°  =  cos 45° cos 30° + sin 45° sin 30°

Using the above trigonometric ratio table, we have

cos 15°  =  (√2/2)  (√3/2) + (√2/2)  (1/2)

cos 15°  =  (√6 / 4)  +  (√2/4)

cos 15°  =  (√6 + √2)/4

Problem 2 :

Find the value of cos 105°.

Solution :

Write the given angle 105° in terms of  sum or difference of two standard angles.

Then,

105°  =  60° + 45°

cos 105°  =  cos (60° + 45°)

cos 105°  =  cos 60° cos 45° - sin 60° sin 45°

Using the above trigonometric ratio table, we have

cos 105°  =  (1/2)  (√2/2) - (√3/2)  (√2/2)

cos 105°  =  (√2 / 4)  -  (√6/4)

cos 105°  =  (√2 - √6) / 4

Problem 3 :

Find the value of sin 75°.

Solution :

Write the given angle 75° in terms of  sum or difference of two standard angles.

Then,

75°  =  45° + 30°

sin 75°  =  sin (45° + 30°)

sin 75°  =  sin 45° cos 30° + cos 45° sin 30°

Using the above trigonometric ratio table, we have

sin 75°  =  (√2/2)  (√3/2) + (√2/2)  (1/2)

sin 75°  =  (√6 / 4) + (√2/4)

sin 75°  =  (√6 + √2) / 4

Problem 4 :

Find the value of tan 15°.

Solution :

Write the given angle 15° in terms of  sum or difference of two standard angles.

Then,

15°  =  45° - 30°

tan 15°  =  tan (45° - 30°)

tan 15°  =  [tan 45° -  tan 30°] / [1 + tan 45° tan 30°]

Using the above trigonometric ratio table, we have

tan 15°  =  [1 - 1/√3] / [1 + 1 ⋅ 1/√3]

tan 15°  =  [1 - 1/√3] / [1 + 1/√3]

tan 15°  =  [√3/√3 - 1/√3] / [√3/√3 + 1/√3]

tan 15°  =  [(√3 - 1)/√3]  /  [(√3 + 1)/√3]

tan 15°  =  [(√3 - 1)/√3]  x  [(√3/(√3 + 1)]

tan 15°  =  (√3 - 1) / (√3 + 1)

By rationalizing the denominator, we get

tan 15°  =  2 - √3

Problem 5 :

If A and B are acute angles, sin A = 3/5, cos B = 12/13,  find cos (A + B).

Solution :

cos (A + B)  =  cos A cos B - sin A sin B -----(1)

To find the value of cos (A + B), we need the values of

cos A, cos B, sin A, sin B

The values of sinA and cos A are given in the question itself.

So, we have to find the value of sin B and cos A.

Finding the value of sin B :

sin2B  =  1 - cos2B

sin2B  =  1 - (12/13)2

sin2B  =  1 - 144/169

sin2B  =  169/169 - 144/169

sin2B  =  (169 - 144)/169

sin2B  =  25/169

sin2B  =  (5/13)2

sin B  =  5/13

Finding the value of cos A :

cos2A  =  1 - sin2A

cos2A  =  1 - (3/5)2

cos2A  =  1 - 9/25

cos2A  =  25/25 - 9/25

cos2A  =  (25 - 9)/25

cos2A  =  16/25

cos2A  =  (4/5)2

cos A  =  4/5

Substitute the values of cosA, cosB, sinA and sinB in (1).

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

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

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

cos (A + B)  =  (48 - 15) / 65

cos (A + B)  =  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. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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