# Trigonometric Ratios of Compound Angles

The branch of mathematics which deals with angles and their measurement is trigonometry. Trigonometric functions include trigonometric ratios, compound angles, multiple angles, which are used to get consistent results. This article explains Trigonometric Ratios of Compound Angles along with solved examples.

An angle is a ray rotated about the initial point. The point where the rotation starts is the initial side and the point where it stops is the terminal side. If a ray is rotated in the anticlockwise direction, it makes a positive angle and the clockwise direction makes a negative angle. The angle can be measured in degree or radians.

## What are Compound Angles?

The algebraic sum of 2 or more angles can be called a compound angle. Trigonometric identities are used to represent compound angles. The basic operations of finding the sum and difference of functions can be computed using the concept of compound angles.

## List of Trigonometric Ratios of Compound Angles

The formulae for trigonometric ratios of compound angles are as follows:

• 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 + tanB] / [1 – tan A tan B]
• tan (A – B) = [tan A – tan B] / [1 + tan A tan B]
• sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A
• cos (A + B) cos (A – B) = cos2 A – sin2 A – sin2 B = cos2 B – sin2 A

Transformation of Products into Sum or Difference of Sines and Cosines:

(a) 2 sin A cos B = sin (A+B) + sin (A-B)

(b) 2 cos A cos B = cos (A+B) + cos (A-B)

(c) 2 cos A sin B = sin (A+B) – sin (A-B)

(d) 2 sin A sin B = cos (A-B) – cos (A+B)

Multiple and sub multiple Angles

(a) sin 2A = 2 sin A cos A = 2 tan A/(1+tan2A)

(b) cos 2A = cos2A – sin2A  = 2cos2A-1 = 1-2 sin2A

(c) tan 2A = 2 tan A/(1-tan2A)

(d) sin 3A = 3 sinA – 4 sin3A

(e) cos 3A = 4 cos3A – 3 cos A

(f) tan 3A = (3 tan A- tan3A)/(1-tan2A)

## Solved Problems

Example 1: If sin A = 1 / √10 and sin B = 1 / √5, where A and B are positive acute angles, then what is A + B ?

Solution:

We know that sin (A + B) = sin A cos B + cos A sin B

= [1 / √10] [√(1 − 1 / 5)] + [1 / √5] √(1 − 1 / 10)

= [1 / √10] [√4 / 5] + [1 / √5] [√9 / 10]

= [1 / √50] * (2 + 3)

= 5 / √50 = 1 / √2

⇒ sin (A + B) = sin π / 4

Hence, A + B = π / 4

Example 2: If sin A + sin B = C, cos A + cos B = D, then the value of sin (A + B) = ?

Solution:

As given [sin A + sin B] / [cos A + cos B] = [C / D]

⇒ {(2 sin [A + B] / 2) * (cos [A − B] / 2)} / {(2 cos [A + B] / 2) * (cos [A − B] / 2)} = C / D

⇒tan ([A + B] / 2) = C / D

Thus, sin (A + B) = {2 (tan [A + B] / 2)} / {(1 + (tan2 [A + B] / 2)}

= {(2 [C / D]) / 1 + [C2 / D2]}

= 2 CD / [C2 / D2]

Example 3: If y = (1 + tan A) (1 − tan B) where A − B = π / 4, then (y + 1)y+1 is equal to __________.

Solution:

A − B = π / 4 ⇒ tan (A − B) = tan π / 4

⇒ [tan A − tan B] / [1 + tan A tan B] = 1

⇒tan A − tan B − tan A tan B = 1

⇒tan A − tan B −tan A tan B + 1 = 2

⇒(1 + tan A) (1 − tan B) = 2

y = 2

Hence, (y + 1)y+1 = (2 + 1)2+1 = (3)3 = 27

Example 4: If A + B = 225o, then find [cot A] / [1 + cotA] * [cot B] / [1 + cot B].

Solution:

[cot A] / [1 + cotA] * [cot B] / [1 + cot B] = 1 / [(1 + tan A) * (1 + tan B)]

=1 / [tan A + tan B + 1 + tan A tan B] [∵ tan (A + B) = tan225o]

⇒ tan A + tan B = 1− tan A tan B

= 1 / [1 − tan A tan B + 1 + tan A tan B]

= 1 / 2

Example 5: If cos (α + β) = 4 / 5, sin (α − β) = 5 / 13 and α, β lie between 0 and π / 4, then find tan 2α.

Solution:

We have cos (α + β) = 4 / 5, sin (α − β) = 5 / 13 ⇒ sin (α + β) = 3 / 5 and cos (α − β) = 12 / 13

⇒ 2α = sin−1 3 / 5 + sin−1 5 / 13

= sin−1 {[[3 / 5] [√(1 − 25 / 169) + [5 / 13] √(1 − 9 / 25)]}

⇒ 2α = sin−1 (56 / 65)

⇒ sin 2α = 56 / 65

Now, tan 2α = sin2α / cos2α = [56/65] / [33 / 65]

= 56 / 33

Example 6: If $\cos \theta =\frac{3}{5}$ and $\cos \varphi =\frac{4}{5},$ where $\theta \text \ and \ \varphi$ are positive acute angles, then $\cos \frac{\theta -\varphi }{2}=$

$A) \frac{7}{\sqrt{2}}\\ B) \frac{7}{5\sqrt{2}}\\ C) \frac{7}{\sqrt{5}}\\ D) \frac{7}{2\sqrt{5}}\\$

Solution:

We have $\cos \theta =\frac{3}{5} \text \ and \ \cos \varphi =\frac{4}{5}$.

Therefore, $\cos (\theta -\varphi )=\cos \theta \cos \varphi +\sin \theta \sin \varphi \\=\frac{3}{5}.\frac{4}{5}+\frac{4}{5}.\frac{3}{5}=\frac{24}{25}$

But $2{{\cos }^{2}}\left( \frac{\theta -\varphi }{2} \right)=1+\cos (\theta -\varphi )=1+\frac{24}{25}=\frac{49}{50} \\ {{\cos }^{2}}\left( \frac{\theta -\varphi }{2} \right)=\frac{49}{50}.$

Hence $\cos \left( \frac{\theta -\varphi }{2} \right)=\frac{7}{5\sqrt{2}}.$

Example 7: If $\cos 2B=\frac{\cos (A+C)}{\cos (A-C)},$ then $\tan A,\ \tan B,\ \tan C$ are in

A) A.P.

B) G.P.

C) H.P.

D) None of these

Solution:

$\cos 2B=\frac{\cos (A+C)}{\cos (A-C)}=\frac{\cos A\cos C-\sin A\sin C}{\cos A\cos C+\sin A\sin C}\\ \frac{1-{{\tan }^{2}}B}{1+{{\tan }^{2}}B}=\frac{1-\tan A\tan C}{1+\tan A\tan C}\\ 1+{{\tan }^{2}}B-\tan A\tan C-\tan A\tan C{{\tan }^{2}}B\\ =1-{{\tan }^{2}}B+\tan A\tan C-\tan A\tan C{{\tan }^{2}}B\\ 2{{\tan }^{2}}B=2\tan A\tan C\Rightarrow {{\tan }^{2}}B=\tan A\tan C$

Hence, tan A, tan B and tan C will be in G.P.

Example 8: If $a\tan \theta =b,$ then $a\cos 2\theta +b\sin 2\theta =$

A) a

B) b

C) −a

D) −b

Solution:

Given that $\tan \theta =\frac{b}{a}.$

Now, $a\cos 2\theta +b\sin 2\theta =a\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)+b\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)$

Putting $\tan \theta =\frac{b}{a},$ we get

$=a\left( \frac{1-\frac{{{b}^{2}}}{{{a}^{2}}}}{1+\frac{{{b}^{2}}}{{{a}^{2}}}} \right)+b\left( \frac{2\frac{b}{a}}{1+\frac{{{b}^{2}}}{{{a}^{2}}}} \right)\\=a\left( \frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \right)+b\left( \frac{2ba}{{{a}^{2}}+{{b}^{2}}} \right)\\ =\frac{1}{({{a}^{2}}+{{b}^{2}})}\{{{a}^{3}}-a{{b}^{2}}+2a{{b}^{2}}\}\\=\frac{a({{a}^{2}}+{{b}^{2}})}{{{a}^{2}}+{{b}^{2}}}\\=a.$

Example 9: If $\tan x=\frac{b}{a},$ then $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=$

$A) \frac{2\sin x}{\sqrt{\sin 2x}}\\ B) \frac{2\cos x}{\sqrt{\cos 2x}}\\ C) \frac{2\cos x}{\sqrt{\sin 2x}}\\ D) \frac{2\sin x}{\sqrt{\cos 2x}}$

Solution:

Given that, $\tan x=\frac{b}{a}$

Now, $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}\\=\sqrt{\frac{1+b/a}{1-b/a}}+\sqrt{\frac{1-b/a}{1+b/a}}\\ =\frac{2}{\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}}=\frac{2}{\sqrt{1-{{\tan }^{2}}x}}\\=\frac{2}{\sqrt{1-\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}}\\=\frac{2\cos x}{\sqrt{\cos 2x}}$