# Problems on Compound Angles

We will learn how to solve different types of problems on compound angles using formula.

We will see step-by-step how to deal with the trigonometrical ratios of compound angles in different questions.

1. An angle θ is divided into two parts so that the ratio of the tangents of the parts is k; if the difference between the parts be ф, prove that, sin ф = (k - 1)/(k + 1) sin θ .

Solution:

Let, α and β be the two parts of the angle θ.

Therefore, θ = α + β.

By question, θ = α - β. (assuming a >β)

and tan α/tan β = k

⇒ sin α cos β/sin β cos α = k/1

⇒ (sin α cos β + cos α sin β)/(sin α cos β - cos α sin β) = (k + 1)/(k - 1), [by componendo and dividendo]

⇒ sin (α + β)/sin (α - β) = (k + 1)/(k - 1)

⇒ (k + 1) sin Ø = (k - 1) sin θ, [Since we know that α + β = θ; α + β = ф]

⇒ sin ф = (k - 1)/(k + 1) sin θ.                                 Proved.

2. If x + y = z and tan x = k tan y, then prove that sin (x - y) = [(k - 1)/(k + 1)] sin z

Solution:

Given tan x = k tan y

⇒ sin x/cos x = k ∙ sin y/cos y

⇒ sin x cos y/cos x sin y = k/1

Applying componendo and dividend, we get

sin x cos y + cos x sin y/ sin x cos y - cos x sin y = k + 1/k - 1

⇒ sin (x + y)/sin (x – y) = k + 1/k - 1

⇒ sin z/sin (x – y) = k + 1/k - 1, [Since x + y = z given]

⇒ sin (x – y) = [k + 1/k – 1] sin z                                  Proved.

3.  If A + B + C = π and cos A = cos B cos C, show that, tan B tan C = 2

Solution:

A + B + C = π

Therefore, B + C = π - A

⇒ cos (B + C) = cos (π - A)

⇒ cos B cos C - sin B sin C = - cos A

⇒ cos B cos C + cos B cos C = sin B sin C,[Since we know, cos A = cos B cos C]

⇒ 2 cos B cos C = sin B sin C

⇒ tan B tan C = 2                                 Proved.

Note: In different problems on compound angles we need to use the formula as required.

4. Prove that cot 2x + tan x = csc 2x

Solution:

L.H.S. = cot 2x + tan x

= cos 2x/sin 2x + sin x/cos x

= cos 2x cos x + sin 2x sin x/sin 2x cos x

= cos (2x - x)/sin 2x cos x

= cos x/sin 2x cos x

= 1/sin 2x

= csc 2x = R.H.S.                                 Proved.

5.  If sin (A + B) + sin (B + C) + cos (C - A) = -3/2 show that,

sin A + cos B + sin C = 0; cos A + sin B + cos C = 0.

Solution:

Since, sin (A + B) + sin (B + C) + cos (C - A) = -3/2

Therefore, 2 (sin A cos B + cos A sin B + sin B cos C + cos B sin C + cos C cos A + sin C sin A) = -3

⇒ 2 (sin A cos B + cos A sin B + sin B cos C + cos B sin C + cos C cos A + sin C sin A) = - (1 + 1 + 1)

⇒ 2 (sin A cos B + cos A sin B + sin B cos C + cos B sin C + cos C cos A + sin C sin A) = - [(sin^2 A + cos^2 A) + (sin^2 B + cos^2 B) + (sin^2 C + cos^2 C)]

⇒ (sin^2 A + cos^2 B + sin^2 C  + 2 sin A sin C + 2 sin A cos B + 2 cos B sin C) + (cos^2 A + sin^2 B + cos^2 C + 2 cos A sin B + 2 sin B cos C + 2 cos A cos C) = 0

⇒ (sin A + sin B + sin C)^2 + (cos A + sin B + cos C)^2

Now the sum of squares of two real quantities is zero if each quantity is separately zero.

Therefore, sin A + cos B + Sin C = 0

and cos A + sin B + cos C = 0.                                 Proved.