# COMPOUND ANGLE PROBLEMS WITH THREE ANGLES

Problem 1 :

Expand cos(A + B + C). Hence prove that

cosAcosB cosC = sinAsinB cosC + sinB sinC cosA + sinC sinAcos B, if A + B + C = π/2

Solution :

cos(A + B + C)  =  cos ( (A + B) + C )

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

=  (cos A cos B - sin A sin B) cos C - (sin A cos B - cos A sin B) sin C

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

If A + B + C = π/2

cos π/2  =  cos A cos B cos C - sin A sin B cos C - sin A cos B sin C  - cos A sin B sin C

The value of cos π/2 is 0.

So, cosAcosB cosC  =  sinAsinB cosC + sinB sinC cosA + sinC sinAcos B.

Hence proved.

Problem 2 :

Prove that

(i) sin(45° + θ) − sin(45° − θ) = √2 sinθ.

(ii) sin(30° + θ) + cos(60° + θ) = cos θ.

Solution :

(i)  sin(45° + θ) − sin(45° − θ) = √2 sinθ :

L.H.S

=  sin(45° + θ) − sin(45° − θ)

=  sin 45°cos θ + cos 45°sinθ-(sin 45°cosθ - cos 45°sin θ)

=  sin 45°cos θ + cos 45°sinθ - sin 45°cosθ + cos 45°sin θ

=  2cos 45°sin θ

=  2(1/√2) sin θ

=  √2 sin θ  ----> R.H.S

Hence proved.

(ii) sin(30° + θ) + cos(60° + θ) = cos θ :

L.H.S

sin(30° + θ) + cos(60° + θ)

=  sin 30°cos θ + cos 30°cos θ + cos 60°cos θ - sin 60°sin θ

=  (1/2) cos θ + (√3/2cos θ + (1/2) cos θ - (√3/2) sin θ

=  (1/2) cos θ + (1/2) cos θ

=  2 cos θ / 2

=  cos θ

Hence proved.

Problem 3 :

If a cos(x + y) = b cos(x − y), show that (a + b) tanx = (a − b) cot y.

Solution :

a cos(x + y) = b cos(x − y)

a (cos x cos y - sin x sin y)  =  b (cos x cos y + sin x sin y)

a cos x cos y - a sin x sin y  =  b cos x cos y + b sin x sin y

a cos x cos y - b cos x cos y  =  b sin x sin y + a sin x sin y

Factor out cos x cos y on the left side, in the right side factor out sin x sin y

(a - b) cos x cos y  =  (a + b) sin x sin y

Divide both sides by cos x sin y

(a - b)(cosx cosy/cosx siny)  =  (a + b)(sinx siny/cosx siny)

(a - b) cot y  =  (a + b) tan x

Hence proved.

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