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, cos A cos B cos C  =  sin A sin B cos C + sin B sin C cos A + sin C sin A cos 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 + (√3/2) cos 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.

Problem 4 :

Let sin A/sin B = sin (A - C) / sin (C - B), where A, B and C are angles of triangle ABC. If the lengths of the sides opposite these angles are a, b and c respectively then

a) b² - a² = a² + c²              b) b² , c², a² are in A.P

c)  c², a², b² are in A.P       d)  a², b², c² are in A.P

Solution :

sin A/sin B = sin (A - C) / sin (C - B) ---(1)

A + B + C = π

A = π - (B + C)

sin A = sin (π - (B + C))

= sin (B + C)

B = π - (A + C)

sin A = sin (π - (A + C))

= sin (A + C)

Applying these values in (1)

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

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

sin² C - sin² B = sin² A - sin² C

sin² C + sin² C = sin² A + sin² B

2sin² C = sin² A + sin² B

By sine rule,

c² = a² + b²

So, a², b² and c² are in arithmetic progression. Option d is correct.

Problem 5 :

cos 10 cos 50 cos 70 = √3/8

Solution :

L.H.S

= cos 10 cos 50 cos 70

Identity to be used :

cos (60 - A) cos (60 + A) cos A = (1/4) cos 3A

50 = 60 - 10

70 = 60 + 10

The angle measures involving in this problem exactly matches with the identity we have.

= (1/4) cos (30)

= 1/4 (√3/2)

= √3/8

Hence it is proved

Problem 6 :

If A + B + C = π, prove that cos² A + cos² B + cos² C = 1 − 2 cos A cos B cos C.

Solution :

cos² A + cos² B + cos² C

= (1/2) [2 cos ²A + 2 cos² B + 2 cos² C] 

= (1/2) [(1 + cos 2A) + (1 + cos 2B) + (1 + cos 2C)]

= 3/2 + 1/2 [(cos 2A + cos 2B) + cos 2C]

= 3/2 + 1/2 [ cos(A + B) cos(A − B) ] + (2 cos² C − 1]

= 3/2 + 1/2[−2 cos C cos(A − B)] + 2 cos² C − 1

(A + B = π − C)

= 3/2 − 1/2 + 1/2 [−2 cos C (cos(A − B) − cos C)]

= 1 − cos C [cos(A − B) − cos C]

= 1 − cos C [cos(A − B) + cos(A + B)]

= 1 − cos C [2 cos A cos B]

= 1 − 2 cos A cos B cos C

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.