GRADE 11 TRIGONOMETRY PROBLEMS INVOLVING ANGLES

Problem 1 :

prove that

sin θ/2 sin 7θ/2 + sin 3θ/2 sin 11θ/2 = sin2θ sin 5θ.

Solution :

L.H.S :

  =  sin θ/2 sin 7θ/2 + sin 3θ/2 sin 11θ/2

  =  (1/2) [2 sin θ/2 sin 7θ/2] + (1/2) [2 sin3θ/2 sin 11θ/2]

2 sin θ/2 sin 7θ/2  =  cos 3θ - cos 4θ   

2 sin3θ/2 sin 11θ/2  =  cos 4θ - cos 7θ

 Applying the those values, we get

  =  (1/2) [cos 3θ - cos 4θ] + (1/2) [cos 4θ - cos 7θ]

  =  (1/2) [cos 3θ - cos 4θ + cos 4θ - cos 7θ]

  =  (1/2) [cos 3θ - cos 7θ]

  =  (1/2) [2 sin 5θ sin 2θ]

  =  sin 5θ sin 2θ

Hence proved.

Problem 2 :

Prove that

cos (30°−A) cos (30°+A) + cos (45°−A) cos (45°+A) = cos2A +  (1/4)

Solution :

L.H.S

  =  cos (30°−A) cos (30°+A) + cos (45°−A) cos (45°+A)

 =  (1/2) (2cos (30°−A) cos (30°+A)) + (1/2) (2cos (45°−A) cos (45°+A))

  =  (1/2) [cos 60 + cos 2A + cos 90 + cos 2A]

  =  (1/2) [(1/2) + cos 2A + 0 + cos 2A]

  =  (1/2) [(1/2) + 2cos 2A]

  =  1/4 + cos 2 A

Hence proved.

Problem 3 :

Prove that

(sin x + sin3x + sin5x + sin7x) / (cos x + cos3x + cos5x + cos7x)  =  tan4x.

Solution :

  =  (sin x + sin3x + sin5x + sin7x) / (cos x + cos3x + cos5x + cos7x) 

sin x + sin3x  =   2 sin 2x cos x -------(1)

sin5x + sin7x  =  2 sin 6x cos x-------(2)

cos x + cos3x  =  2 cos 2x cos x -------(3)

cos5x + cos7x  =  2 cos 6x cos x -------(4)

(1) + (2)

  =  2 sin 2x cos x + 2 sin 6x cos x

  =  2 cos x (sin 2x + sin 6x)  -----(A)

(3) + (4)  

  =   2 cos 2x cos x + 2 cos 6x cos x

  =  2 cos x (cos 2x + cos 6x)  -----(B)

(A)/(B)

  =  2 cos x (sin 2x + sin 6x) / 2 cos x (cos 2x + cos 6x)

  =  (sin 2x + sin 6x) / (cos 2x + cos 6x)

Again using the formula for sin C + sin D and cos C + cos D, we get

  =  2 sin 4x cos 2x / 2 cos 4x cos 2x

  =  tan 4x

Hence proved. 

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.