Question

Prove that: 4sin(pi/3-theta)sin(pi/3+theta)=3-4sin^2 theta

Answer 1

Answer:

1/4 cos 3 alpha

Step-by-step explanation:

LHS,

= cos alpha × cos [90° × 2 - (60° - alpha)] × cos [90° × 2 + (60° + alpha)]

= cos alpha × [- cos (60° - alpha)] × [- cos (60° + alpha)]

= cos alpha × cos (60° - alpha) × cos (60° + alpha)

= cos alpha × ( cos^2 alpha - sin^2 60°) [since, cos^2 b - sin^2 a = cos (a - b) × cos (a + b)]

= cos alpha × [cos^2 alpha - ( root 3 / 2)^2]

= cos alpha × [(4 cos^2 alpha - 3) / 4]

= (4 cos^3 alpha - 3 cos alpha) / 4

= 1 / 4 cos 3 alpha

proved