Question

if α = π/9, prove that

(i) 16sinα. sin 2α. sin 3α. sin 4α = 3

(ii) 16cosα. cos 2α. cos 3α. cos 4α = 1

Answer 1

Given

α = π/9 = 20°

(i) L.H.S = 16 sinα. sin 2α. sin 3α. sin 4α

= 16 . sin 20°. sin 40°. sin 60°. sin 80°

= 16(√3/2).sin 20° .sin (60° + 20°) sin 60° - 20°)

= 8√3(1/4) sin3(20°)  (since,sinθ . sin(60°+θ). sin(60°-θ) = 1/4sin 3θ)

= 2√3. sin60° = 2.√3.(√3/2)

= 3 = R.H.S

(ii) L.H.S 

= 16. cosα. cos 2α. cos3αcos4α

= 16. cos20°. cos40°.cos60° cos80°

= 16. (1/2). cos20°. cos(60°+20°). cos(60°-20°)

= 8(1/4) cos 3(20°)  [ since,cosθ.cos(60°+θ).cos(60°-θ)=(1/4) cos3θ]

= 2.cos60°

= 2(1/2)

= 1 = R.H.S