Question

Sin 6 theta +sin 4 theta -sin 2 theta =4cos theta ×sin 2 theta ×cos 3 theta

Answer 1

We have to prove that, sin6θ + sin4θ - sin2θ = 4cosθ sin2θ cos3θ

Proof : LHS = sin6θ + sin4θ - sin2θ

Using formula,

sinC - sinD = 2cos(C + D)/2 sin(C - D)/2

so, sin6θ - sin2θ = 2cos(6θ - 2θ)/2 sin(6θ + 2θ)/2

= 2cos2θ sin4θ

LHS = 2cos2θ sin4θ + sin4θ

Using sin2x = 2sinx cosx

so, sin4θ = 2sin2θ cos2θ

= 2cos2θsin4θ + 2sin2θ cos2θ

= 2cos2θ(sin4θ + sin2θ)

= 2cos2θ [2sin(4θ + 2θ)/2 cos(4θ - 2θ)/2]

= 4cos2θ sin3θ cosθ = RHS

Hence proved.

Answer 2

Answer:

sin6theta+sin4theeta