Verified answer

use sinC + sinD = 2sin(C+D)/2 *cos(C - D)/2 and cosC - cosD = 2 sin (C +D)/2 *sin (D - C)/2

=>sinθ[sinθ+sin(3θ)] = cosθ[cosθ-cos(3θ)]

=>sinθ [2sin(θ+3θ)/2 *cos(θ-3θ)/2] = cosθ [2sin(θ+3θ)/2 *sin(3θ-θ)/2]

=>sinθ[2sin(2θ) *cos(-θ)] = cosθ[2sin(2θ)*sin(θ)]

=>sinθ*2sin(2θ)*cos(θ) = cosθ*2sin(2θ)*sin(θ) ; as we kno cos(-θ) = cosθ

=>2sin(θ)*cos(θ)*sin(2θ) = 2sin(θ)*cos(θ)*sin(2θ) ; or

=>sin(2θ)*sin(2θ) = sin(2θ)*sin(2θ) ; or

=>sin²(2θ) = sin²(2θ)

LHS = RHS

Hope this helped

Vick

sin ? [ sin(3?) + sin (5?) ] = sin ? [ sin (4? - ?) + sin(4? + ?) ] use the id sin(A + B) + sin(A - B) = 2 sin A cos B = sin ? [ 2sin (4?) cos ? ] = cos ? [ 2 sin (4?) sin ? ] use the id 2 sin A sin B = cos (A - B) - cos (A + B) = cos ? [ cos(4? - ?)- cos(4? + ?) ] = cos ? [ cos (3?) - cos (5?) ]