Question

Prove the following:
cot 4x ( sin 5x + sin 3x) = cot x (sin 5x- sin 3x)

Answer 1

We have L.H.S. = cot 4x (sin 5x+ sin 3x)

= $\frac{\text{cos 4x}}{\text{sin 4x}}\left[2\text{sin}\left(\frac{5x+3x}{2}\right)\text{cos}\left(\frac{5x-3x}{2}\right)\right]$

= $\frac{\text{cos 4x}}{\text{sin 4x}}\left[\text{2 sin 4x cos x}\right]$

= 2 cos 4x cos x

We have R.H.S = cot x [sin 5x- sin 3x]

= $\frac{\text{cos x}}{\text{sin x}}\left[\text{2 cos}\left(\frac{5x+3x}{2}\right)\text{sin}\left(\frac{5x-3x}{2}\right)\right]$

$[\because \text{sin C- sin D = 2 cos}\frac{C+D}{2}.\text{sin}\frac{C-D}{2}]$

= $\frac{\text{cos x}}{\text{sin x}}\left[\text{2 cos 4x sin x}\right]$

= 2 cos 4x cos x

Hence L.H.S. = R.H.S.