Question

Prove that cot 4x (sin 5x+sin 3x)=cot x (sin 5x-sin 3x)

Answer 1

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

$=\frac{cos\\.4x}{sin\,4x}[2\,sin(\frac{5x+3x}{2})cos(\frac{5x-3x}{2})]$

$[∵sin\,A+sin\,B=2\,sin(\frac{A+B}{2})cos(\frac{A-B}{2})]$

$=\frac{cos\\.4x}{sin\,4x}[2\,sin\,4x\,cos\,x]$

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x-sin 3x)

$=\frac{cos\\.x}{cos\,x}[2\,cos(\frac{5x+3x}{2})cos(\frac{5x-3x}{2})]$

$[∵sin\,A+sin\,B=2\,sin(\frac{A+B}{2})cos(\frac{A-B}{2})]$

$=\frac{cos\\.x}{sin\,x}[2\,cos\,4x\,sin\,x]$

= 2 cos 4x. cos x

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