Question
Prove the following:
$\sin2\text{x}+2\sin4\text{x}+\sin6\text{x}=4\cos^2\text{x}\sin4\text{x}$
$\sin2\text{x}+2\sin4\text{x}+\sin6\text{x}=4\cos^2\text{x}\sin4\text{x}$
$=[\sin2\text{x}+\sin6\text{x}]+2\sin4\text{x}$
$=\Big[2\sin\Big(\frac{2\text{x}+6\text{x}}{2}\Big)\Big(\frac{2\text{x}-6\text{x}}{2}\Big)\Big]+2\sin4\text{x}$
$=\Big[\therefore\sin\text{A}+\sin\text{B}=2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=2\sin4\text{x}\cos(-2\text{x})+2\sin4\text{x}$
$=2\sin4\text{x}\cos2\text{x}+2\sin4\text{x}$ $=2\sin4\text{x}(\cos2\text{x}+1)$ $=2\sin4\text{x}(2\cos^2\text{x}-1+1)$ $=2\sin4\text{x}(2\cos^2\text{x})$ $=4\cos^2\text{x}\sin4\text{x}$ $= \text{R.H.S.}$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.