Question

Sinx +sin3x+sin5x+sin7x=4cosxcos2xsin4x

Answer 1

this is the answer to your question

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Left hand side :-

sinx + sin3x + sin5x + sin7x

${\boxed{\sf\:{Identity=sinA+sinB=2sin\dfrac{A+B}{2}cos\dfrac{A-B}{2}}}}$

$\tt{\rightarrow 2sin\dfrac{x+5x}{2}cos\dfrac{x-5x}{2}+2sin\dfrac{3x+7x}{2}cos\dfrac{3x-7x}{2}}$

= 2sin3xcos(-2x) + 2sin5xcos(-2x)

= 2sin3xcos2x + 2sin5xcos2x

= 2cos2x(sin3x + sin5x)

$\tt{\rightarrow 2cos2x(2sin\dfrac{3x+5x}{2}cos\dfrac{3x-5x}{2})}$

= 2cos2x{2sin4x.cos(-x)}

= 4cosxcos2xsin4x

LHS = RHS

$\boxed{\begin{minipage}{11 cm} Fundamental Trignometric Indentities \\ \\ \tan (90 - A) = cotA \\ \\ cot (90 - A) = tanA \\ \\ sec (90 - A) = cosecA \\ \\ tan\theta =\dfrac{sin\theta}{cos\theta} \\ \\ cot\theta =\dfrac{cos\theta}{sin\theta} \\ \\ cosec (90 - A) = secA \\ \\ sin^{2}\theta+\cos^{2}\theta =1\\ \\ 1+tan^{2}\theta=\sec^{2}\theta \\ \\ 1 + cot^{2}\theta=\text{cosec}^2\theta \end{minipage}}$