Prove that $\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$
Prove that $\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$
$L.H.S$ $=\cot 4 x(\sin 5 x+\sin 3 x)$
$=\frac{\cot 4 x}{\sin 4 x}\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right]$
$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=\left(\frac{\cos 4 x}{\sin 4 x}\right)[2 \sin 4 x \cos x]$
$=2 \cos 4 x \cos x$
$R.H.S.$ $=\cot x(\sin 5 x-\sin 3 x)$
$=\frac{\cos x}{\sin x}\left[2 \cos \left(\frac{5 x+3 x}{2}\right) \sin \left(\frac{5 x-3 x}{2}\right)\right]$
$\left[\because \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right]$
$=\frac{\cos x}{\sin x}[2 \cos 4 x \sin x]$
$=2 \cos 4 x \cdot \cos x$
$L.H.S.$ $=$ $R.H.S.$
Similar Questions
$\frac{{\sin 3\theta - \cos 3\theta }}{{\sin \theta + \cos \theta }} + 1 = $
$\frac{{\sin 3\theta - \cos 3\theta }}{{\sin \theta + \cos \theta }} + 1 = $
The value of $sin\,10^o$ $sin\,30^o$ $sin\,50^o$ $sin\,70^o$ is
The value of $sin\,10^o$ $sin\,30^o$ $sin\,50^o$ $sin\,70^o$ is
- [JEE MAIN 2019]
If $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ and $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,where $0 \le \alpha ,\beta \le \frac{\pi }{4}$ . Then $\tan 2\alpha =$
If $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ and $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,where $0 \le \alpha ,\beta \le \frac{\pi }{4}$ . Then $\tan 2\alpha =$
- [IIT 1979]
$\frac{{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta }}{{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta }} = $
$\frac{{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta }}{{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta }} = $
If $\sin \theta = \frac{1}{2}\left( {\sqrt {\frac{x}{y}\,} + \,\sqrt {\frac{y}{x}} } \right)\,,\,\left( {x,y \in R\, - \{ 0\} } \right)$. Then
If $\sin \theta = \frac{1}{2}\left( {\sqrt {\frac{x}{y}\,} + \,\sqrt {\frac{y}{x}} } \right)\,,\,\left( {x,y \in R\, - \{ 0\} } \right)$. Then