Prove that $\sin (n+1) x \sin (n+2) x+\cos (n+1) x \cos (n+2) x=\cos x$
Prove that $\sin (n+1) x \sin (n+2) x+\cos (n+1) x \cos (n+2) x=\cos x$
$L.H.S.$ $=\sin (n+1) x \sin (n+2) x+\cos (n+1) x \cos (n+2) x$
$=\frac{1}{2}[2 \sin (n+1) x \sin (n+2) x+2 \cos (n+1) x \cos (n+2) x]$
$=\frac{1}{2}\left[\begin{array}{c}\cos \{(n+1) x-(n+2) x\}-c i s\{(n+1) x+(n+2) x\} \\ +\cos \{(n+1) x+(n+2) x\}+\cos \{(n+1) x-(n+2) x\}\end{array}\right]$
$\left[\begin{array}{c}\because-2 \sin A \sin B=\cos (A+B)-\cos (A-B) \\ 2 \cos A \cos B=\cos (A+B)+\cos (A-B)\end{array}\right]$
$=\frac{1}{2} \times 2 \cos \{(n+1) x-(n+2) x\}$
$=\cos (-x)=\cos x= R . H.S$
Similar Questions
Which of the following relations is correct
Which of the following relations is correct
If $\sin \theta = \frac{{ - 4}}{5}$ and $\theta $ lies in the third quadrant, then $\cos \frac{\theta }{2} = $
If $\sin \theta = \frac{{ - 4}}{5}$ and $\theta $ lies in the third quadrant, then $\cos \frac{\theta }{2} = $
If $\sin x = \frac{{ - 24}}{{25}},$ then the value of $\tan \, x$ is
If $\sin x = \frac{{ - 24}}{{25}},$ then the value of $\tan \, x$ is
If $a\,{\cos ^3}\alpha + 3a\,\cos \alpha \,{\sin ^2}\alpha = m$ and $a\,{\sin ^3}\alpha + 3a\,{\cos ^2}\alpha \sin \alpha = n,$ then ${(m + n)^{2/3}} + {(m - n)^{2/3}}$ is equal to
If $a\,{\cos ^3}\alpha + 3a\,\cos \alpha \,{\sin ^2}\alpha = m$ and $a\,{\sin ^3}\alpha + 3a\,{\cos ^2}\alpha \sin \alpha = n,$ then ${(m + n)^{2/3}} + {(m - n)^{2/3}}$ is equal to
If ${\rm{cosec }}A + \cot A = \frac{{11}}{2},$ then $\tan A = $
If ${\rm{cosec }}A + \cot A = \frac{{11}}{2},$ then $\tan A = $