Question

If $x=(2\cos \theta -\cos 2\theta )and$$y=(2\sin \theta -\sin 2\theta ),$find ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{\theta =\frac{\pi }{2}}$

Answer 1

Consider the given function,

$y=\cos x-\cos 2x$

Differentiate both side with respect to x,

$\frac{dy}{dx}=-\sin x+2\sin 2x$

Differentiate again both side with respect to x,

$\frac{d^{2}y}{d{x}^2}=-\cos x+2\cos 2x$

Put,$x=\frac{\pi }{2}$ we get

${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=\frac{\pi }{2}}=-\cos \frac{\pi }{2}+2\cos 2.\frac{\pi }{2}$

$=0+2\cos \pi$

$=2(-1)$

$=-2$

Hence, this is the answer.