# How do you show that 4cos^2x- 2 = 2sin(2x+ pi/2)?

##### 1 Answer
Dec 12, 2016

$\implies 2 \left(2 {\cos}^{2} x - 1\right) = 2 \sin x \left[2 \left(x + \frac{\pi}{4}\right)\right]$

$\implies 2 {\cos}^{2} x - 1 = \sin \left[2 \left(x + \frac{\pi}{4}\right)\right]$

Make $x + \frac{\pi}{4} = t$.

$\implies 2 {\cos}^{2} x - 1 = \sin \left(2 t\right)$

By the formula $\sin 2 x = 2 \sin x \cos x$:

$\implies 2 {\cos}^{2} x - 1 = 2 \sin t \cos t$

$\implies 2 {\cos}^{2} x - 1 = 2 \sin \left(x + \frac{\pi}{4}\right) \cos \left(x + \frac{\pi}{4}\right)$

Expand using the sum formulae $\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B$ and $\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$.

$\implies 2 {\cos}^{2} x - 1 = 2 \left(\sin x \cos \left(\frac{\pi}{4}\right) + \cos x \sin \left(\frac{\pi}{4}\right)\right) \left(\cos x \cos \left(\frac{\pi}{4}\right) - \sin x \sin \left(\frac{\pi}{4}\right)\right)$

=>2cos^2x - 1 = 2(sinx(1/sqrt(2)) + cosx(1/sqrt(2))(cosx(1/sqrt(2)) - sinx(1/sqrt(2)))

$\implies 2 {\cos}^{2} x - 1 = 2 \left(\frac{\sin x + \cos x}{\sqrt{2}}\right) \left(\frac{\cos x - \sin x}{\sqrt{2}}\right)$

$\implies 2 {\cos}^{2} x - 1 = 2 \frac{\left({\cos}^{2} x - {\sin}^{2} x\right)}{2}$

$\implies 2 {\cos}^{2} x - 1 = {\cos}^{2} x - {\sin}^{2} x$

Use ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$:

$\implies 2 {\cos}^{2} x - 1 = {\cos}^{2} x + {\cos}^{2} x - 1$

$\implies 2 {\cos}^{2} x - 1 = 2 {\cos}^{2} x - 1$

$L H S = R H S$

Identity proved!!

Hopefully this helps!