# Solve cos2A=sqrt(2)(cosA-sinA)?

Oct 21, 2017

#### Explanation:

$\cos 2 A = \sqrt{2} \left(\cos A - \sin A\right)$
$\implies \cos 2 A \left(\cos A + \sin A\right) = \sqrt{2} \left({\cos}^{2} A - {\sin}^{2} A\right)$
$\implies \cos 2 A \left(\cos A + \sin A\right) = \sqrt{2} \cdot \cos 2 A$
=>cancel(cos2A)(cosA+sinA)=sqrt2 cdot cancel(cos2A
$\implies \left(\cos A + \sin A\right) = \sqrt{2}$
$\implies {\sin}^{2} A + {\cos}^{2} A + 2 \sin A \cos A = 2$[squared both side]
$\implies 1 + \sin 2 A = 2$
$\implies \sin 2 A = 1 = \sin {90}^{\circ}$
$\implies 2 A = {90}^{\circ}$
$\implies A = {45}^{\circ}$

THANK YOU...

Nov 9, 2017

$\cos 2 A = \sqrt{2} \left(\cos A - \sin A\right)$

$\implies {\cos}^{2} A - {\sin}^{2} A - \sqrt{2} \left(\cos A - \sin A\right) = 0$

$\implies \left(\cos A - \sin A\right) \left(\cos A + \sin A\right) - \sqrt{2} \left(\cos A - \sin A\right) = 0$

$\implies \left(\cos A - \sin A\right) \left(\cos A + \sin A - \sqrt{2}\right) = 0$

When

$\cos A + \sin A = 0$

$\implies \tan A = 1 = \tan \left(\frac{\pi}{4}\right)$

$\implies A = n \pi + \frac{\pi}{4} \text{ where } n \in \mathbb{Z}$

$\cos A + \sin A = \sqrt{2}$

$\implies \frac{1}{\sqrt{2}} \cos A + \frac{1}{\sqrt{2}} \sin A = 1$

$\implies \cos \left(\frac{\pi}{4}\right) \cos A + \sin \left(\frac{\pi}{4}\right) \sin A = 1$

$\implies \cos \left(A - \frac{\pi}{4}\right) = 1$

$\implies A = 2 m \pi + \frac{\pi}{4} \text{ where } m \in \mathbb{Z}$