Question

Find $\left(\cos \right(\frac{\pi }{4}+x)+\cos (\frac{\pi }{4}-x)-\sqrt{2}\cos x)$$\left(\cos \right(\frac{3\pi }{4}+x)-\cos (\frac{3\pi }{4}-x)+\sqrt{2}\sin x)$

Answer 1

$(\cos (\frac{\pi }{4}+x)+\cos (\frac{\pi }{4}-x)-\sqrt{2}cosx)(\cos (\frac{3\pi }{4}+x)-\cos (\frac{3\pi }{4}-x)+\sqrt{2}sinx)$

$\because \cos (A+B)+\cos (A-B)=2cosAcosB$

$cos(A+B)-cos(A-B)=-2sinAsinB$

$\Rightarrow (2\cos \left(\frac{\pi }{4}\right)\cos \left(x\right)-\sqrt{2}cosx)(2\cos \left(\frac{3\pi }{4}\right)\cos \left(x\right)+\sqrt{2}sinx)$

$\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\&\cos \left(\frac{3\pi }{4}\right)=\frac{-1}{\sqrt{2}}$

$\Rightarrow (2\times \frac{1}{\sqrt{2}}cosx-\sqrt{2}cosx)(-2\times \frac{1}{\sqrt{2}}cosx+\sqrt{2}sinx)$

$\Rightarrow (\sqrt{2}cosx-\sqrt{2}cosx)(-\sqrt{2}cosx+\sqrt{2}sinx)$

$=0$