Question

Prove that:
(i) $\cos \left(\frac{3\mathrm{\pi }}{4}+x\right)-\cos \left(\frac{3\mathrm{\pi }}{4}-x\right)=-\sqrt{2}\sin x$

(ii) $\cos \left(\frac{\mathrm{\pi }}{4}+x\right)+\cos \left(\frac{\mathrm{\pi }}{4}-x\right)=\sqrt{2}\cos x$

Answer 1

(i)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \cos \left(\frac{3\mathrm{\pi }}{4}+x\right)-\cos \left(\frac{3\mathrm{\pi }}{4}-x\right) \\ =-2\sin \left\{\frac{\left({\displaystyle \frac{3\mathrm{\pi }}{4}}+x\right)+\left({\displaystyle \frac{3\mathrm{\pi }}{4}}-x\right)}{2}\right\}\sin \left\{\frac{\left({\displaystyle \frac{3\mathrm{\pi }}{4}}+x\right)-\left({\displaystyle \frac{3\mathrm{\pi }}{4}}-x\right)}{2}\right\}\left\{\because \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)\right\} \\ =-2\sin \frac{3\mathrm{\pi }}{4}\sin x \\ =-2\sin \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)\sin x \\ =-2\sin \frac{\mathrm{\pi }}{4}\sin x \\ =-\sqrt{2}\sin x \end{gathered}$$


(ii)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \cos \left(\frac{\mathrm{\pi }}{4}+x\right)+\cos \left(\frac{\mathrm{\pi }}{4}-x\right) \\ =2\cos \left\{\frac{\left(\frac{\mathrm{\pi }}{4}+x\right)+\left(\frac{\mathrm{\pi }}{4}-x\right)}{2}\right\}\cos \left\{\frac{\left(\frac{\mathrm{\pi }}{4}+x\right)-\left(\frac{\mathrm{\pi }}{4}+x\right)}{2}\right\}\left\{\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right\} \end{gathered}$$

$$\begin{gathered} =2\cos \left\{\frac{\frac{\mathrm{\pi }}{4}+x+\frac{\mathrm{\pi }}{4}-x}{2}\right\}\cos \left\{\frac{\frac{\mathrm{\pi }}{4}+x-\frac{\mathrm{\pi }}{4}+x}{2}\right\} \\ =2\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)\cos x \\ =2\times \frac{1}{\sqrt{2}}\times \cos x \\ =\sqrt{2}\cos x \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$