# Prove that

Question:

Prove that

(i) $\cos \left(\frac{\pi}{3}+x\right)=\frac{1}{2}(\cos x-\sqrt{3} \sin x)$

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

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

(iv) $\cos \mathrm{x}+\cos \left(\frac{2 \pi}{3}+\mathrm{x}\right)+\cos \left(\frac{2 \pi}{3}-\mathrm{x}\right)=0$

Solution:

(i) $\cos \left(\left(\frac{\pi}{3}+x\right)=\cos \frac{\pi}{3} \cdot \cos x-\sin \frac{\pi}{3} \cdot \sin x\right.$

$\Rightarrow \frac{1}{2} \cdot \cos x-\frac{\sqrt{3}}{2} \cdot \sin x=\frac{1}{2}(\cos x-\sqrt{3} \sin x)$

(ii) $\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{\pi}{4}-x\right)$

$=\sin \frac{\pi}{4} \cdot \cos x+\cos \frac{\pi}{4} \cdot \sin x+\sin \frac{\pi}{4} \cdot \cos x-\cos \frac{\pi}{4} \cdot \sin x$

$=2 \cdot \sin \frac{\pi}{4} \cdot \cos x \Rightarrow 2 \cdot \frac{1}{\sqrt{2}} \cdot \cos x=\sqrt{2} \cdot \cos x$

(iii) $\frac{1}{\sqrt{2}} \cdot \cos \left(\frac{\pi}{4}+x\right)=\frac{1}{\sqrt{2}} \cdot\left(\cos \frac{\pi}{4} \cdot \cos x-\sin \frac{\pi}{4} \cdot \sin x\right)$

$\Rightarrow \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}} \cdot \cos x-\frac{1}{\sqrt{2}} \cdot \sin x\right)=\frac{1}{2}(\cos x-\sin x)$

(iv) $\cos x+\cos \left(\frac{2 \pi}{3}+x\right)+\cos \left(\frac{2 \pi}{3}-x\right)$

$=\cos x+\cos \frac{2 \pi}{3} \cdot \cos x-\sin \frac{2 \pi}{3} \cdot \sin x+\cos \frac{2 \pi}{3} \cdot \cos x+\sin \frac{2 \pi}{3} \cdot \sin x$

$=\cos x+2 \cdot \cos \left(\pi-\frac{\pi}{3}\right) \cdot \cos x$

$=\cos x+2 \cdot\left(-\frac{1}{2}\right) \cdot \cos x$

$=\cos x-\cos x \Rightarrow 0$