Question

Prove that $\sin 40°-\cos 70°=\sqrt{3}\cos 80°$

Answer 1

As we have $\sin 40°-\cos 70°=\sqrt{3}\cos 80°$

Use $\sin A–\sin B=2\cos \frac{(A+B)}{2}\sin \frac{(A-B)}{2}$

Also $\cos (90–x)=\sin x$

So $\cos 70=\cos (90–20)=\sin 20$

$\sin 40°–\cos 70°=\sin 40°–\sin 20°$

$$\begin{gathered} =2\cos \left(\frac{40+20}{2}\right)\sin \left(\frac{40-20}{2}\right) \\ =2\cos \left(\frac{60}{2}\right)\sin \left(\frac{20}{2}\right) \\ =2\cos 30\sin 10° \\ =2(\surd 3/2)\sin 10° \\ =\surd 3\cos 80°(\because \sin 10=\cos 80) \end{gathered}$$

Hence proved.