Question

Prove that $\cos A=\frac{\cos 3A}{2\cos 2A-1}$ and hence deduce that $\cos {15}^o=\frac{\sqrt{3}+1}{2\sqrt{2}}$

Answer 1

$\cos A=\frac{\cos 3A}{2\cos 2A-1}$

$\Rightarrow R.H.S=\frac{\cos 3A}{2\cos 2A-1}=\frac{4{\cos }^{3}A-3\cos A}{2\left(2{\cos }^{2}A-1\right)-1}$

$=\frac{4{\cos }^{3}A-3\cos A}{4{\cos }^{2}A-2-1}$

$=\frac{4{\cos }^{3}A-3\cos A}{4{\cos }^{2}A-3}$

$=\frac{\cos A(4{\cos }^{2}A-3)}{4{\cos }^{2}A-3}$

$=\cos A=L.H.S$

$\Rightarrow \cos 15°=\cos (60°-45°)$

$=\cos 60°\cos 45°+\sin 60°\sin 45°$

$=\frac{1}{2}\times \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}$

$=\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}$

$=\frac{\sqrt{3}+1}{2\sqrt{2}}$

Hence, the answer is proved.