Question

Find the value of the following.
$\cos {72}^{\circ }$

Answer 1

Let $A=18°$

$\Rightarrow 5A=90°=2A+3A=90°=2A=90°-3A$

$\Rightarrow \sin 2A=\sin (90°-3A)=\cos 3A$

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

$\Rightarrow 2\sin A\cos A-4{\cos }^{2}A+3\cos A=0$

$\Rightarrow \cos A(2\sin A-4{\cos }^{2}A+3)=0$

$\Rightarrow \cos A\ne 0$ $\therefore 2\sin A-4{\cos }^{2}A+3=0$

$2\sin A-4(1-{\sin }^{2}A)+3=0$

$4{\sin }^{2}A+2\sin A-1=0$

$\therefore \sin A=\frac{-2\pm \sqrt{4-4\left(4\right)(-1)}}{2\times 4}=\frac{-2\pm \sqrt{4+16}}{8}=\frac{-2\pm \sqrt{20}}{8}$

$\sin A=\frac{-2\pm 2\sqrt{5}}{2\times 4}=\frac{-1\pm \sqrt{5}}{4}$

$\sin 18°$ is positive as $18°$ lies in $1st$ quadrant.

$\therefore \sin 18°=\sin A=\frac{\sqrt{5}-1}{4}$

Now, $\cos 72°=\cos (90°-18°)=\sin 18°=\frac{\sqrt{5}-1}{4}.$