Question

Prove that
i) $\sin {18}^o=\frac{\sqrt{5}-1}{4}$
ii) $\cos {36}^o=\frac{\sqrt{5}+1}{4}$

Answer 1

(i)

Let $\theta =18$. Then

$5\theta =90$

$\therefore 2\theta +3\theta =90$

$\Rightarrow 2\theta =90-3\theta$

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

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

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

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

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

$\Rightarrow \sin \theta =\frac{-2\pm \sqrt{2^2-4\times 4\times (-1)}}{2\times 4}$

$\Rightarrow \sin \theta =\frac{-2\pm \sqrt{20}}{8}$

$\Rightarrow \sin \theta =\frac{\sqrt{5}-1}{4}$

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

(ii)

$\cos 36=1-2{\sin }^{2}18$

$\therefore \cos 36=1-2{\left(\frac{\sqrt{5}-1}{4}\right)}^2=1-\left(\frac{5+1-2\sqrt{5}}{8}\right)=\frac{\sqrt{5}+1}{4}$

$\cos 36=\frac{\sqrt{5}+1}{4}$