Question

Show that : $\sin {18}^o=\frac{1}{4}(\surd 5-1)$

Answer 1

$A={18}^o$

$5A={90}^o$

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

$\sin 2A=\sin ({90}^o-3A)=\cos 3A$

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

$\Rightarrow \cos A(2\sin A-4{\cos }^2\alpha A+3)=0$ $\cos A=\cos {18}^o\ne 0$

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

$\Rightarrow 4\sin A+2\sin A-1=0$

$\sin A=\frac{-2\pm 2\sqrt{5}}{8}=\frac{-1\pm \sqrt{5}}{4}$

$\sin {18}^o=\frac{-1\pm \sqrt{5}}{4}$