Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

We know $\sin 36°=\frac{\sqrt{10-2\sqrt{5}}}{4}$

Let $x=\frac{\sqrt{10-2\sqrt{5}}}{4}$

On squaring both sides, we get

${\left(4x\right)}^2=10-2\sqrt{5}$$\Rightarrow 16x^2=10-2\sqrt{5}$

Again squaring, we get 

${\left(\sqrt{5}\right)}^2={\left(5-8x^2\right)}^2$ $\Rightarrow 5=25+64x^4-80x^2$

$\Rightarrow 16x^4-20x^2+5=0$