cos 18^{circ} = | vedclass

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(B) We know that,$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we have:$\cos 18^{\circ} = \sqrt{1 -

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{2}} \sqrt{5+\sqrt{5}}$

Vedclass · Answer

(B) We know that,$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we have:$\cos 18^{\circ} = \sqrt{1 - \sin^2 18^{\circ}}$$= \sqrt{1 - \left(\frac{\sqrt{5}-1}{4}ight)^2}$$= \sqrt{1 - \frac{5+1-2\sqrt{5}}{16}}$$= \sqrt{\frac{16 - 6 + 2\sqrt{5}}{16}}$$= \sqrt{\frac{10 + 2\sqrt{5}}{16}}$$= \frac{\sqrt{10 + 2\sqrt{5}}}{4}$$= \frac{\sqrt{2(5 + \sqrt{5})}}{4} = \frac{\sqrt{2} \cdot \sqrt{5+\sqrt{5}}}{2 \cdot 2} = \frac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}}$.

$\cos 18^{\circ} = $

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