cos 18^{circ} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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} = $

Similar Questions

$\tan \theta \sin \left( \frac{\pi }{2} + \theta \right) \cos \left( \frac{\pi }{2} - \theta \right) = $

Find the value of the trigonometric function $\cot \left(-\frac{15 \pi}{4}\right)$.

$\sin 15^\circ + \cos 105^\circ = $

If the arcs of the same lengths in two circles subtend angles $65^{\circ}$ and $110^{\circ}$ at the centre,find the ratio of their radii.

If $\sin A = \frac{-7}{25}$,$\cos B = \frac{8}{17}$,$A$ does not lie in the $3^{\text{rd}}$ quadrant and $B$ does not lie in the $1^{\text{st}}$ quadrant,then $8 \tan A - 5 \cot B =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module