cos ^2 76^{circ}+sin ^2 46^{circ}+sin 76^{cir | vedclass

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

Question

(C) Let $E = \cos ^2 76^{\circ}+\sin ^2 46^{\circ}+\sin 76^{\circ} \cos 46^{\circ}$.Using the identity $2\cos^2 	heta = 1 + \cos 2	heta$ and $2\sin^

Vedclass · Accepted Answer

$\frac{5}{4}$

Vedclass · Answer

(C) Let $E = \cos ^2 76^{\circ}+\sin ^2 46^{\circ}+\sin 76^{\circ} \cos 46^{\circ}$.Using the identity $2\cos^2 	heta = 1 + \cos 2	heta$ and $2\sin^2 	heta = 1 - \cos 2	heta$,we have:$E = \frac{1 + \cos 152^{\circ}}{2} + \frac{1 - \cos 92^{\circ}}{2} + \frac{1}{2} (2 \sin 76^{\circ} \cos 46^{\circ})$$E = 1 + \frac{1}{2} (\cos 152^{\circ} - \cos 92^{\circ}) + \frac{1}{2} (\sin(76^{\circ} + 46^{\circ}) + \sin(76^{\circ} - 46^{\circ}))$$E = 1 + \frac{1}{2} (-2 \sin 122^{\circ} \sin 30^{\circ}) + \frac{1}{2} (\sin 122^{\circ} + \sin 30^{\circ})$Since $\sin 30^{\circ} = \frac{1}{2}$,we get:$E = 1 - \frac{1}{2} \sin 122^{\circ} + \frac{1}{2} \sin 122^{\circ} + \frac{1}{4}$$E = 1 + \frac{1}{4} = \frac{5}{4}$.

$\cos ^2 76^{\circ}+\sin ^2 46^{\circ}+\sin 76^{\circ} \cos 46^{\circ} = $

Similar Questions

$\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}$ is equal to

The value of $\tan 9^{\circ} - \tan 27^{\circ} - \tan 63^{\circ} + \tan 81^{\circ}$ is $............$.

$\frac{\cos 12^\circ - \sin 12^\circ}{\cos 12^\circ + \sin 12^\circ} + \frac{\sin 147^\circ}{\cos 147^\circ} = $

If $\cos A+\cos (A+B)+\cos (A+2 B)+\ldots$ up to $n$ terms $=$ $\cos \left(\frac{2 A+(n-1) B}{2}\right) \sin \frac{n B}{2} \operatorname{cosec} \frac{B}{2}$,then $\cos \frac{\pi}{19}+\cos \frac{3 \pi}{19}+\cos \frac{5 \pi}{19}+\ldots+\cos \frac{17 \pi}{19} = $

If $x \cos \theta + y \sin \theta = 5$ and $x \sin \theta - y \cos \theta = 3$,then the value of $x^{2} + y^{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module