JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

The value of

\cos ^2 \frac{\pi}{16}+\cos ^2 \frac{3 \pi}{16}+\cos ^2 \frac{5 \pi}{16}+\cos ^2 \frac{7 \pi}{16}

View answer and explanation

Correct option: A

Option A (2) is correct because by using the identity

\cos(\frac{\pi}{2} - x) = \sin x

, we can pair the terms and simplify using

\sin^2 x + \cos^2 x = 1

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\tan \alpha=\frac{p}{q}

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\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

\triangle A B C

C

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=