The value of $\mathop \sum \limits_{k = 1}^3 \,\,{\cos ^2}\left( {2k - 1} \right)\frac{\pi }{{12}}$, is

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

The value of

\sum \limits_{k = 1}^3 \,\,{\cos ^2}\left( {2k - 1} \right)\frac{\pi }{{12}}

, is

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Correct option: D

3/2

Option D (3/2) is correct because by substituting the values of k and using trigonometric identities, the summation simplifies to

\frac{3}{2}

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

\alpha=6 \beta

\alpha

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

\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)=