JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

\sin ^2 5^{\circ}+\sin ^2 10^{\circ}+\sin ^2 15^{\circ}+\ldots .+\sin ^2 90^{\circ}

is equal to

View answer and explanation

Correct option: C

9 \frac{1}{2}

Option C (

9 \frac{1}{2}

) is correct because pairing

\sin^2 x

with

\sin^2 (90-x) = \cos^2 x

allows us to use the identity

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

, and accounting for the remaining terms.

More Questions from TRIGONOMETRY RATIOS

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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)=