JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

The minimum value of

\sin \theta \cos \theta

View answer and explanation

Correct option: C

-\frac{1}{2}

Option C (-

\frac{{1}}{{2}}

) is correct because the minimum value of

\sin \theta \cos \theta

is -

\frac{{1}}{{2}}

, which can be found using trigonometric identities or calculus.

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)=