JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Hard

1 of 1

Application +4 | -1

The minimum value of

{27^{\cos 2x}}{.81^{\sin 2x}}

, is

View answer and explanation

Correct option: A

\frac{1}{{243}}

Option A (

\frac{1}{{243}}

) is correct because the minimum value is found by minimizing the exponent of 3 using trigonometric identities.

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

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\alpha

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

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\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)=