JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

\sin x+\sin ^2 x+\sin ^3 x=1

then evaluate the value of

\cos ^6 x-4 \cos ^4 x+8 \cos ^2 x

View answer and explanation

Correct option: A

Option A (4) is correct because by substituting

\sin x + \sin^2 x + \sin^3 x = 1

and using trigonometric identities, the expression

\cos^6 x - 4 \cos^4 x + 8 \cos^2 x

simplifies to 4.

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)=