If $\sin \,\,x + {\sin ^2}\,\,x = 1$, then ${\cos ^8}\,\,x + 2{\cos ^6}\,\,x + {\cos ^4}\,\,x = $

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

\sin \,\,x + {\sin ^2}\,\,x = 1

, then

{\cos ^8}\,\,x + 2{\cos ^6}\,\,x + {\cos ^4}\,\,x =

View answer and explanation

Correct option: D

\mathrm{1}

Option D (1) is correct because by substituting

\sin x = \cos^2 x

into the expression

{\cos ^8}\\ x + 2{\cos ^6}\\ x + {\cos ^4}\\ x

, we can simplify it to

(\sin x + \sin^2 x)^2

, which equals 1.

Practice All Chapter MCQs →

\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)=