JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

The product

\left(\cos \frac{x}{2}\right) \cdot\left(\cos \frac{x}{4}\right) \cdot\left(\cos \frac{x}{8}\right) \cdots \cdots \cdot\left(\cos \frac{x}{256}\right)

is equal to:

View answer and explanation

Correct option: B

\frac{\sin x}{256 \sin \frac{x}{256}}

Option B (

\frac{{\sin x}}{{256 \sin \frac{{x}}{{256}}}}

) is correct because repeatedly multiplying and dividing by sine terms and using the identity

\sin 2\theta = 2 \sin \theta \cos \theta

leads to the given expression.

More Questions from TRIGONOMETRY RATIOS

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)=