JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Medium

1 of 1

Application +4 | -1

The numerical value of

\sin \frac{\pi}{18} \cdot \sin \frac{5 \pi}{18} \cdot \sin \frac{7 \pi}{18}

is equal to

View answer and explanation

Correct option: B

1 / 8

Option B (

1 / 8

) is correct because the given expression can be simplified using the identity

\sin x \cdot \sin (60^\circ - x) \cdot \sin (60^\circ + x) = \frac{{1}}{{4}} \sin 3x

, leading to the solution

1/8

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)=