Test Index

UPSC NDA I 2017 Mathematics Solved Paper

© examsnet.com

Question : 46 of 120

Marks: +1, -0

If

K=sin(\frac{\pi }{18})sin(\frac{5\pi }{18})sin(\frac{7\pi }{18})

, then what is the value of K?

$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{8}$
$\frac{1}{16}$

Solution:

Concept:

Use the trigonometric identity

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

.

Explanation:

First, convert the radian measures to degrees:

\frac{\pi}{18} = 10^\circ

,

\frac{5\pi}{18} = 50^\circ

,

\frac{7\pi}{18} = 70^\circ

.

Thus,

K = \sin 10^\circ \sin 50^\circ \sin 70^\circ

.

Notice that

50^\circ = 60^\circ - 10^\circ

and

70^\circ = 60^\circ + 10^\circ

.

Apply the identity with

\theta = 10^\circ

:

\sin 10^\circ \sin(60^\circ - 10^\circ) \sin(60^\circ + 10^\circ) = \frac{1}{4}\sin(3 \times 10^\circ)

.

This simplifies to

\frac{1}{4}\sin 30^\circ = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}

.

Answer:

K = \frac{1}{8}

.

© examsnet.com

Go to Question:

Prev Question

Next Question