Test Index

UPSC NDA 2 2016 Mathematics Solved Paper

© examsnet.com

Question : 46 of 120

Marks: +1, -0

What is the value of

\cos (2 \cos^{-1} 0.8)

?

0.81
0.56
0.78
0.28

Solution:

Concept:

Use the identity

1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right)

and the property

\cos(\cos^{-1} x) = x

.

Explanation:

Let

\theta = 2 \cos^{-1} 0.8

.

Then

\cos^{-1} 0.8 = \frac{\theta}{2}

, so

\cos\left(\frac{\theta}{2}\right) = 0.8

.

Apply the identity:

1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right) = 2 \times (0.8)^2 = 1.28

.

Thus

\cos \theta = 1.28 - 1 = 0.28

.

Therefore

\theta = \cos^{-1}(0.28)

.

Now

\cos(2 \cos^{-1} 0.8) = \cos(\theta) = \cos( \cos^{-1}(0.28) ) = 0.28

.

Answer:

0.28

(option D).

© examsnet.com

Go to Question:

Prev Question

Next Question