If tanα =1/7,sinβ ={1}{{10}}; 0

Correct Answer is : 12{1}{{2}}21

Correct Answer is : 12{1}{{2}}21

Test Index

UPSC NDA II 2023 Mathematics Solved Paper

Question : 30 of 120

Marks: +1, -0

tan\alpha =\frac{1}{7},sin\beta =\frac{1}{\sqrt{10}}

0 < \alpha ,\beta < \frac{\pi }{2}

, then what is the value of cos (α + 2β) ?

Solution:

Concept:

Use compound angle formula for

\cos(\alpha+2\beta)

and known values of trigonometric ratios.

Explanation:

Given:

\tan\alpha = \frac{1}{7}

and

\sin\beta = \frac{1}{\sqrt{10}}

with

0 < \alpha,\beta < \frac{\pi}{2}

From a right triangle,

\tan\alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{7}

. So

\sin\alpha = \frac{1}{\sqrt{1^2+7^2}} = \frac{1}{\sqrt{50}}

and

\cos\alpha = \frac{7}{\sqrt{50}}

For

\beta

\sin\beta = \frac{1}{\sqrt{10}}

gives

\cos\beta = \sqrt{1-\frac{1}{10}} = \frac{3}{\sqrt{10}}

Use formula:

\cos(\alpha+2\beta) = \cos\alpha \cos 2\beta - \sin\alpha \sin 2\beta

Recall

\cos 2\beta = 2\cos^2\beta - 1

and

\sin 2\beta = 2\sin\beta \cos\beta

Substitute values:

\cos(\alpha+2\beta) = \frac{7}{\sqrt{50}}\left[2\left(\frac{3}{\sqrt{10}}\right)^2-1\right] - \frac{1}{\sqrt{50}}\left(2\cdot\frac{1}{\sqrt{10}}\cdot\frac{3}{\sqrt{10}}\right)

Simplify inside brackets:

2\cdot\frac{9}{10} - 1 = \frac{9}{5} - 1 = \frac{4}{5}

; the second term:

2\cdot\frac{3}{10} = \frac{3}{5}

Now:

\cos(\alpha+2\beta) = \frac{7}{\sqrt{50}}\cdot\frac{4}{5} - \frac{1}{\sqrt{50}}\cdot\frac{3}{5} = \frac{28}{5\sqrt{50}} - \frac{3}{5\sqrt{50}} = \frac{25}{5\sqrt{50}} = \frac{5}{\sqrt{50}}

Simplify:

\frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}

Thus final value is

\frac{1}{\sqrt{2}}

Answer:

\frac{1}{\sqrt{2}}

(Option C).

Go to Question:

Prev Question

Next Question