If sin x={1}{{5}}, sin y={1}{{10}}, where 0

Correct Answer is : π/4π/4π/4

Correct Answer is : π/4π/4π/4

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UPSC NDA I 2018 Mathematics Solved Paper

Question : 56 of 120

Marks: +1, -0

If sin

x=\frac{1}{\sqrt{5}},

y=\frac{1}{\sqrt{10}}

0 < x < \frac{\pi }{2}, 0 < y < \frac{\pi }{2},

then what is (x + y) equal to?

Solution:

Concept:

Use the sine addition formula:

\sin(x+y) = \sin x \cos y + \cos x \sin y

and

\cos \theta = \sqrt{1 - \sin^2 \theta}

for angles in the first quadrant.

Explanation:

Given

\sin x = \frac{1}{\sqrt{5}}

and

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

, with

0 < x, y < \frac{\pi}{2}

Compute

\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}

Compute

\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}

Now apply the addition formula:

\sin(x+y) = \frac{1}{\sqrt{5}} \cdot \frac{3}{\sqrt{10}} + \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{10}} = \frac{3+2}{\sqrt{50}} = \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}}

Since

\frac{1}{\sqrt{2}} = \sin\frac{\pi}{4}

, we get

\sin(x+y) = \sin\frac{\pi}{4}

Because

0 < x, y < \frac{\pi}{2}

, the sum

x+y

lies in

(0, \pi)

. The principal value of

\arcsin(\frac{1}{\sqrt{2}})

\frac{\pi}{4}

, which lies in that range.

Thus

x+y = \frac{\pi}{4}

Answer:

\frac{\pi}{4}

(Option C).

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