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If $\sin x=\frac{1}{\sqrt{5}}, \sin y=\frac{1}{\sqrt{10}}$, where $0 < x < \frac{\pi}{2}, 0 < y < \frac{\pi}{2}$,
then what is $(\mathrm{x}+\mathrm{y})$ equal to?
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then what is $(\mathrm{x}+\mathrm{y})$ equal to?
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Verified Answer
The correct answer is:
$\frac{\pi}{4}$
$\sin x=\frac{1}{\sqrt{5}}, \sin y=\frac{1}{\sqrt{10}}, 0 < x < \frac{\pi}{2}, 0 < y < \frac{\pi}{2}$
$\cos x=\sqrt{1-\sin ^{2} x} \quad \cos y=\sqrt{1-\sin ^{2} y}$
$=\sqrt{1-\frac{1}{5}} \quad=\sqrt{1-\frac{1}{10}}$
$=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}} \quad=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}} .$
$\sin (x+y)=\sin x \cos y+\cos x \sin y$
$=\frac{1}{\sqrt{5}} \cdot \frac{3}{\sqrt{10}}+\frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{10}}$
$=\frac{5}{\sqrt{5} \cdot \sqrt{10}}=\frac{\sqrt{5} \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{10}}=\sqrt{\frac{5}{10}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} .$
$\therefore \quad x+y=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4} .$
$\cos x=\sqrt{1-\sin ^{2} x} \quad \cos y=\sqrt{1-\sin ^{2} y}$
$=\sqrt{1-\frac{1}{5}} \quad=\sqrt{1-\frac{1}{10}}$
$=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}} \quad=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}} .$
$\sin (x+y)=\sin x \cos y+\cos x \sin y$
$=\frac{1}{\sqrt{5}} \cdot \frac{3}{\sqrt{10}}+\frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{10}}$
$=\frac{5}{\sqrt{5} \cdot \sqrt{10}}=\frac{\sqrt{5} \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{10}}=\sqrt{\frac{5}{10}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} .$
$\therefore \quad x+y=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4} .$
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