How do you evaluate $sin (2 arctan (\sqrt{2}))$ $Sin(2 arctan(sqrt 2))$ ?

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How do you evaluate $sin (2 arctan (\sqrt{2}))$ $Sin(2 arctan(sqrt 2))$ ?

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Answer 1 · 2016-05-07T12:26:22

$\frac{\sqrt{2}}{3}$ $sqrt 2/3$ .

Explanation:

Let $a = a r c tan \sqrt{2}$ $a = arc tan sqrt 2$ . Then $tan a = \sqrt{2}$ $tan a = sqrt 2$ . Tangent is positive.

Both sine and cosine have the same sign. So,

either $sin a = \frac{\sqrt{2}}{\sqrt{3}} and cos a = \frac{1}{\sqrt{3}}$ $sin a = sqrt2/ sqrt 3 and cos a = 1/sqrt 3$

or $sin a = - \frac{\sqrt{2}}{\sqrt{3}} and cos a = - \frac{1}{\sqrt{3}}$ $sin a = - sqrt 2/ sqrt 3 and cos a = -1/sqrt 3$ .

Now, $sin (2 a r c tan \sqrt{2}) = sin 2 a = 2 sin a cos a = \frac{\sqrt{2}}{3}$ $sin (2 arc tan sqrt 2)=sin 2a = 2 sin a cos a = sqrt2 /3$ , in both the cases.

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How do you evaluate $sin (2 arctan (\sqrt{2}))$ $Sin(2 arctan(sqrt 2))$ ?

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