How do you evaluate $cot [{sin}^{- 1} (- \frac{\sqrt{2}}{3})]$ $cot [sin^-1 (- sqrt2 / 3)]$ ?

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How do you evaluate $cot [{sin}^{- 1} (- \frac{\sqrt{2}}{3})]$ $cot [sin^-1 (- sqrt2 / 3)]$ ?

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Answer 1 · 2016-05-20T02:51:53

$\pm \sqrt{\frac{7}{2}}$ $+-sqrt (7 / 2)$

Explanation:

Let $a = {sin}^{- 1} - \frac{\sqrt{2}}{3}$ $a = sin^(-1)-sqrt 2 / 3$ . Then, $sin a = - \frac{\sqrt{2}}{3} < 0$ $sin a=-sqrt 2 / 3<0$ .

Accordingly, a is in the 3rd quadrant or in the 4th.

So, $cos a = \pm \sqrt{1 - {sin}^{2} a} = \pm \frac{\sqrt{7}}{3}$ $cos a = +-sqrt(1-sin^2a)=+-sqrt 7 / 3$ .
The answer is $cot a = \frac{cos a}{sin a} = \pm \sqrt{\frac{7}{2}}$ $cot a = cos a/sin a = +- sqrt (7/2)$ .

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How do you evaluate $cot [{sin}^{- 1} (- \frac{\sqrt{2}}{3})]$ $cot [sin^-1 (- sqrt2 / 3)]$ ?

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