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

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

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Answer 1 · 2016-02-13T12:46:40

Explanation is given below.

Explanation:

A good idea is to know the trigonometric ratios of the standard angles.

A glance would help you note that $sin (45^{o}) = \frac{\sqrt{2}}{2}$ $sin(45^o)=sqrt(2)/2$

${sin}^{- 1} (\frac{\sqrt{2}}{2}) = 45^{o}$ $sin^-1(sqrt(2)/2) =45^o$

You need to take care that the range of ${sin}^{- 1} (x)$ $sin^-1(x)$ is $[- 90^{o}, 90^{o}]$ $[-90^o,90^o]$ That is why we can safely say
${sin}^{- 1} (\frac{\sqrt{2}}{2})$ $sin^-1(sqrt(2)/2)$ is $45^{o}$ $45^o$

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

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Explanation:

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