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

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Answer 1 · 2016-04-11T06:04:45

${sin}^{- 1} (\frac{\sqrt{2}}{2}) = \frac{π}{4}$ $sin^(-1)((sqrt2)/2)=pi/4$

${sin}^{- 1} a$ $sin^(-1)a$ means an angle whose sine is $a$ $a$ .

Now $\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{1}{\sqrt{2}}$ $sqrt2/2=sqrt2/(sqrt2xxsqrt2)=1/sqrt2$

and we know $sin (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $sin(pi/4)=1/sqrt2$

Hence

${sin}^{- 1} (\frac{\sqrt{2}}{2}) = {sin}^{- 1} (\frac{1}{\sqrt{2}}) = \frac{π}{4}$ $sin^(-1)((sqrt2)/2)=sin^(-1)(1/(sqrt2))=pi/4$

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