How do you evaluate ${sin}^{- 1} (sin (\frac{13 π}{10}))$ $sin^-1(sin((13pi)/10))$ ?

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Answer 1 · 2016-07-30T23:20:51

$- \frac{3 π}{10}$ $-(3pi)/10$

The inverse sine function has domain $[- 1, 1]$ $[-1,1]$ which means it will have range $- \frac{π}{2} \leq y \leq \frac{π}{2}$ $-pi/2<=y<=pi/2$

This means that any solutions we obtain must lie in this interval.

As a consequence of double angle formulae, $sin (x) = sin (π - x)$ $sin(x) = sin(pi-x)$ so

$sin (\frac{13 π}{10}) = sin (- \frac{3 π}{10})$ $sin((13pi)/(10)) = sin(-(3pi)/10)$

Sine is $2 π$ $2pi$ periodic so we can say that

$sin^(-1)(sin(x)) = x + 2npi, n in ZZ$

However any solutions must lie in the interval $-pi/2 <= y <= pi/2$ .

There is no integer multiple of $2pi$ we can add to $(13pi)/10$ to get it within this interval so the only solution is $-(3pi)/10$ .

How do you evaluate sin^-1(sin((13pi)/10))sin−1(sin(13π10))sin^-1(sin((13pi)/10))?