Find the exact value of:

Solution (a) We seek an angle whose sine is $\dfrac{\sqrt{3}}{2} .$ Since $-1{<}\dfrac{\sqrt{3}}{2}{<}1,$ let $y=\sin ^{-1}\dfrac{\sqrt{3}}{2}$. Then by definition, $\sin y=\dfrac{\sqrt{3}}{2}$, where $-\dfrac{\pi }{2} \leq y\leq \dfrac{\pi }{2}$. Although $\sin y=\dfrac{\sqrt{3}}{2}$ has infinitely many solutions, the only number in the interval $\left[ -\dfrac{ \pi }{2},\dfrac{\pi }{2}\right]$, for which $\sin y=\dfrac{\sqrt{3}}{2}$, is $\dfrac{\pi }{3}.$ So $\sin ^{-1}\dfrac{\sqrt{3}}{2}=\dfrac{\pi }{3}$.

(b) We seek an angle whose sine is $-\dfrac{\sqrt{2}}{2}.$ Since $-1<-\dfrac{\sqrt{2}}{2}<1$, let $y=\sin ^{-1}\left( -\dfrac{\sqrt{2}}{2}\right)$. Then by definition, $\sin y=-\dfrac{\sqrt{2}}{2}$, where $-\dfrac{\pi }{2}\leq y\leq \dfrac{\pi }{2}$. The only number in the interval $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right] ,$ whose sine is $-\dfrac{ \sqrt{2}}{2},$ is $-\dfrac{\pi }{4}$. So $\sin ^{-1}\left( -\dfrac{\sqrt{2}}{ 2}\right) =-\dfrac{\pi }{4}$.

(c) Since the number $\dfrac{\pi }{8}$ is in the interval $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right]$, we use (1). $$\sin ^{-1}\left( \sin \dfrac{\pi }{8}\right) =\dfrac{\pi }{8}$$

Figure 84

(d) Since the number $\dfrac{5\pi }{8}$ is not in the closed interval $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right] ,$ we cannot use (1). Instead, we find a number $x$ in the interval $\left[ -\dfrac{\pi }{2} ,\dfrac{\pi }{2}\right]$ for which $\sin x=\sin \dfrac{5\pi }{8}.$ Using Figure 84, we see that $\sin \dfrac{5\pi }{8}=y=\sin \dfrac{3\pi }{8}.$ The number $\dfrac{3\pi }{8}$ is in the interval $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right]$, so $$\sin ^{-1}\left( \sin \dfrac{5\pi }{8}\right) =\sin ^{-1}\left( \sin \dfrac{ 3\pi }{8}\right) =\dfrac{3\pi }{8}$$