Inverse Trigonometric Functions

Sol A a) $\sin(\sin^{-1}4/9)=4/9$ (since $-1\le 4/9\le 1$).

b)$\sin^{-1}5/3$ is undefined, since 5/3 is not in $[-1,1]$.

Sol B a) $\tan(\tan^{-1}4)=4$ (since $\tan(\tan^{-1}x)=x$ for any $x$).

b) $3\pi/5-\pi=-2\pi/5$, so $\tan(-2\pi/5)=\tan3\pi/5$ since the tangent has period $\pi$, and $-2\pi/5$ is in $(-\pi/2,\pi/2)$; so $\tan^{-1}(\tan3\pi/5)=-2\pi/5$.

Sol 1 a) $\tan^{-1}1=\pi/4$, since $\tan(\pi/4)=1$ and $-\pi/2<\pi/4<\pi/2$.

b) $\tan^{-1}\sqrt{3}=\pi/3$, since $\tan(\pi/3)=\sqrt{3}$ and $-\pi/2<\pi/3<\pi/2$.

Sol 2 a) $\sin^{-1}(\sqrt{3}/2)=\pi/3$, since $\sin(\pi/3)=\sqrt{3}/2$ and $-\pi/2\le \pi/3\le \pi/2$.

b) $\sin^{-1}(\sin(3\pi/2))=\sin^{-1}(-1)=-\pi/2$, since $\sin(-\pi/2)=-1$ and $-\pi/2\le -\pi/2\le \pi/2$.

Sol 3 a) $\cos^{-1}0=\pi/2$ since $\cos(\pi/2)=0$ and $0\le \pi/2\le \pi$.

b) $\cos^{-1}(-1/2)=\pi-\pi/3=2\pi/3$ since $\cos(2\pi/3)=-1/2$ and $0\le 2\pi/3 \le \pi$.

Sol 4 Let $\theta=\tan^{-1}3$, so $\tan\theta=3$ From the right triangle shown below, $\sin\theta=3/\sqrt{10}$; so $\sin(\tan^{-1}3)=3/\sqrt{10}$.

Sol 5 Let $\theta=\sin^{-1}(5/13)$, so $\sin\theta=5/13$. From the right triangle shown below, $\tan\theta=5/12$; so $\tan(\sin^{-1}(5/13))=5/12$.

Sol 6 Let $\theta=\arcsin(3/4)$, so $\sin\theta=3/4$. Then $\cos(2\arcsin(3/4))=\cos2\theta=1-2\sin^2\theta=1-2(3/4)^2=1-2(9/16)$ $=1-9/8=-1/8$.

Sol 7 Let $\theta=\arctan(3/4)$, so $\tan\theta=3/4$. Then $\sin\theta=3/5$ and $\cos\theta=4/5$ using the right triangle below, so $\sin(2\arctan(3/4))=\sin 2\theta=2\sin\theta\cos\theta=2(3/5)(4/5)=24/25$.

Sol 8 Since $18\pi/5-2(2\pi)=18\pi/5-20\pi/5=-2\pi/5$, $\sin(18\pi/5)=\sin(-2\pi/5)$ where $-\pi/2\le -2\pi/5\le \pi/2$; so $\sin^{-1}(\sin(18\pi/5))=-2\pi/5$.

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