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Section 3.2 #9: Find the exact value of $\cos \left( \sin^{-1} \left( \dfrac{\sqrt{2}}{2} \right) \right)$.
Solution: Let $\theta=\sin^{-1} \left( \dfrac{\sqrt{2}}{2} \right)$ --- since the range of $\sin^{-1}$ is $\left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]$, we can deduce that $\theta$ lies in quadrant I or quadrant IV. Then $\sin(\theta) = \dfrac{\sqrt{2}}{2}$ -- since $\sin(\theta) > 0$ we can deduce that $\theta$ lies in quadrant I or quadrant II. From this we conclude that $\theta$ lies in quadrant I.

Now since, generally, $\sin(\theta) = \dfrac{y}{r} = \dfrac{\sqrt{2}}{2}$ we take $y=\sqrt{2}$ and $r=2$. Recall the formula $x^2+y^2=r^2$ (i.e. the circle with radius $r$ centered at the origin). We need the value of $x$ (to find the cosine). Plug in our values for $y$ and $r$ to get $$x^2 + (\sqrt{2})^2 = 2^2,$$ or $$x^2 + 2 = 4,$$ or $$x^2 = 2,$$ or $$x = \pm \sqrt{2}.$$ But we know that $\theta$ lies in quadrant I and so $\cos(\theta)$ will be positive -- so we take the positive solution for $x$ --- i.e. we take $x=\sqrt{2}$. Now we are prepared to calculate $$\cos \left( \sin^{-1} \left( \dfrac{\sqrt{2}}{2} \right) \right) = \cos(\theta) = \dfrac{x}{r} = \dfrac{\sqrt{2}}{2}.$$ (Note: some algebra systems like Wolfram Alpha report the answer to be $\dfrac{1}{\sqrt{2}}$. This is of course the same number because $$\dfrac{1}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \left( \dfrac{\sqrt{2}}{\sqrt{2}} \right) = \dfrac{\sqrt{2}}{2},$$ matching our solution.)

Section 3.2 #15: Find the exact value of $\csc \left(\tan^{-1}(1) \right)$.
Solution: Let $\theta=\tan^{-1}(1)$, and so since the range of $\tan^{-1}$ is $\left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]$ we may conclude that $\theta$ lies in quadrant I or quadrant IV. Then $\tan(\theta)=1$ and so we may conclude, since $\tan(\theta)$ is positive, that $\theta$ lies in quadrant I or quadrant III. Therefore $\theta$ lies in quadrant I.

Since, generally, $\tan(\theta) = \dfrac{y}{x}=\dfrac{1}{1}$ we may take $y=1$ and $x=1$. To find $\csc(\theta)$ we will have to find $\dfrac{r}{y}$. To find $r$ use the formula $x^2+y^2=r^2$ and plug in our values for $x$ and $y$ to get $$1^2 + 1^2 = r^2,$$ or $$1+1=r^2,$$ or $$2=r^2,$$ or $$\pm \sqrt{2}=r.$$ But $r$ represents the radius of a circle and so we must take the positive solution $r=\sqrt{2}$. We can now calculate $$\csc \left( \tan^{-1}(1) \right) = \csc(\theta) = \dfrac{r}{y} = \dfrac{\sqrt{2}}{1} = \sqrt{2}.$$

Section 3.2 #31: Find the exact value of $\sin \left( \tan^{-1} (-3) \right)$.
Solution: Let $\theta = \tan^{-1}(-3)$ so we may conclude that $\theta$ lies in quadrant I or quadrant IV. Now see that $\tan(\theta) = -3$ so since $\tan(\theta)$ is negative, we may conclude that $\theta$ lies in quadrant I or quadrant III. We must conclude that $\theta$ lies in quadrant I.

So $\tan(\theta)=\dfrac{y}{x} = -3 = \dfrac{-3}{1}$ means we will take $y=-3$ and $x=1$. From this we find $r$ using $x^2+y^2=r^2$: $$1^2 + (-3)^2 = r^2,$$ or $$10=r^2,$$ or $$\sqrt{10}=r,$$ because we must take the nonnegative solution ($r$ is a radius). Now we may compute $$\sin\left( \tan^{-1}(-3) \right)=\sin(\theta) = \dfrac{y}{r} = \dfrac{-3}{\sqrt{10}}.$$

Section 3.2 #40: Find the exact value of $\csc^{-1}(\sqrt{2})$.
Solution: Since $\sqrt{2} = \dfrac{2}{\sqrt{2}} = \dfrac{1}{\frac{\sqrt{2}}{2}}$ it follows from the unit circle that $$\csc^{-1}(\sqrt{2})=\dfrac{\pi}{4}.$$

Section 3.2 #59: Write $\tan(\sin^{-1}(u))$ as an algebraic expresison of $u$.
Solution: We take $\theta=\sin^{-1}(u)$ implying that $\theta$ lies in quadrant I or quadrant IV. Also, $\sin(\theta)=u$, but we cannot comment on what this means for the quadrant of $\theta$.

To find $\tan(\sin^{-1}(u))=\tan(\theta)$ we must make $\sin(\theta)$ "appear". The way this is done is to recall the Pythagorean identity $$\sin^2(\theta)+\cos^2(\theta)=1,$$ so that $$\cos^2(\theta) = 1-\sin^2(\theta),$$ implying $$\cos(\theta) =\pm \sqrt{1-\sin^2(\theta)}.$$ Since we know that $\theta$ lies in quadrant I or quadrant IV, we must conclude that $\cos(\theta)$ is positive (cosine is always positive in those quadrants!) and so using the formula $\sin(\theta)=u$ from above, we have $$\cos(\theta) = \sqrt{1-\sin^2(\theta)} = \sqrt{1-u^2}.$$ Now back to the matter at hand: compute $$\tan(\sin^{-1}(\theta)) = \tan(\theta) =\dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{u}{\sqrt{1-u^2}}.$$