.

Back to the class
Section 3.1 #16: Find the exact value of $\cos^{-1}(-1)$.
Solution: Recall the domain of $\cos^{-1}$ is $[-1,1]$ and the range of $\cos^{-1}$ is $[0,\pi]$. Therefore we can say, from the unit circle, that $$\cos^{-1}(-1)=\pi.$$
Section 3.1 #40: Find the exact value of $\sin^{-1} \left[ \sin \left( - \dfrac{3\pi}{7} \right) \right]$.
Solution: The angle $-\dfrac{3\pi}{7}$ is in quadrant 4. The range of $\sin^{-1}$ is $\left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]$, which corresponds to quadrants $1$ and $4$. This means we may just use the cancellation property of the inverse to calculate $$\sin^{-1} \left[ \sin \left( -\dfrac{3\pi}{7} \right) \right] = -\dfrac{3\pi}{7}.$$
Section 3.1 #42: Find the exact value of $\cos^{-1} \left[ \cos \left( - \dfrac{5\pi}{3} \right) \right]$.
Solution: The angle $-\dfrac{5\pi}{3}$ is in quadrant $1$. However, the range of $\cos^{-1}$ is $[0,\pi]$ (corresponding to quadrants $1$ and $2$), and so $-\dfrac{5\pi}{3}$ is out of that range. To fix this, we will find an equivalent angle to $-\dfrac{5\pi}{3}$ in the appropriate bounds. To do this we will add $2\pi$ to the angle (i.e. a "full revolution" around the unit circle) to get $$-\dfrac{5\pi}{3} + 2\pi = -\dfrac{5\pi}{3} + \dfrac{6\pi}{3} = \dfrac{\pi}{3},$$ which is in the interval $[0,\pi]$. Now we calculate $$\cos^{-1} \left[ \cos \left( -\dfrac{5\pi}{3} \right) \right] = \cos^{-1} \left[ \cos \left( \dfrac{\pi}{3} \right) \right] = \dfrac{\pi}{3}.$$

Section 3.1 #56: Let $f(x)=3\sin(2x)$. Find the inverse function $f^{-1}$, its domain, and its range.
Solution: Write $y=3\sin(2x)$ and swap $x$ and $y$ to get $x=3\sin(2y)$. Now we solve for y: to start divide by $3$ to get $$\dfrac{x}{3} = \sin(2y).$$ Now take $\sin^{-1}$ of both sides to get $$\sin^{-1} \left( \dfrac{x}{3} \right) = \sin^{-1}(\sin(2y))=2y.$$ Now divide by $2$ to get $$\dfrac{1}{2} \sin^{-1} \left( \dfrac{x}{3} \right) = y.$$ This means we have found the inverse function $$f^{-1}(x) = \dfrac{1}{2} \sin^{-1} \left( \dfrac{x}{3} \right).$$ To determine its domain, we recall that the domain of $\sin^{-1}$ is $[-1,1]$, i.e. if we want to compute $\sin^{-1}(\mathrm{something})$ then $-1 \leq \mathrm{something} \leq 1$. For us, we want to compute $\sin^{-1} \left( \dfrac{x}{3} \right)$ so we consider the inequality $$-1 \leq \dfrac{x}{3} \leq 1,$$ and multiplying it by $3$ $$-3 \leq x \leq 3,$$ showing the domain of $f^{-1}$ is $[-3,3]$. To find the range, recall that the rnage of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]$ and so we know that $-\dfrac{\pi}{2} \leq \sin^{-1}(\mathrm{anything}) \leq \dfrac{\pi}{2}$. We will want to make our function, $f^{-1}$ appear in the middle. So we use the range of $\sin^{-1}$ to write $$-\dfrac{\pi}{2} \leq \sin \left( \dfrac{x}{3} \right) \leq \dfrac{\pi}{2},$$ but we want our inverse function we found, $f^{-1}(x)=\dfrac{1}{2}\sin^{-1} \left( \dfrac{x}{2} \right)$, to appear in the middle. So, divide the inequality by $2$ and we get $$-\dfrac{\pi}{4} \leq \dfrac{1}{2} \sin^{-1} \left( \dfrac{x}{3} \right) \leq \dfrac{\pi}{4},$$ which shows us that the range of $f^{-1}$ is $\left[ -\dfrac{\pi}{4}, \dfrac{\pi}{4} \right]$.

Section 3.1 #64: Find the exact solution of the equation $-6\sin^{-1}(3x)=\pi$.
Solution: We want to solve for $x$ in the equation. First divide by $-6$ to get $$\sin^{-1} (3x) = -\dfrac{\pi}{6}.$$ Now take $\sin$ of both sides, recalling that $\sin(\sin^{-1}(x))=x$: $$3x = \sin \left( - \dfrac{\pi}{6} \right).$$ We can evaluate the right hand side with the unit circle to get $$3x = -\dfrac{1}{2}.$$ Finally solve for $x$ by dividing by $3$ to get $$x = -\dfrac{1}{6}.$$