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Section 8.3 #33: Calculate $\sin \left( \cos^{-1} \left(\dfrac{3}{5} \right) \right)$.
Solution: Let $\theta = \cos^{-1} \left( \dfrac{3}{5} \right)$, and so $\theta$ must lie in QI or QII (because the range of $\cos^{-1}$ is $[0,\pi]$). Taking the cosine of each side yields $\cos(\theta) = \dfrac{3}{5}$, and so $\theta$ must lie in QI or QIV (because the cosine of $\theta$ is a positive number). Therefore $\theta$ lies in QI -- $\sin(\theta)$ is positive.

Draw a right triangle including $\theta$:
Solve for $?$ using the Pythagorean theorem: $$3^2+?^2=5^2,$$ and hence $$? = \sqrt{25-9} = \sqrt{16}=4.$$ Now we compute $$\sin(\theta) = \dfrac{4}{5}.$$

Section 8.3 #34: Calculate $\sin \left( \tan^{-1} \left( \dfrac{4}{3} \right) \right)$.
Solution: Let $\theta=\tan^{-1}\left( \dfrac{4}{3} \right)$, and so $\theta$ must lie in QI or QIV since the range of $\tan^{-1}$ is $\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right)$. Taking tangent of both sides yields $\tan(\theta) = \dfrac{4}{3}$, and so $\theta$ must lie in QI or QIII because tangent is positive. Therefore $\theta$ must lie in QI and so $\sin(\theta)$ is positive.

Draw a right triangle including $\theta$:
Solve for the unknown side $?$ using the Pythagorean theorem: $$3^2+4^2 = ?^2,$$ and get $?=5$. Now we compute $$\sin\left( \tan^{-1} \left( \dfrac{4}{3} \right) \right)=\sin(\theta)=\dfrac{4}{5}.$$

Section 8.3 #58: The line $y=\dfrac{3}{5}x$ passes through the origin in the $xy$-plane. What is the measure of the angle that the line makes with the positive $x$-axis?
Solution: Draw this graph and place a point on the graph. For example, if $x=5$ then $y=3$ giving us the point $(5,3)$:
The right triangle with bottom leg $5$ and right leg $3$ is clear. If $\theta$ is the angle at the origin, then we see $$\tan(\theta) = \dfrac{3}{5},$$ hence $$\theta = \tan^{-1} \left( \dfrac{3}{5} \right) \approx 0.5404 \mathrm{\hspace{2pt} radians}.$$

Section 8.3 #61: A $20$-foot ladder leands up against the side of a building so that the foot of the ladder is $10$ feet from the base of the building. If specifications call for the ladder's angle of elevation to be between $35^{\circ}$ and $45^{\circ}$, does the placement of this ladder satisfy safety specifications?
Solution: Draw this:
From this picture, we see $$\cos(\theta) = \dfrac{10}{20} = \dfrac{1}{2}.$$ Therefore $$\theta = \cos^{-1} \left( \dfrac{1}{2} \right) = \dfrac{\pi}{3}=60^{\circ}.$$ Therefore, no, the ladder placement does not obey the specifications.

Section 9.1 #6: Simplify $$\sin(-x)\cos(-x)\csc(-x).$$ Solution: Using the oddness of sine, i.e. $\sin(-x)=-\sin(x)$, and the evenness of cosine, i.e. $\cos(-x)=\cos(x)$, we see that $$\begin{array}{ll} \sin(-x)\cos(-x)\csc(-x) &= \dfrac{\sin(-x)\cos(-x)}{\sin(-x)} \\ &= \dfrac{-\sin(x)\cos(x)}{-\sin(x)} \\ &= \dfrac{\cos(x)}{1} \\ &= \cos(x). \end{array}$$

Section 9.1 #9: Simplify $$\dfrac{\cot(t) + \tan(t)}{\sec(-t)}.$$ Solution: Using that cosine is even, calculate $$\begin{array}{ll} \dfrac{\cot(t) + \tan(t)}{\sec(-t)} &= \dfrac{\frac{\cos(t)}{\sin(t)} + \frac{\sin(t)}{\cos(t)}}{\frac{1}{\cos(-t)}} \\ &= \dfrac{\frac{\cos^2(t)+\sin^2(t)}{\sin(t)\cos(t)}}{\frac{1}{\cos(t)}} \\ &= \dfrac{1}{\sin(t)\cos(t)} \dfrac{\cos(t)}{1} \\ &= \dfrac{1}{\sin(t)} \\ &= \csc(t). \end{array}$$

Section 9.1 #10: Simplify $$3\sin^3(t)\csc(t) + \cos^2(t) + 2\cos(-t)\cos(t).$$ Solution: Using the evenness of cosine, i.e. $\cos(-t)=\cos(t)$, and the Pythagorean identity $\cos^2(t)+\sin^2(t)=1$, calculate $$\begin{array}{ll} 3\sin^3(t)\csc(t) + \cos^2(t) + 2\cos(-t)\cos(t) &= 3\dfrac{\sin^3(t)}{\sin(t)} + \cos^2(t) + 2\cos^2(t) \\ &= 3\sin^2(t) + \cos^2(t) + 2\cos^2(t) \\ &= 3 \sin^2(t) + 3\cos^2(t) \\ &= 3 ( \cos^2(t) + \sin^2(t)) \\ &= 3. \end{array}$$

Section 9.1 #18: Simplify the first expression in terms of the second expression:
$$\dfrac{\cos(x)}{1+\sin(x)} + \tan(x); \cos(x).$$ Solution: Calculate $$\begin{array}{ll} \dfrac{\cos(x)}{1+\sin(x)} + \tan(x) &= \dfrac{\cos(x)}{1+\sin(x)} + \dfrac{\sin(x)}{\cos(x)} \\ &=\dfrac{\cos^2(x) + \sin(x)(1+\sin(x))}{\cos(x)(1+\sin(x))} \\ &= \dfrac{\cos^2(x)+\sin^2(x)+\sin(x)}{\cos(x)(1+\sin(x))} \\ &= \dfrac{1+\sin(x)}{\cos(x)(1+\sin(x))} \\ &= \dfrac{1}{\cos(x)}. \end{array}$$

Section 9.1 #19: Simplify the first expression in terms of the second exp ression:
$$\dfrac{1}{\sin(x)\cos(x)} - \cot(x); \cot(x).$$ Solution: Recall the Pythagorean identity $\cos^2(x)+\sin^2(x)=1$ and rearrange it to get $1-\cos^2(x)=\sin^2(x)$. Now calculate $$\begin{array}{ll} \dfrac{1}{\sin(x)\cos(x)} - \cot(x) &= \dfrac{1}{\sin(x)\cos(x)} - \dfrac{\cos(x)}{\sin(x)} \\ &= \dfrac{1-\cos^2(x)}{\sin(x)\cos(x)} \\ &= \dfrac{\sin^2(x)}{\sin(x)\cos(x)} \\ &= \dfrac{\sin(x)}{\cos(x)} \\ &= \tan(x) \\ &= \dfrac{1}{\cot(x)}. \end{array}$$