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\begin{coverpages}
\begin{center}{\huge MATH 1115 - EXAM 3 FALL 2016 \\
SOLUTION}
\end{center}

\begin{flushleft}\vspace{0.5in}
Thursday 17 November 2016 \\
Instructor: Tom Cuchta
\end{flushleft}

\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}


\end{coverpages}

\begin{questions}
\question[15] Solve the following trigonometric equations as specified.
\begin{parts}
\part[5] Find the general solution of $\sin(\theta) = \dfrac{1}{2}$. \\
\textit{Solution:} From the unit circle, the solution is
$$\left\{ \begin{array}{ll}
\theta = \dfrac{\pi}{6} + 2\pi k \\
\theta = \dfrac{5\pi}{6} + 2\pi k.
\end{array} \right.$$
\part[5] Solve $4\cos^2(\theta)=1$ for $0 \leq \theta \leq 2\pi$. \\
\textit{Solution:} First isolate $\cos^2(\theta)$ to get
$$\cos^2(\theta) = \dfrac{1}{4}.$$
Now take the square root of both sides yielding
$$\cos(\theta)  =\pm \dfrac{1}{\sqrt{2}} = \pm \dfrac{\sqrt{2}}{2}.$$
From the unit circle, we see that the solution is
$$\theta = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}.$$
\part[5] Solve $\tan(2\theta)=1$ for $0 \leq \theta \leq 2\pi$. \\
\textit{Solution:} By the unit circle, the general solution of  $\tan(2\theta)=1$ is
$$\left\{ \begin{array}{ll}
2\theta = \dfrac{\pi}{4} + 2\pi k \\
2\theta = \dfrac{5\pi}{4} + 2\pi k.
\end{array} \right.$$
Solving each for $\theta$ by dividing by $2$ yields
$$\left\{ \begin{array}{ll}
\theta = \dfrac{\pi}{8} + \pi k \\
\theta = \dfrac{5\pi}{8} + \pi k.
\end{array} \right.$$
For which $k$ do we get values of $\theta$ between $0$ and $2\pi$? Some calculation shows that we can take $k=0$ and $k=1$ yielding four total solutions
$$\theta = \dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{9\pi}{8}, \dfrac{13\pi}{8}.$$
\end{parts}
\newpage
\question[15] Solve the following right triangle: \\
\begin{center}\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (0,4) -- (0,0);
\draw (0,0.4) -- (0.4,0.4) -- (0.4,0);
\draw (0.3,3.3) node {$10^{\circ}$};
\draw (2,-0.5) node {$1$};
\draw (-0.4,2) node {$a$};
\draw (2.2,2.2) node {$c$};
\draw (3.3,0.3) node {$A$};
\end{tikzpicture}\end{center}
\textit{Solution:} \underline{\textbf{Find $A$}} \\
Using the fact that sum of the angles in a triangle is $180^{\circ}$ we get
$$10^{\circ}+90^{\circ}+A=180^{\circ}.$$
Therefore
$$A = 80^{\circ}.$$
\underline{\textbf{Find $a$}} \\
There are a number of ways to do this. We will proceed by noting that
$$\tan(10^{\circ}) = \dfrac{1}{a},$$
and so
$$a = \dfrac{1}{\tan(10^{\circ})}.$$
\underline{\textbf{Find $c$}}
There are a number of ways to do this. We will proceed by noting that
$$\sin(10^{\circ}) = \dfrac{1}{c},$$
so
$$c = \dfrac{1}{\sin(10^{\circ})}.$$
\newpage
\question[15] Solve the following triangle: \\
\begin{center} \begin{tikzpicture}
\draw (-2,0) -- (2,1) -- (0,4) -- (-2,0);
\draw (-1.5,0.35) node {$A$};
\draw (1.5,1.1) node {$C$};
\draw (0.05,3.3) node {$20^{\circ}$};
\draw (1.5,2.5) node {$a$};
\draw (-1.3,2.3) node {$6$};
\draw (0,0.2) node {$4$};
\end{tikzpicture}\end{center}
\textit{Solution:} \underline{\textbf{Find $C$}} \\
Using the law of sines, 
$$\dfrac{\sin(20^{\circ})}{4} = \dfrac{\sin(C)}{6}.$$
Isolate $\sin(C)$ to get
$$\dfrac{6\sin(20^{\circ})}{4} = \sin(C).$$
Now solve for $C$ by taking $\sin^{-1}$ of both sides to get
$$C=\sin^{-1} \left( \dfrac{6\sin(20^{\circ})}{4} \right) \stackrel{\mathrm{table}}{=} 30.9^{\circ}.$$
{\color{blue} Be careful! You must also check to see if there is a second solution. The other possibility would be for
$$C = 180^{\circ}-30.9^{\circ} = 149.1^{\circ}.$$
This angle is not too big, so we must solve a second triangle associated with this value for $C$. We will do that solution in blue after we do the solution for the first value of $C$.} \\
\underline{\textbf{Find $A$}} \\
Using the fact that the sum of the angles of a triangle is $180^{\circ}$ to write
$$20^{\circ} + 30.9^{\circ} + A =180^{\circ}.$$
This yields
$$A = 129.1^{\circ}.$$
{\color{blue} Proceed similarly using the other value for $C$:
$$20^{\circ}+149.1^{\circ}+A=180^{\circ}.$$
Solving for $A$ we get
$$A=10.9^{\circ}.$$} \\
\underline{\textbf{Find $a$}} \\
Using the law of sines (you could also use the law of cosines),
$$\dfrac{\sin(129.1^{\circ})}{a} = \dfrac{\sin(20^{\circ})}{4}.$$
Therefore
$$a = \dfrac{4\sin(129.1^{\circ})}{\sin(20^{\circ})}.$$
{\color{blue} In the second triangle, use the law of sines to find $a$:
$$\dfrac{\sin(10.9^{\circ})}{a} = \dfrac{\sin(20^{\circ})}{4}$$
and solving for $a$ yields
$$a = \dfrac{4\sin(10.9^{\circ})}{\sin(20^{\circ})}.$$}
\newpage
\question[15] Solve the following triangle: \\
\begin{center}\begin{tikzpicture}
\draw (-2,0) -- (2,1) -- (0,4) -- (-2,0);
\draw (-1.5,0.35) node {$95^{\circ}$};
\draw (1.5,1.1) node {$A$};
\draw (0.05,3.3) node {$C$};
\draw (1.5,2.5) node {$4$};
\draw (-1.3,2.3) node {$a$};
\draw (0,0.2) node {$5$};
\end{tikzpicture}\end{center}
\textit{Solution:} \underline{\textbf{Find $C$}} \\
Using the law of sines,
$$\dfrac{\sin(C)}{5} = \dfrac{\sin(95^{\circ})}{4}.$$
Isolate $\sin(C)$ to get
$$\sin(C) = \dfrac{5\sin(95^{\circ})}{4}.$$
Now take $\sin^{-1}$ of both sides to get
$$C = \sin^{-1} \left( \dfrac{5\sin(95^{\circ})}{4} \right),$$
BUT from the table, we see that this quantity \textbf{does not exist}. Therefore this triangle has no solution.
\newpage
\question[15] Solve the following triangle: \\
\begin{center}\begin{tikzpicture}
\draw (-2,0) -- (2,1) -- (0,4) -- (-2,0);
\draw (-1.5,0.35) node {$A$};
\draw (1.5,1.1) node {$B$};
\draw (0.05,3.3) node {$C$};
\draw (1.5,2.5) node {$5$};
\draw (-1.3,2.3) node {$8$};
\draw (0,0.2) node {$4$};
\end{tikzpicture}\end{center} 
\underline{\textbf{Find $A$}} \\
Using the law of cosines,
$$5^2 = 8^2 + 4^2 - 2(8)(4) \cos(A).$$
Simplify both sides with arithmetic to get
$$25 = 80-64\cos(A).$$
Isolating $\cos(A)$ yields
$$\dfrac{25-80}{-64} = \cos(A).$$
Use arithmetic to simplify the numerator to get
$$\dfrac{55}{64}=\cos(A).$$
To find $A$ take $\cos^{-1}$ of both sides to get
$$A = \cos^{-1} \left( \dfrac{55}{64} \right) \stackrel{\mathrm{table}}{=} 30.8^{\circ}.$$
\underline{\textbf{Find $B$}} \\
Using the law of cosines,
$$8^2 = 4^2+5^2 - 2(4)(5) \cos(B).$$
Isolate $\cos(B)$ to get
$$-\dfrac{23}{40} = \cos(B).$$
Take $\cos^{-1}$ to get
$$B = \cos^{-1} \left( - \dfrac{23}{40} \right) \stackrel{\mathrm{table}}{=} 125.1^{\circ}$$
\underline{\textbf{Find $C$}} \\
We could use the law of cosines here. But we will find $C$ using the sum of angles equaling $180^{\circ}$:
$$30.8^{\circ}+125.1^{\circ}+C=180^{\circ}.$$
Solving for $C$ yields
$$C = 180^{\circ}-30.8^{\circ}-125.1^{\circ} = 24.1^{\circ}.$$
\newpage
\question[15] Find the area of the given triangle: \\
\begin{center}\begin{tikzpicture}
\draw (-2,0) -- (2,1) -- (0,4) -- (-2,0);
\draw (-1.5,0.35) node {$A$};
\draw (1.5,1.1) node {$B$};
\draw (0.05,3.3) node {$C$};
\draw (1.5,2.5) node {$9$};
\draw (-1.3,2.3) node {$6$};
\draw (0,0.2) node {$4$};
\end{tikzpicture}\end{center}
\textit{Solution:} We will proceed using Heron's formula. Find the semiperimeter:
$$s= \dfrac{1}{2}(6+4+9) = \dfrac{19}{2}.$$
Now using Heron's formula, the area is
$$\mathrm{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{ \dfrac{19}{2} \left( \dfrac{19}{2}-9 \right) \left( \dfrac{19}{2} - 6 \right) \left( \dfrac{19}{2} - 4 \right)}.$$
\question[10] \begin{parts}
\part[5] Convert the polar coordinates $(r,\theta)=(-3,\pi)$ into rectangular coordinates $(x,y)$. \\
\textit{Solution:} Recall polar coordinates
$$\left\{ \begin{array}{ll}
x = r\cos(\theta) \\
y = r\sin(\theta) 
\end{array} \right.$$
Plugging in the known information $r=-3$ and $\theta=\pi$ yields
$$\left\{ \begin{array}{ll}
x = -3 \cos(\pi) = 3 \\
y = -3 \sin(\pi) = 0.
\end{array} \right.$$
Therefore the given polar point is $(3,0)$ in rectangular coordinates.
\part[5] Convert the rectangular coordinates $(x,y)=(\sqrt{3},1)$ into polar coordinates $(r,\theta)$. \\
\textit{Solution:} Using the formulas for polar coordinates 
$$\left\{ \begin{array}{ll}
r^2 = x^2+y^2 \\
\theta = \tan^{-1} \left( \dfrac{y}{x} \right),
\end{array} \right.$$
we plug in the values $x=\sqrt{3}$ and $y=1$ to get
$$\left\{ \begin{array}{ll}
r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2 \\
\theta = \tan^{-1} \left( \dfrac{1}{\sqrt{3}} \right) = \dfrac{\pi}{6}.
\end{array} \right.$$
Therefore the given rectangular point $(\sqrt{3},1)$ can be expressed using the polar coordinates $\left( 2 , \dfrac{\pi}{6} \right)$. 
\end{parts}
\end{questions}
\newpage
\begin{minipage}{0.45\textwidth}
$$\begin{array}{l|l}
\sin^{-1} \left( \dfrac{4\sin(20^{\circ})}{6} \right) = 13.2^{\circ} & \sin^{-1} \left( \dfrac{6}{4\sin(20^{\circ})} \right) \bf{DNE} \\
\sin^{-1}\left( \dfrac{6\sin(20^{\circ})}{4} \right) = 30.9^{\circ} & \sin^{-1} \left( \dfrac{4}{6\sin(20^{\circ})} \right) \bf{DNE} \\
\sin^{-1} \left( \dfrac{4}{5 \sin(95^{\circ})} \right) 53.4^{\circ} & \sin^{-1} \left( \dfrac{5\sin(95^{\circ})}{4} \right) \bf{DNE} \\
\sin^{-1} \left( \dfrac{4\sin(95^{\circ})}{5} \right) = 52.8^{\circ} & \sin^{-1} \left( \dfrac{5}{4\sin(95^{\circ})} \right) \bf{DNE}  \\
\cos^{-1} \left( \dfrac{73}{80} \right) = 24.2^{\circ} & \cos^{-1} \left( \dfrac{9}{64} \right)=81.9^{\circ} \\
\sin^{-1} \left( \dfrac{4}{5\sin(24.2^{\circ})} \right) \bf{DNE} & \sin^{-1} \left( \dfrac{5\sin(81.9^{\circ})}{4} \right) \bf{DNE} \\
\sin^{-1} \left( \dfrac{4\sin(24.2^{\circ})}{5} \right) = 19.1^{\circ} & \sin^{-1} \left( \dfrac{4\sin(81.9^{\circ})}{5} \right) = 52.8^{\circ} \\
\sin^{-1} \left( \dfrac{5}{4\sin(24.2^{\circ})} \right) \bf{DNE} & \sin^{-1} \left( \dfrac{5}{4\sin(81.9^{\circ})} \right) \bf{DNE} \\
\sin^{-1} \left( \dfrac{5\sin(24.2^{\circ})}{4} \right) = 30.8^{\circ} & \sin^{-1} \left( \dfrac{4}{5\sin(81.9^{\circ})} \right) = 53.9^{\circ} \\
\sin^{-1} \left( 2\sin(24.2^{\circ}) \right) = 55.1^{\circ} & \sin^{-1} \left( 2\sin(81.9^{\circ}) \right) \bf{DNE} \\
\sin^{-1} \left( \dfrac{2}{\sin(24.2^{\circ})} \right) \bf{DNE} & \sin^{-1} \left( \dfrac{\sin(81.9^{\circ})}{2} \right) = 29.7^{\circ} \\
\sin^{-1} \left( \dfrac{\sin(24.2^{\circ})}{2} \right)=11.83^{\circ} & \sin^{-1} \left( \dfrac{2}{\sin(81.9^{\circ})} \right) \bf{DNE} \\
\sin^{-1} \left( \dfrac{1}{2\sin(24.2^{\circ})} \right) \bf{DNE}  & \sin^{-1} \left( \dfrac{1}{2\sin(81.9^{\circ})} \right) = 30.3^{\circ} \\
		\cos^{-1}\left(-\dfrac{23}{40}\right)=125.1^{\circ} & \cos^{-1}\left(\dfrac{23}{40}\right)=54.9^{\circ}\\
		\sin^{-1}\left(\dfrac{5\sin(125.1^{\circ})}{8}\right)=30.8^{\circ} & \sin^{-1}\left(\dfrac{8\sin\left(54.9^{\circ}\right)}{5}\right) \textbf{DNE}\\
		\sin^{-1}\left(\dfrac{5}{8\sin(125.1^{\circ})}\right)=49.8^{\circ} & \sin^{-1}\left(\dfrac{5}{8\sin(54.9^{\circ})}\right)=49.8^{\circ}\\
		\sin^{-1}\left(\dfrac{8}{5\sin(125.1^{\circ})}\right) \textbf{DNE} & \sin^{-1}\left(\dfrac{5\sin(54.9^{\circ})}{8}\right)=30.8^{\circ}\\
		\sin^{-1}\left(\dfrac{8\sin(125.1^{\circ})}{5}\right) \textbf{DNE} & \sin^{-1}\left(\dfrac{8}{5\sin(54.9^{\circ})}\right) \textbf{DNE}\\
		\sin^{-1}\left(\dfrac{4}{8\sin(125.1^{\circ})}\right)=37.7^{\circ} & \sin^{-1}\left(\dfrac{4\sin(54.9^{\circ})}{8}\right)=24.1^{\circ}\\
		\sin^{-1}\left(\dfrac{8\sin(125.1^{\circ})}{4}\right) \textbf{DNE} & \sin^{-1}\left(\dfrac{4}{8\sin(54.9^{\circ})}\right)=37.7^{\circ}\\
		\sin^{-1}\left(\dfrac{8}{4\sin(125.1^{\circ})}\right) \textbf{DNE} & \sin^{-1}\left(\dfrac{8\sin(54.9^{\circ})}{4}\right) \textbf{DNE}\\
		\sin^{-1}\left(\dfrac{4\sin(125.1^{\circ})}{8}\right)=24.1^{\circ} & \sin^{-1}\left(\dfrac{8}{4\sin(54.9^{\circ})}\right) \textbf{DNE}
		\end{array}$$
\end{minipage}

\begin{minipage}{0.45\textwidth}
$$\begin{array}{l|l}
\cos^{-1} \left( \dfrac{23}{400} \right) = 93.3^{\circ} & \cos^{-1} \left( \dfrac{55}{64} \right) = 30.8^{\circ} \\
\sin^{-1} \left( \dfrac{4}{5\sin(93.3^{\circ})} \right) = 53.3^{\circ} & \sin^{-1} \left( \dfrac{4}{5\sin(30.8^{\circ})} \right) \bf{DNE} \\
\sin^{-1} \left(\dfrac{4\sin(93.3^{\circ})}{5} \right) = 53.0^{\circ} & \sin^{-1} \left( \dfrac{4\sin(30.8^{\circ})}{5} \right) = 24.2^{\circ} \\
\sin^{-1} \left( \dfrac{5}{4\sin(93.3^{\circ})}\right) \bf{DNE} & \sin^{-1} \left( \dfrac{5}{4\sin(30.8^{\circ})} \right) \bf{DNE} \\
\sin^{-1} \left( \dfrac{5\sin(93.3^{\circ})}{4} \right) \bf{DNE} & \sin^{-1} \left( \dfrac{5\sin(30.8^{\circ})}{4} \right) = 39.8^{\circ} \\
\sin^{-1} \left( \dfrac{8}{5\sin(93.3^{\circ})} \right) \bf{DNE} & \sin^{-1} \left( \dfrac{8}{5\sin(30.8^{\circ})} \right) \bf{DNE} \\
\sin^{-1} \left( \dfrac{8\sin(93.3^{\circ})}{5} \right) \bf{DNE} & \sin^{-1} \left( \dfrac{8\sin(30.8^{\circ})}{5} \right) = 55.0^{\circ} \\
\sin^{-1} \left( \dfrac{5}{8\sin(93.3^{\circ})} \right) = 38.8^{\circ} & \sin^{-1} \left( \dfrac{5}{8\sin(30.8^{\circ})} \right) \bf{DNE} \\
\sin^{-1} \left( \dfrac{5\sin(93.3^{\circ})}{8} \right) = 38.6^{\circ} & \sin^{-1} \left( \dfrac{5\sin(30.8^{\circ})}{8} \right) = 18.7^{\circ} \\
\end{array}$$ 
\end{minipage}
\end{document}