\documentclass[10pt,addpoints]{exam}
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%\header{MATH 11}{Exam 1}{Fall 2016}
\begin{document}
\begin{coverpages}
\begin{center}\huge MATH 1540 - EXAM 2 - SPRING 2018\\
SOLUTION
\end{center}
\begin{flushleft}\vspace{0.5in}
8 March 2018 \\
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[25] Graph the function. Include all relevant labels on the $x$-axis and $y$-axis for full credit.
\begin{parts}
\part[5] Graph $y=\tan(x)$. \\
\textit{Solution}: There are asymptotes at $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$. Therefore \\
\textbf{Anchor points}: $-\dfrac{\pi}{2}, -\dfrac{\pi}{4}, 0, \dfrac{\pi}{4}, \dfrac{\pi}{2}$ \\
\begin{center}
\begin{tikzpicture}[scale=2]
\draw (0,-3) -- (0,3);
\draw (-3,0) -- (3,0);
\draw[dashed,thick] (1.5707,-3) -- (1.5707,3);
\draw[dashed,thick] (-1.5707,-3) -- (-1.5707,3);
\draw[white,fill=white] (1.5707,-0.3) circle (0.27);
\draw (1.5707,-0.3) node {$\dfrac{\pi}{2}$};
\draw[white,fill=white] (-1.5707,-0.3) circle (0.27);
\draw (-1.707,-0.3) node {$-\dfrac{\pi}{2}$};

\draw[white,fill=white] (0.785,-0.3) circle (0.27);
\draw (0.785,-0.3) node {$\dfrac{\pi}{4}$};
\draw[white,fill=white] (-0.785,-0.3) circle (0.27);
\draw (-0.785,-0.3) node {$-\dfrac{\pi}{4}$};

\draw (-0.3,1) node {$1$};
\draw (-0.3,-1) node {$-1$};

\draw (0.1,-0.3) node {$0$};

\draw[<->,scale=1,domain=-1.25:1.25,smooth,variable=\x,black,thick] plot ({\x},{tan(deg(\x))});
\end{tikzpicture}
\end{center}
\newpage
\part[10] Graph $y=\sec\left(x+\dfrac{\pi}{4}\right)$. \\
\textit{Solution}: Secant has asymptotes at $-\dfrac{\pi}{2}, \dfrac{\pi}{2}, \dfrac{3\pi}{2}$, so:  \\
\begin{center}\textbf{Anchor points}: $-\dfrac{\pi}{2}, 0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}$\end{center}
$$\downarrow \text{(h.shift)}$$
$$-\dfrac{3\pi}{4}, -\dfrac{\pi}{4}, \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}$$

\begin{center}
\begin{tikzpicture}[scale=2]
\draw (0,-3) -- (0,3);
\draw (-3,0) -- (5,0);
\draw[dashed,thick] (-2.356,-3) -- (-2.356,3);
\draw[dashed,thick] (0.7853,-3) -- (0.7853,3);
\draw[dashed,thick] (3.9269,-3) -- (3.9269,3);


\draw[white,fill=white] (-2.356,-0.3) circle (0.27);
\draw (-2.356,-0.3) node {$-\dfrac{3\pi}{4}$};
\draw[white,fill=white] (0.7853,-0.3) circle (0.27);
\draw (0.7853,-0.3) node {$\dfrac{\pi}{4}$};
\draw[white,fill=white] (3.9269,-0.3) circle (0.27);
\draw (3.9269,-0.3) node {$\dfrac{5\pi}{4}$};


\draw (2.3561,-0.3) node {$\dfrac{3\pi}{4}$};
\draw (-0.7853,-0.3) node {$-\dfrac{\pi}{4}$};

\draw (-0.3,1) node {$1$};
\draw (-0.3,-1) node {$-1$};

\draw[<->,scale=1,domain=-2:0.45,smooth,variable=\x,black,thick] plot ({\x},{sec(deg(\x+0.7853))});
\draw[<->,scale=1,domain=1.1:3.6,smooth,variable=\x,black,thick] plot ({\x},{sec(deg(\x+0.7853))});

\end{tikzpicture}
\end{center}
\newpage
\part[10] Graph $y=7\cos \left( 3 \left(x - \dfrac{\pi}{2}\right) \right)+5$. \\
\textit{Solution}: Start with the anchor points: \\
\begin{center}\textbf{Anchor points}: $0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}, 2\pi$\end{center}
$$\downarrow \text{(h.compr)}$$
$$0, \dfrac{\pi}{6}, \dfrac{\pi}{3}, \dfrac{\pi}{2}, \dfrac{2\pi}{3}$$
$$\downarrow \text{(h.shift)}$$
$$\dfrac{\pi}{2}, \dfrac{2\pi}{3}, \dfrac{5\pi}{6}, \pi, \dfrac{7\pi}{6}$$
\begin{center}\hspace*{-150pt}
\begin{tikzpicture}[scale=3]
\draw (0,-0.1) -- (0,3.5);
%\draw (-0.1,0) -- (3,0);

\draw[dashed,thick] (-0.1,2) -- (3,2);


\draw[white,fill=white] (-2.356,-0.3) circle (0.27);

\draw[white,fill=white] (0.7853,-0.3) circle (0.27);
\draw[white,fill=white] (3.9269,-0.3) circle (0.27);



\draw (0.5239,-0.3) node {$\dfrac{\pi}{2}$};
\draw (1.0471,-0.3) node {$\dfrac{2\pi}{3}$};
\draw (1.5707,-0.3) node {$\dfrac{5\pi}{6}$};
\draw (2.0943,-0.3) node {$\pi$};
\draw (2.6179,-0.3) node {$\dfrac{7\pi}{6}$};

\draw (-0.3,3) node {$12$};
\draw (-0.3,2) node {$5$};
\draw (-0.3,1) node {$-2$};

\draw[<->,scale=1,domain=0.5235:2.6179,smooth,variable=\x,black,thick] plot ({\x},{2-cos(deg(3*(\x-1.5707))});

\end{tikzpicture}
\end{center}
\end{parts}

\newpage
\question[24] Do each part.
\begin{parts} 
\part[8] Find the exact value of $\tan^{-1}(1)$. \\
\textit{Solution}: $\tan^{-1}(1) = \dfrac{\pi}{4}$
\part[8] Find the exact value of $\sin(\cos^{-1}(x))$. \\
\textit{Solution}: Let $\theta=\cos^{-1}(x)$, then $\cos(\theta)=x$. Now draw a triangle for this situation:
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (4,4) -- cycle;
\draw (4.5,1.7) node {$?$};
\draw (1.8,2.6) node {$1$};
\draw (2,-0.5) node {$x$};
%\draw (3.7,3.3) node {$A$};
\draw (0.6,0.2) node {$\theta$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}
\end{center}
By the Pythagorean theorem, we know that $x^2+?^2=1^2$, hence $?=\sqrt{1-x^2}$. Therefore, we may compute
$$\sin \left(\cos^{-1}(x) \right) = \sin(\theta) = \sqrt{1-x^2}.$$

\part[8] Find the exact value of $\csc\left(\tan^{-1}\left(-\dfrac{3}{7} \right)\right)$. \\
\textit{Solution}: Let $\theta = \tan^{-1} \left( -\dfrac{3}{7} \right)$, so we may conclude that $\theta$ lives in $QI$ or $QIV$. Then $\tan(\theta) = -\dfrac{3}{7}$, so we may conclude that $\theta$ lives in $QII$ or $QIV$. Therefore $\theta$ is in $QIV$ (so the cosecant of $\theta$ will be negative). Draw a triangle that agrees with $\tan(\theta) = -\dfrac{3}{7}$ (except for the sign):
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (4,4) -- cycle;
\draw (4.5,1.7) node {$3$};
\draw (1.8,2.6) node {$?$};
\draw (2,-0.5) node {$7$};
%\draw (3.7,3.3) node {$A$};
\draw (0.6,0.2) node {$\theta$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}
\end{center}
To find $?$ use the Pythagorean theorem: $7^2+3^2=?^2$, hence $?=\sqrt{49+9}=\sqrt{58}$. Therefore compute
$$\csc \left( \tan^{-1} \left( - \dfrac{3}{7} \right) \right) = \csc(\theta) = -\dfrac{\sqrt{58}}{3}.$$
\end{parts}

\question[24] 
\begin{parts}
\part[8] Simplify $\dfrac{\cos(-x)\sec(-x)}{\sin(-x)\csc(-x)}$ \\
\textit{Solution}: Calculate
$$\begin{array}{ll}
\dfrac{\cos(-x)\sec(-x)}{\sin(-x)\csc(-x)} &= \dfrac{\cos(x) \frac{1}{\cos(-x)}}{-\sin(x)\frac{1}{\sin(-x)}} \\
&=\dfrac{\cos(x) \frac{1}{\cos(x)}}{-\sin(x)\frac{1}{-\sin(x)}} \\
&=\dfrac{1}{(-1)(-1)} \\
&= 1.
\end{array}$$
\part[8] Write the first expression in terms of the second expression: $\dfrac{\cos(x)}{1+\sin(x)} + \tan(x); \cos(x)$ \\
\textit{Solution}: Calculate
$$\begin{array}{ll}
\dfrac{\cos(x)}{1+\sin(x)} + \tan(x) &= \dfrac{\cos(x)}{1+\sin(x)} + \dfrac{\sin(x)}{\cos(x)} \\
&= \dfrac{\cos^2(x) + \sin(x)(1+\sin(x))}{\cos(x)(1+\sin(x))} \\
&= \dfrac{\sin(x) + \overbrace{\cos^2(x) + \sin^2(x)}^{=1}}{\cos(x)(1+\sin(x))} \\
&= \dfrac{\sin(x)+1}{\cos(x)(1+\sin(x))} \\
&= \dfrac{1}{\cos(x)}.
\end{array}$$
\part[8] Prove the following identity: $\cos(x)(\tan(x)-\sec(-x))=\sin(x)-1.$ \\
\textit{Solution}: Start with the left:
$$\begin{array}{ll}
\cos(x)(\tan(x)-\sec(-x)) &= \cos(x) \left[ \dfrac{\sin(x)}{\cos(x)} - \dfrac{1}{\cos(-x)} \right] \\
&= \cos(x) \left[ \dfrac{\sin(x)}{\cos(x)} - \dfrac{1}{\cos(x)} \right] \\
&= \sin(x) - 1. \\
\end{array}$$
\vfill
\end{parts}


\newpage

\question[3] Calculate, if possible. If not possible, explain why not (``my calculator says so" is not an explanation). Express your answer correct to two decimal points. 
\begin{parts}
\part[1] $\sin^{-1}(0.783)$ \\
\textit{Solution}: $\sin^{-1}(0.783) \approx 0.899 \text{ radians}=51.54^{\circ}$
\part[1] $\cos^{-1}(2)$ \\
\textit{Solution}: This is not possible, because $2$ is not in the domain of $\cos^{-1}$. Asking the question ``what is $\cos^{-1}(2)$?" is the same as asking the question ``what angle in the unit circle gives us a cosine equal to $2$?" --- there isn't one!
\part[1] $\tan^{-1}\left( \dfrac{\pi}{2} \right)$ \\
\textit{Solution}: $\arctan\left( \dfrac{\pi}{2} \right) \approx 1.003 \text{ radians}=57.52^{\circ}$
\end{parts}

\question[12] The line $y=-\dfrac{2}{7}x$ passes through the origin in  the $xy$-plane. What is the angle that the line makes with the negative $x$-axis? Express your answer correct to two decimal places. \\
\textit{Solution}: 
\begin{center}
\begin{tikzpicture}[scale=2]
\draw (0,-1) -- (0,1);
\draw (-2,0) -- (2,0);

\draw[<->,scale=1,domain=-2:2,smooth,variable=\x,black,thick] plot ({\x},{(-2/7)*\x});

\draw[fill=black] (-1,2/7) circle (0.08em);
\draw (-1,3/7) node {$\left(-1,\frac{2}{7} \right)$};
\draw[fill=black] (0,0) circle (0.08em);
\draw (-0.2,-0.2) node {$(0,0)$};
\end{tikzpicture}
\end{center}
There is a triangle here -- draw it:
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (-4,0) -- (-4,4) -- cycle;
\draw (-4.5,1.7) node {$\dfrac{2}{7}$};
%\draw (1.8,2.6) node {$1$};
\draw (-2,-0.5) node {$1$};
%\draw (3.7,3.3) node {$A$};
\draw (-0.6,0.2) node {$\theta$};
\draw (-3.8,0) -- (-3.8,0.2) -- (-4,0.2);
\end{tikzpicture}
\end{center}
From this picture, we observe that 
$$\tan \left( \theta \right ) =\dfrac{\frac{2}{7}}{1} = \dfrac{2}{7}.$$
Therefore,
$$\theta = \tan^{-1} \left( \dfrac{2}{7} \right) \approx 0.27 \text{ radians}=15.95^{\circ}.$$
\question[12] An isosceles triangle has two sides of length $5$ inches. The third side has a length $3$ inches. Find the angle that a side of length $5$ makes with the side of length $3$. Express your answer correct to two decimal places. \\
\textit{Solution}: Draw the described triangle:
\begin{center}
\begin{tikzpicture}[scale=0.5]
\draw (-4,0) -- (4,0) -- (0,8) -- cycle;
\draw (0,-0.4) node {$3$};
\draw (2.3,4.2) node {$5$};
\draw (-2.3,4.2) node {$5$};
\end{tikzpicture}
\end{center}

\begin{center}
\begin{tikzpicture}[scale=0.5]
\draw (-4,0) -- (4,0) -- (0,8) -- cycle;
\draw (0,-2) -- (0,10);
\draw (2,-0.4) node {$1.5$};
\draw (-2,-0.4) node {$1.5$};
\draw (2.3,4.2) node {$5$};
\draw (-2.3,4.2) node {$5$};
\end{tikzpicture}
\end{center}
Take, say, the right triangle.
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (-4,0) -- (-4,4) -- cycle;
\draw (-2,-0.5) node {$1.5$};
\draw (-0.6,0.2) node {$\theta$};
\draw (-1.5,2) node {$5$};
\draw (-3.8,0) -- (-3.8,0.2) -- (-4,0.2);
\end{tikzpicture}
From this we see that $\cos(\theta) = \dfrac{1.5}{5}$, therefore
$$\theta = \cos^{-1} \left( \dfrac{1.5}{5} \right) \approx 1.266 \text{ radians} = 72.54^{\circ}.$$
\newpage
\end{center}
\question[4] (\textbf{Bonus}): Draw $y=\csc(x)$. Indicate an interval in the $x$-axis which could serve as a domain restriction of $\csc(x)$ that could be used to define an inverse cosecant function. \\
(\textit{hint: recall ``one-to-one" and what it has to do with inverse functions}) \\
\textit{Solution}: \\
\begin{center}
\begin{tikzpicture}[scale=2]
\draw (0,-3) -- (0,3);
\draw (-3,0) -- (3,0);
\draw[dashed,very thick] (0,-3) -- (0,3);
\draw[dashed,thick] (-3.14159,-3) -- (-3.14159,3);
\draw[dashed,thick] (3.14159,-3) -- (3.14159,3);


\draw[white,fill=white] (0,-0.3) circle (0.27);
\draw (0,-0.3) node {$0$};
\draw[white,fill=white] (-3.14159,-0.3) circle (0.27);
\draw (-3.14159,-0.3) node {$-\pi$};
\draw[white,fill=white] (3.14159,-0.3) circle (0.27);
\draw (3.14159,-0.3) node {$\pi $};


\draw (1.5707,-0.3) node {$\dfrac{\pi}{2}$};
\draw (-1.5707,-0.3) node {$-\dfrac{\pi}{2}$};

\draw (-0.3,1) node {$1$};
\draw (-0.3,-1) node {$-1$};

\draw[color=red] (-1.5707,0) -- (1.5707,0);
\draw[color=red,fill=white] (0,0) circle (0.15em);

\draw[<->,scale=1,domain=-2.8:-0.35,smooth,variable=\x,black,thick] plot ({\x},{1/sin(deg(\x))});
\draw[->,scale=1,domain=-1.5707:-0.35,smooth,variable=\x,red,thick] plot ({\x},{1/sin(deg(\x))});
\draw[<->,scale=1,domain=0.35:2.8,smooth,variable=\x,black,thick] plot ({\x},{1/sin(deg(\x))});
\draw[<-,scale=1,domain=0.35:1.5707,smooth,variable=\x,red,thick] plot ({\x},{1/sin(deg(\x))});
\end{tikzpicture}
\end{center}
In order to be one-to-one, a domain restriction of cosecant like {\color{red} $\left[ -\dfrac{\pi}{2}, 0 \right) \cup \left( 0, \dfrac{\pi}{2} \right]$} is necessary. This defines a $\csc^{-1}$ function whose domain is $(-\infty,-1] \cup [1,\infty) = \mathbb{R} \setminus \{(-1,1)\}$ and whose range is $\left[ -\dfrac{\pi}{2}, 0 \right) \cup \left( 0, \dfrac{\pi}{2} \right]$.
\end{questions}
\end{document} 