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%\header{MATH 11}{Exam 1}{Fall 2016}
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\begin{center}\huge MATH 1115 - EXAM 2 - FALL 2016\\
SOLUTION \\
\end{center}
\begin{flushleft}\vspace{0.5in}
Thursday 20 October 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] Find the exact value.
	\begin{parts}
		\part[5] $\sin^{-1}(-1)$ \\
		\textit{Solution:} \textbf{This problem resembles problems \#16 from Section 3.1 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw6.php}{HW6}}.} It may seem clear from the unit circle that the answer should be $\dfrac{3\pi}{2}$, but since the range of $\sin^{-1}$ is $\left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]$, we must express $\dfrac{3\pi}{2}$ {\color{blue}\href{http://www.wolframalpha.com/input/?i=arcsin(-1)}{as}} $-\dfrac{\pi}{2}$. 
		\part[5] $\cos^{-1} \left( \cos \left( \dfrac{5\pi}{4} \right) \right)$ \\
		\textit{Solution:} \textbf{This problem resembles \#42 Section 3.1 in {\color{blue}\href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw6.php}{HW6}}.} Unit circle at $\dfrac{5\pi}{4}$; it has coordinates $\left( -\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2} \right)$ there. But the range of $\cos^{-1}$ is $[0,\pi]$, so we must find the angle in the interval $[0,\pi]$ who has the same cosine as as $\dfrac{5\pi}{4}$. Moving vertically along the unit circle starting at $\dfrac{5\pi}{4}$, we arrive at the angle $\dfrac{3\pi}{4}$ which has coordinates $\left( -\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right)$ (i.e. it has the same cosine). Therefore we {\color{blue} \href{http://www.wolframalpha.com/input/?i=arccos(cos(5pi%2F4))}{see}}
		$$\cos^{-1} \left( \cos \left( \dfrac{5\pi}{4} \right) \right)=\dfrac{3\pi}{4}.$$
    \scalebox{0.91}{\begin{tikzpicture}[scale=5.3,cap=round,>=latex]
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        \draw[->] (0cm,-1.5cm) -- (0cm,1.5cm) node[above,fill=white] {$y$};
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%                \draw[gray] (0cm,0cm) -- (\x:1cm);
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                % draw each angle in degrees
                %\draw (\x:0.6cm) node[fill=white] {$\x^\circ$};
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        % draw each angle in radians
        \foreach \x/\xtext in {
%            30/\frac{\pi}{6},
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%            60/\frac{\pi}{3},
            90/\frac{\pi}{2},
%            120/\frac{2\pi}{3},
            135/\frac{3\pi}{4},
%            150/\frac{5\pi}{6},
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%            210/\frac{7\pi}{6},
            225/\frac{5\pi}{4},
%            240/\frac{4\pi}{3},
            270/\frac{3\pi}{2},
%            300/\frac{5\pi}{3},
            315/\frac{7\pi}{4},
%            330/\frac{11\pi}{6},
            360/2\pi}
                \draw (\x:0.85cm) node[fill=white] {$\xtext$};
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            % the coordinates for the first quadrant
%            30/\frac{\sqrt{3}}{2}/\frac{1}{2},
            45/\frac{\sqrt{2}}{2}/\frac{\sqrt{2}}{2},
%            60/\frac{1}{2}/\frac{\sqrt{3}}{2},
            % the coordinates for the second quadrant
%            150/-\frac{\sqrt{3}}{2}/\frac{1}{2},
            135/-\frac{\sqrt{2}}{2}/\frac{\sqrt{2}}{2},
%            120/-\frac{1}{2}/\frac{\sqrt{3}}{2},
            % the coordinates for the third quadrant
%            210/-\frac{\sqrt{3}}{2}/-\frac{1}{2},
            225/-\frac{\sqrt{2}}{2}/-\frac{\sqrt{2}}{2},
%            240/-\frac{1}{2}/-\frac{\sqrt{3}}{2},
            % the coordinates for the fourth quadrant
%            330/\frac{\sqrt{3}}{2}/-\frac{1}{2},
            315/\frac{\sqrt{2}}{2}/-\frac{\sqrt{2}}{2}}
%            300/\frac{1}{2}/-\frac{\sqrt{3}}{2}}
                \draw (\x:1.25cm) node[fill=white] {$\left(\xtext,\y\right)$};
        % draw the horizontal and vertical coordinates
        % the placement is better this way
        \draw (-1.25cm,0cm) node[above=1pt] {$(-1,0)$}
              (1.25cm,0cm)  node[above=1pt] {$(1,0)$}
              (0cm,-1.25cm) node[fill=white] {$(0,-1)$}
              (0cm,1.25cm)  node[fill=white] {$(0,1)$};
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    \newpage
		\part[5] $\cot^{-1} \left( \sqrt{3} \right)$ \\
		\textit{Solution:} \textbf{This problem resembles \#40 Section 3.2 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw7.php}{HW7}}.} Which angle has cotangent equal to $\sqrt{3}$? Recall that $\cot(\theta)=\dfrac{x}{y}$. Looking in quadrant $1$, we see that the angle $\dfrac{\pi}{6}$ has coordinates $\left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)$, notice that $\dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\sqrt{3}$. Hence we {\color{blue} \href{http://www.wolframalpha.com/input/?i=arccot(sqrt(3))}{see}} that
		$$\cot^{-1} \left( \sqrt{3} \right)=\dfrac{\pi}{6}.$$
	\end{parts}
	\question[10] Let $f(x)=5\cos(3x)$. 
	\begin{parts}
		\part[4] Find $f^{-1}(x)$. \\
		\textit{Solution:} \textbf{This whole problem resembles \#56 Section 3.1 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw6.php}{HW6}}.} Write $x=5\cos(3y)$. We must solve for $y$ (which is $f^{-1}(x)$). To do this first divide by $5$ to get
		$$\dfrac{x}{5} = \cos(3y).$$
		{\color{red} Note: at this point, many people tried to divide by $3$, which is not permitted. Numbers do not divide out from the inside of trigonometric functions!} \\
		Recall that the inverse cosine function $\cos^{-1}$ obeys the property that $\cos^{-1}(\cos(x))=x$ (for all $x$!). So what we do now is take $\cos^{-1}$ of both sides yielding
		$$\cos^{-1} \left( \dfrac{x}{5} \right) = \cos^{-1} ( \cos(3y)),$$
		and simplifying the right-hand-side with the inverse cosine property, we get
		$$\cos^{-1} \left( \dfrac{x}{5} \right) = 3y.$$
		Finally, we solve for $y$ by dividing by $3$:
		$$\dfrac{1}{3} \cos^{-1} \left( \dfrac{x}{5} \right) = y.$$
		This means that
		$$f^{-1}(x) = \dfrac{1}{3} \cos^{-1} \left( \dfrac{x}{5} \right).$$
		\part[3] Find the domain of $f^{-1}$. \\
		\textit{Solution:} The domain of $f^{-1}$ depends on the domain of $\cos^{-1}$. The domain of $\cos^{-1}$ is $[-1,1]$ and so if we ever write $\cos^{-1}(\mathrm{anything})$ then the following inequality must hold: 
		$$-1 \leq \mathrm{anything} \leq 1.$$
		Our goal is to find an interval of the form $a \leq x \leq b$ for our domain. \\~\\
		\noindent Since $f^{-1}$ contains $\cos^{-1} \left( \dfrac{x}{5} \right)$, we consider the inequality
		$$-1 \leq \dfrac{x}{5} \leq 1.$$
		To get $x$ in the middle, multiply all sides of the inequality by $5$ to get
		$$-5 \leq x \leq 5.$$
		This is the domain; it could also be expressed as $[-5,5]$.
		\newpage
		\part[3] Find the range of $f^{-1}$. \\
		\textit{Solution:} The range of $f^{-1}$ is the set of possible outputs of $f^{-1}$. Since the range of $\cos^{-1}$ is $[0,\pi]$, we may always write $0 \leq \cos^{-1}\left(\mathrm{anything} \right) \leq \pi$. Our goal is to make $f^{-1}(x) = \dfrac{1}{3} \cos^{-1} \left( \dfrac{x}{5} \right)$ appear in the middle. Start with
		$$0 \leq \cos^{-1} \left( \dfrac{x}{5} \right) \leq \pi,$$
		and then divide by $3$ on all sides of the inequality to get
		$$\dfrac{0}{3} \leq \dfrac{1}{3} \cos^{-1} \left( \dfrac{x}{5} \right) \leq \dfrac{\pi}{3},$$
		or more simply
		$$0 \leq f^{-1}(x) \leq \dfrac{\pi}{3}.$$
		The range could also be expressed as $\left[ 0, \dfrac{\pi}{3} \right]$. 
	\end{parts}
	\question[15] Prove the identity.
	\begin{parts}
		\part[3] $\dfrac{\csc(\theta)}{\sec(\theta)}=\cot(\theta)$ \\
		\textit{Solution:} \textbf{This problem is similar to \#20 Section 3.4 in {\color{blue}\href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw5.php}{HW5}}.} Starting with the left-hand side,
		$$\dfrac{\csc(\theta)}{\sec(\theta)} = \dfrac{\frac{1}{\sin(\theta)}}{\frac{1}{\cos(\theta)}} = \dfrac{\cos(\theta)}{\sin(\theta)}=\cot(\theta).$$
		\part[4] $\dfrac{\sin(\theta)\cos(\theta)}{\sin^2(\theta)-\cos^2(\theta)} = \dfrac{\cot(\theta)}{1-\cot^2(\theta)}$ \\
		\textit{Solution:} \textbf{This problem resembles \#58 Section 3.4 in {\color{blue}\href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw5.php}{HW5}.}} \\
		\textbf{Proof 1 (starting with the left-hand side)}:
		$$\begin{array}{ll}
		\dfrac{\sin(\theta)\cos(\theta)}{\sin^2(\theta)-\cos^2(\theta)} &= \dfrac{\sin(\theta)\cos(\theta)}{\sin^2(\theta)-\cos^2(\theta)} \left( \dfrac{\frac{1}{\sin^2(\theta)}}{\frac{1}{\sin^2(\theta)}} \right) \\
		&=	\dfrac{\frac{\cos(\theta)}{\sin(\theta)}}{1-\frac{\cos^2(\theta)}{\sin^2(\theta)}} \\
		&= \dfrac{\cot(\theta)}{1-\cot^2(\theta)},
		\end{array}$$
		as was to be shown. \\~\\
		\noindent \textbf{Proof 2 (starting with the right-hand side):} 
		$$\begin{array}{ll}
		\dfrac{\cot(\theta)}{1-\cot^2(\theta)} &= \dfrac{\frac{\cos(\theta)}{\sin(\theta)}}{1-\frac{\cos^2(\theta)}{\sin^2(\theta)}} \\
		&= \dfrac{\frac{\cos(\theta)}{\sin(\theta)}}{\frac{\sin^2(\theta)-\cos^2(\theta)}{\sin^2(\theta)}} \\
		&= \dfrac{\cos(\theta)}{\sin(\theta)} \dfrac{\sin^2(\theta)}{\sin^2(\theta)-\cos^2(\theta)} \\
		&= \dfrac{ \cos(\theta)\sin(\theta)}{\sin^2(\theta)-\cos^2(\theta)},
		\end{array}$$
		as was to be shown.
		\newpage
		\part[4] $\csc(\theta)-\cot(\theta) = \dfrac{\sin(\theta)}{1+\cos(\theta)}$ \\
		\textit{Solution:} \textbf{This problem resembles \#37 Section 3.4 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw5.php}{HW5}}.} Recall that if we want to move a sum in a trigonometric identity from one side of a fraction bar to the other, we multiply top and bottom by its algebraic conjugate (in this problem, the $1+\cos(\theta)$ below is the algebraic conjugate of $1-\cos(\theta)$). Starting with the left-hand side,
		$$\begin{array}{ll}
		\csc(\theta)-\cot(\theta) &= \dfrac{1}{\sin(\theta)} - \dfrac{\cos(\theta)}{\sin(\theta)} \\
		&= \dfrac{1-\cos(\theta)}{\sin(\theta)} \\
		&= \dfrac{(1-\cos(\theta))}{\sin(\theta)} \left( \dfrac{1+\cos(\theta)}{1+\cos(\theta)} \right) \\
		&=\dfrac{1-\cos^2(\theta)}{\sin(\theta)(1+\cos(\theta))} \\
		&= \dfrac{\sin^2(\theta)}{\sin(\theta)(1+\cos(\theta))} \\
		&=\dfrac{\sin(\theta)}{1+\cos(\theta)},
		\end{array}$$
		as was to be shown.
		\part[4] $\cos(2\theta) = \dfrac{\cot(\theta)-\tan(\theta)}{\cot(\theta)+\tan(\theta)}$ \\
		\textit{hint: it may be easier to start with the right-hand side} \\
		\textit{Solution:} \textbf{This problem was assigned in HW8 -- it was problem 48 in Section 3.6. Also, a similar problem, problem 49 in Section 3.6, was graded in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw8.php}{HW8}}.} Recall that $\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$ and the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$. Starting with the right-hand side,
		$$\begin{array}{ll}
		\dfrac{\cot(\theta)-\tan(\theta)}{\cot(\theta)+\tan(\theta)} &= \dfrac{\frac{\cos(\theta)}{\sin(\theta)} - \frac{\sin(\theta)}{\cos(\theta)}}{\frac{\cos(\theta)}{\sin(\theta)}+\frac{\sin(\theta)}{\cos(\theta)}} \\
		&= \left( \dfrac{\frac{\cos(\theta)}{\sin(\theta)} - \frac{\sin(\theta)}{\cos(\theta)}}{\frac{\cos(\theta)}{\sin(\theta)}+\frac{\sin(\theta)}{\cos(\theta)}} \right) \left( \dfrac{\sin(\theta)\cos(\theta)}{\sin(\theta)\cos(\theta)} \right) \\
		&= \dfrac{\cos^2(\theta) - \sin^2(\theta)}{\cos^2(\theta)+\sin^2(\theta)} \\
		&=\dfrac{\cos(2\theta)}{1} \\
		&= \cos(2\theta),
		\end{array}$$
		as was to be shown.
		\vfill
	\end{parts}


\newpage

	\question[20] Find an exact value for the expression.
	\begin{parts}
		\part[5] $\tan \left( \tan^{-1} \left( \dfrac{6}{7} \right) \right)$ \\
		\textit{Solution:} Since $\tan(\tan^{-1}(x))=x$ for all $x$, the {\color{blue} \href{http://www.wolframalpha.com/input/?i=tan(arctan(6%2F7))}{answer}} is $\dfrac{6}{7}$.
		\part[5]  $\sec \left( \sin^{-1} \left( \dfrac{3}{5} \right) \right)$ \\
		\textit{Solution:} \textbf{This problem resembles problems \#9, \#15, \#31  in Section 3.2 in {\color{blue}\href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw7.php}{HW7}}. } Let $\theta=\sin^{-1} \left( \dfrac{3}{5} \right)$ and so $\theta$ lives in quadrant $I$ or $IV$ (because of the range of $\sin^{-1}$). Also $\sin(\theta)=\dfrac{3}{5}$ and so $\theta$ lives in quadrant $I$ or $II$. Therefore $\theta$ lives in quadrant $I$. \\~\\~\\
		\noindent We understand $\sin(\theta)=\dfrac{y}{r}$ and so we have $y=3$ and $r=5$. Always: $x^2+y^2=r^2$. We must find $x$ because $\sec(\theta)=\dfrac{r}{x}$. Therefore plugging $r=3$ and $y=5$ into $x^2+y^2=r^2$ yields the equation $x^2+9=25$. Solving for $x$ yields $x^2=16$ and so $x = \pm \sqrt{16} = \pm 4$. But since $\theta$ lies in quadrant $I$, we must take $x=4$ (not the negative value). From this we {\color{blue} \href{http://www.wolframalpha.com/input/?i=sec(arcsin(3%2F5))}{conclude}}
		$$\sec \left( \sin^{-1} \left( \dfrac{3}{5} \right) \right) = \sec( \theta) = \dfrac{r}{x} = \dfrac{5}{4}.$$
		\part[5] $\sin \left( \cos^{-1}\left( -\dfrac{1}{2} \right) \right)$ \\
		\textit{Solution:} \textbf{This problem resembles problems \#9, \#15, \#31  in Section 3.2 in {\color{blue}\href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw7.php}{HW7}}. } \\
		\textbf{Proof 1:} {\color{blue} \href{http://www.wolframalpha.com/input/?i=arccos(-1%2F2)}{Notice that}}
		$$\cos^{-1} \left( - \dfrac{1}{2} \right) =\dfrac{2\pi}{3}.$$
		Now {\color{blue} \href{http://www.wolframalpha.com/input/?i=sin(2pi%2F3)}{compute}} 
		$$\sin \left( \dfrac{2\pi}{3} \right) = \dfrac{\sqrt{3}}{2},$$
		finishes the calculation. \\
		\textbf{Proof 2 (a different way):}  \\
		Let $\theta=\cos^{-1} \left( -\dfrac{1}{2} \right)$, so $\theta$ lies in quadrant $I$ or quadrant $II$ (due to the range of $\cos^{-1}$). Also, $\cos(\theta)=-\dfrac{1}{2}$, so $\theta$ lies in quadrant $II$ or quadrant $III$. From this we conclude that $\theta$ lies in quadrant $II$ (and so the $y$-value we find will be positive). \\~\\~\\
		We understand $\cos(\theta) = \dfrac{x}{r} = \dfrac{-1}{2}$ and so we have $x=-1$ and $r=2$. Always: $x^2+y^2=r^2$. We must find $y$ because $\sin(\theta)=\dfrac{y}{r}$. Therefore plugging $r=2$ and $x=-1$ into $x^2+y^2=r^2$ yields the equation $(-1)^2+y^2=4$, or equivalently $y^2=3$, so $y=\pm \sqrt{3}$. But since $\theta$ is in quadrant $II$, we take the positive solution so that $y=\sqrt{3}$. {\color{blue} \href{http://www.wolframalpha.com/input/?i=sin(arccos(-1%2F2))}{Therefore}}
		$$\sin \left( \cos^{-1} \left( - \dfrac{1}{2} \right) \right) = \sin \left( \theta \right)=\dfrac{y}{r} = \dfrac{\sqrt{3}}{2}$$
		\part[5] $\sin \left( 2\cos^{-1} \left( \dfrac{1}{3} \right) \right)$ \\
		\textit{Solution:} \textbf{This problem resembles \#79 Section 3.6 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw8.php}{HW8}}.} Firstly, the $2$ inside the sine function means we should use the the double angle identity. Let $\theta=\cos^{-1} \left( \dfrac{1}{3} \right)$. Thus applying the double angle identity,
		$$\begin{array}{ll}
		\sin \left(2\theta \right) &= 2\sin(\theta)\cos(\theta)\\
		&=2\sin \left(\cos^{-1} \left( \dfrac{1}{3} \right) \right)\cos \left(\cos^{-1} \left( \dfrac{1}{3} \right) \right) \\
		&= \dfrac{2}{3} \sin \left( \cos^{-1} \left( \dfrac{1}{3} \right) \right).
		\end{array}$$
		Now we must evaluate $\sin \left( \cos^{-1} \left( \dfrac{1}{3} \right) \right)$. Notice that $\theta$ lies in quadrant $I$ or $II$, $\cos(\theta)=\dfrac{1}{3}$, and so $\theta$ lies in quadrant $I$ or $IV$ --- from this conclude $\theta$ lies in quadrant $I$ (so the $y$ value we find will be positive). Since $\cos(\theta)=\dfrac{x}{r}$ we get $x=1$ and $r=3$, and so we find $y$ using $x^2+y^2=r^2$: $1^2+y^2=9$ ,or $y=\pm \sqrt{8}$, and since $\theta$ is in quadrant $I$, we get $y=\sqrt{8}$. From this we can deduce
		$$\sin \left( \cos^{-1} \left( \dfrac{1}{3} \right) \right) = \sin(\theta)=\dfrac{y}{r}=\dfrac{\sqrt{8}}{3}.$$
	\end{parts}	
	\question[10] Express $\cos \left( \sin^{-1}(u) \right)$ as an algebraic expression of $u$. \\
	\textit{Solution:} \textbf{This problem resembles \#59 Section 3.2 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw7.php}{HW7}} and \#85 Section 3.5 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw8.php}{HW8}}.} Let $\theta=\sin^{-1}(u)$, so $\theta$ lies in quadrant $I$ or quadrant $II$ (hence the value of $x$ we find will be positive). Then $\sin(\theta)=u$. Recalling that $\sin(\theta) = \dfrac{y}{r}$ we take $u=y$ and $r=1$. From this deduce that $x=\pm\sqrt{1-u^2}$, and take the positive value because $\theta$ lies in $QI$ or $QIV$: $x=\sqrt{1-u^2}$. We may now conclude
	$$\cos(\sin^{-1}(u))=\dfrac{x}{r} = \dfrac{\sqrt{1-u^2}}{1} = \sqrt{1-u^2}.$$
	\question[10] Solve the equation. \textbf{These problems resemble \#64 Section 3.1 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw6.php}{HW6}}.}
	\begin{parts}
		\part[5] $2\sin^{-1}(9x)=\dfrac{\pi}{2}$ \\
		\textit{Solution:} Divide by $2$ to get
		$$\sin^{-1}(9x) = \dfrac{\pi}{4}.$$
		Taking $\sin$ of both sides yields
		$$\sin \left( \sin^{-1} ( 9x ) \right) = \sin \left( \dfrac{\pi}{4} \right).$$
		On the left side remember that $\sin^{-1}$ and $\sin$ cancel each out and on the right evaluate $\sin \left( \dfrac{\pi}{4} \right)$ to get
		$$9x = \dfrac{\sqrt{2}}{2}.$$
		From here solve for $x$ by dividing by $9$:
		$$x = \dfrac{\sqrt{2}}{18}.$$
		\part[5] $4\tan^{-1}(2x)=\pi$ \\
		\textit{Solution:} Divide by $4$ to get
		$$\tan^{-1}(2x) = \dfrac{\pi}{4}.$$
		Now take tangent of both sides to get
		$$\tan(\tan^{-1}(2x))=\tan \left( \dfrac{\pi}{4} \right).$$
		On the left side, $\tan$ and $\tan^{-1}$ cancel each other and on the right side evaluate $\tan \left( \dfrac{\pi}{4} \right)$ to get
		$$2x = 1,$$
		and now divide by $2$ to get
		$$x = \dfrac{1}{2}.$$
		\vfill
	\end{parts}
	
	\question[20] Find the exact value of the expression.
	\begin{parts}
		\part[6] $\cos \left( \dfrac{\pi}{8} \right)$ \\
		\textit{Solution:} \textbf{This problem resembles \#27 Section 3.6 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw8.php}{HW8}}.} Recall that 
		$$\cos \left( \dfrac{\theta}{2} \right)=\pm \sqrt{ \dfrac{1+\cos(\theta)}{2} },$$
		where the sign depends on which quadrant $\dfrac{\theta}{2}$ lies in. Since $\dfrac{\pi}{8}$ lies in quadrant $I$, we take the positive value in the following half-angle identity:
		$$\begin{array}{ll}
		\cos \left( \dfrac{\pi}{8} \right) &= \cos \left( \dfrac{1}{2} \dfrac{\pi}{4} \right)  \\
		&= \cos \left( \dfrac{\frac{\pi}{4}}{2} \right) \\
		&= + \sqrt{ \dfrac{1+\cos(\frac{\pi}{4})}{2}} \\
		&= \sqrt{ \dfrac{1+\frac{\sqrt{2}}{2}}{2}} \\	
		\end{array}$$
		\part[7] $\sin \left( \dfrac{5\pi}{12} \right)$ \\
		\textit{Solution:} \textbf{This problem resembles \#13 Section 3.5 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw8.php}{HW8}}.} Notice that 
		$$\dfrac{5\pi}{12} = \dfrac{9\pi}{12} - \dfrac{4\pi}{12} = \dfrac{3\pi}{4} - \dfrac{\pi}{3}.$$
		Recall the difference identity for sine:
		$$\sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta).$$
		Using $\alpha=\dfrac{3\pi}{4}$ and $\beta=\dfrac{\pi}{3}$, we compute
		$$\begin{array}{ll}
		\sin \left( \dfrac{5\pi}{12} \right) &= \sin \left( \dfrac{3\pi}{4} - \dfrac{\pi}{3} \right) \\
		&= \sin \left( \dfrac{3\pi}{4} \right) \cos \left( \dfrac{\pi}{3} \right) - \cos \left( \dfrac{3\pi}{4} \right) \sin \left( \dfrac{\pi}{3} \right) \\
		&= \left( - \dfrac{\sqrt{2}}{2} \right) \left( \dfrac{1}{2} \right) - \left( - \dfrac{\sqrt{2}}{2} \right) \left( \dfrac{\sqrt{3}}{2} \right) \\
		&= \dfrac{\sqrt{6}-\sqrt{2}}{4}.
		\end{array}$$
		\part[7] $\cos \left( \dfrac{5\pi}{12} \right) \cos \left( \dfrac{7\pi}{12} \right) - \sin \left( \dfrac{5\pi}{12} \right) \sin \left( \dfrac{7\pi}{12} \right)$ \\
		\textit{Solution:} \textbf{This problem resembles \#29 Section 3.5 in {\color{blue} \href{http://tomcuchta.com/teach/classes/2016/MATH-1115-Fall-2016-FairmontState/homework/hw8.php}{HW8}}.} The key to this problem is to realize that we have the right-hand side of the sum identity for cosine:
		$$\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta),$$
		with $\alpha = \dfrac{5\pi}{12}$ and $\beta = \dfrac{7\pi}{12}$. So we see
		$$\begin{array}{ll}
		\cos \left( \dfrac{5\pi}{12} \right) \cos \left( \dfrac{7\pi}{12} \right) - \sin \left( \dfrac{5\pi}{12} \right) \sin \left( \dfrac{7\pi}{12} \right) &= \cos \left( \dfrac{5\pi}{12} - \dfrac{7\pi}{12} \right) \\
		&= \cos \left( -\dfrac{2\pi}{12} \right) \\
		&= \cos \left( - \dfrac{\pi}{6} \right) \\
		&= \dfrac{\sqrt{3}}{2}.
		\end{array}$$
		\vfill
	\end{parts}
	
	\vfill
\end{questions}
\end{document}