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%\header{MATH 11}{Exam 1}{Fall 2016}
\begin{document}
\begin{coverpages}
\begin{center}\huge MATH 1540 - EXAM 3 - FALL 2017\\
\vspace*{2in}
\large Name: \makebox[3in]{\hrulefill}\\
\end{center}
\begin{flushleft}\vspace{0.5in}
2 November 2017\\
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 Write the first expression in terms of the second expression.
%\begin{parts}
%\part $\csc(x)\tan(x);\cos(x)$
%\end{parts}

\question[24] Prove the identity.
\begin{parts}
\part[8] $\tan(x)\cos(x)=\sin(x)$ \\
\textit{Solution:} Recall that $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$. Start on the left-hand side and compute
$$\begin{array}{ll}
\tan(x)\cos(x) &= \dfrac{\sin(x)}{\cancel{\cos(x)}} \cancel{\cos(x)} \\
&= \sin(x),
\end{array}$$
as was to be proved.
\part[8] $\sin(x)-\sin^3(x)=\sin(x)\cos^2(x)$ \\
\textit{Solution:} Recall the Pythagorean identity $\cos^2(x)+\sin^2(x)=1$ and rearrange it so that $\cos^2(x)=1-\sin^2(x)$. Start on the left-hand side and compute
$$\begin{array}{ll}
\sin(x) - \sin^3(x) &= \sin(x) \left[ 1 - \sin^2(x) \right] \\
&= \sin(x) \cos^2(x),
\end{array}$$
as was to be proved. 
\part[8] $\dfrac{\sin(x+h)-\sin(x)}{h} = \sin(x) \left( \dfrac{\cos(h)-1)}{h} \right) + \cos(x) \left( \dfrac{\sin(h)}{h} \right)$ \\
\textit{Solution:} Recall the sum identity for sine: $\sin(x+h)=\sin(x)\cos(h)+\cos(x)\sin(h)$. Now compute
$$\begin{array}{ll}
\dfrac{\sin(x+h)-\sin(x)}{h} &= \dfrac{(\sin(x)\cos(h)+\cos(x)\sin(h))-\sin(x)}{h} \\
&= \dfrac{\sin(x)[\cos(h)-1] + \cos(x) \sin(h)}{h} \\
&= \sin(x) \dfrac{\cos(h)-1}{h} + \cos(x) \dfrac{\sin(h)}{h},
\end{array}$$
as was to be proved.
\end{parts}

\question[14] Find the exact value.
\begin{parts}
\part[7] $\sin \left( \dfrac{5\pi}{12} \right)$ \\
\textit{Solution:} Note that $\dfrac{5\pi}{12}=\dfrac{\pi}{4}+\dfrac{\pi}{6}$. Therefore use the sum identity for sine to \href{http://www.wolframalpha.com/input/?i=sin(5pi{\%}2F12)}{compute}
$$\begin{array}{ll}
\sin\left( \dfrac{\pi}{4} + \dfrac{\pi}{6} \right) &= \sin \left( \dfrac{\pi}{4} \right) \cos \left( \dfrac{\pi}{6} \right) + \cos \left( \dfrac{\pi}{4} \right) \sin \left( \dfrac{\pi}{6} \right) \\
&= \left( \dfrac{\sqrt{2}}{2} \right) \left( \dfrac{1}{2} \right) + \left( \dfrac{\sqrt{2}}{2} \right) \left( \dfrac{\sqrt{3}}{2} \right) \\
&= \dfrac{\sqrt{2}+\sqrt{6}}{4}.
\end{array}$$
\part[7] $\sin \left( \dfrac{7\pi}{8} \right)$ \\
\textit{Solution:} Recall the half-angle identity for sine: $\sin \left( \dfrac{\alpha}{2} \right) = \pm \sqrt{ \dfrac{1-\cos(\alpha)}{2}}$. Therefore since $\dfrac{7\pi}{8}$ is in quadrant II and if $\alpha = \dfrac{7\pi}{4}$ then $\dfrac{\alpha}{2} = \dfrac{7\pi}{8}$, \href{http://www.wolframalpha.com/input/?i=sin(7pi{\%}2F8)}{compute}
$$\begin{array}{ll}
\sin \left( \dfrac{7\pi}{8} \right) &= \sin \left( \dfrac{\frac{7\pi}{4}}{2} \right) \\
&= +\sqrt{ \dfrac{1-\cos(\frac{7\pi}{4})}{2}} \\
&= \sqrt{ \dfrac{1-\frac{\sqrt{2}}{2}}{2}}.
\end{array}$$
\end{parts}

\question[14] Consider the following triangle:
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (4,4) -- cycle;
\draw (4.5,1.7) node {$b$};
\draw (1.8,2.6) node {$3$};
\draw (2,-0.5) node {$2$};
\draw (3.7,3.3) node {$\theta$};
\draw (0.6,0.2) node {$\alpha$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}
\end{center}
Now find
\begin{parts}
\part[7] $\sin \left(\dfrac{\alpha}{2} \right)$ \\
\textit{Solution:} First find $b$: using the Pythagorean theorem, we write $2^2+b^2=3^2$ and so $b=\sqrt{9-4}=\sqrt{5}$. Therefore, using the half-angle identity for sine:
$$\begin{array}{ll}
\sin \left( \dfrac{\alpha}{2} \right) &= +\sqrt{ \dfrac{1-\cos(\alpha)}{2} } = \sqrt{ \dfrac{1-\frac{2}{3}}{2} } \\
\end{array}$$
\part[7] $\tan(2\alpha)$ \\
\textit{Solution:} Recall the double angle identities for sine and cosine: $\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ and $\cos(2\alpha)=\cos^2(\alpha)-\sin^2(\alpha)$ (using the other double angle identities for cosine are also OK). Now compute
$$\begin{array}{ll}
\tan(2\alpha) &= \dfrac{\sin(2\alpha)}{\cos(2\alpha)} \\
&= \dfrac{2\sin(\alpha)\cos(\alpha)}{\cos^2(\alpha)-\sin^2(\alpha)}   \\
&= \dfrac{2(\frac{\sqrt{5}}{3})(\frac{2}{3})}{\left(\frac{2}{3} \right)^2-\left(\frac{\sqrt{5}}{3}\right)^2} \\
&= \dfrac{\frac{4\sqrt{5}}{9}}{\frac{4}{9}-\frac{5}{9}} \\
&= \dfrac{\frac{4\sqrt{5}}{9}}{\frac{1}{9}} \\
&= 4\sqrt{5}
\end{array}$$
\end{parts}

\question[16] If $\tan(x)=-\dfrac{1}{3}$ and $x$ is in quadrant IV, find
\begin{parts}
\part[8] $\sin(2x)$ \\
\textit{Solution:} We need to find $\sin(x)$ and $\cos(x)$. To do so, draw a triangle that agrees with $\tan(x)=\dfrac{1}{3}$:
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (4,4) -- cycle;
\draw (4.5,1.7) node {$1$};
\draw (1.8,2.6) node {$?$};
\draw (2,-0.5) node {$3$};
\draw (0.6,0.2) node {$x$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}
\end{center}
To find the hypotenuse (labeled $?$) we use the Pythagorean theorem: $3^2+1^2=?^2$ so that $?=\sqrt{10}$. Therefore $\cos(x)=\dfrac{3}{\sqrt{10}}$ and $\sin(x) = -\dfrac{1}{\sqrt{10}}$ (note: the negative is because the angle $x$ is in quadrant IV!). Therefore using the double angle identity for sine,
$$\sin(2x)=2\sin(x)\cos(x)=2 \left( - \dfrac{1}{\sqrt{10}} \right) \left( \dfrac{3}{\sqrt{10}} \right).$$
\part[8] $\cos \left( \dfrac{x}{2} \right)$ \\
\textit{Solution:} Using the half-angle identity for cosine,
$$\cos \left( \dfrac{x}{2} \right) = -\sqrt{ \dfrac{1-\cos(x)}{2}} = -\sqrt{ \dfrac{1-\frac{3}{\sqrt{10}}}{2} }.$$
\end{parts}

\question[32] Solve the equation on $[0,2\pi)$.
\begin{parts}
\part[8] $2\cos(\theta)-1=0$ \\
\textit{Solution:} Solve for $\cos(\theta)$ to get $\cos(\theta) = \dfrac{1}{2}$. From the unit circle, we observe that $\theta = \dfrac{\pi}{3}, \dfrac{5\pi}{3}$. 
\part[8] $\sin^2(\theta)=\dfrac{3}{4}$ \\
\textit{Solution:} Take the square root of both sides of the equation (recall: when doing this, you \textbf{must} introduce a ``$\pm$"):
$$\sin(\theta) = \pm \sqrt {\dfrac{3}{4}} = \pm \dfrac{\sqrt{3}}{2}.$$
From the unit circle we see 
$$\theta = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}.$$
\part[8] $\cos(x)-\cos(x)\tan(x)=0$ \\
\textit{Solution:} Factor $\cos(x)$ out of the left-hand side to get 
$$\cos(x)(1-\tan(x)) = 0.$$
Using the property of real numbers that $a \cdot b = 0$ implies $a=0$ or $b=0$, we break the trig equation down into $\cos(x)=0$ and $1-\tan(x)=0$ (i.e. $\tan(x)=1$). Therefore by the unit circle, $\cos(x)=0$ has solution $x=\dfrac{\pi}{2}, \dfrac{3\pi}{2}$ and $\tan(x)=1$ has solution $x=\dfrac{\pi}{4}, \dfrac{5\pi}{4}$. 
\part[8] $\sin(2x)=\dfrac{1}{2}$ \\
\textit{Solution:} Let $\psi=2x$. Then solving $\sin(\psi)=\dfrac{1}{2}$ yields the general solutions
$$\left\{ \begin{array}{lll}
2x &= \psi &=  \dfrac{\pi}{6} + 2n \pi = \dfrac{\pi}{6} + \dfrac{12n\pi}{6} \\
2x &= \psi &= \dfrac{5\pi}{6} + 2n\pi = \dfrac{5\pi}{6} + \dfrac{12n\pi}{6}
\end{array} \right.$$
Dividing by $2$ yields
$$\left\{ \begin{array}{ll}
x &= \dfrac{\pi}{12} + \dfrac{12n\pi}{12} \\
x &= \dfrac{5\pi}{12} + \dfrac{12n\pi}{12} \\
\end{array} \right.$$
Now we seek the values of $n=\ldots,-2,-1,0,1,2,\ldots$ for which $0 \leq x < 2\pi$, or in other words $0 \leq x < \dfrac{24\pi}{12}$. 
$$n=-1: \left\{ \begin{array}{lll}
x = \dfrac{\pi}{12} - \dfrac{12\pi}{12} = -\dfrac{11\pi}{12} & \times \\
x = \dfrac{5\pi}{12} - \dfrac{12\pi}{12} = -\dfrac{7\pi}{12} & \times
\end{array} \right.$$

$$n=0: \left\{ \begin{array}{lll}
x = \dfrac{\pi}{12} & \checkmark \\
x = \dfrac{5\pi}{12} & \checkmark
\end{array} \right.$$

$$n=1: \left\{ \begin{array}{lll}
x = \dfrac{\pi}{12} + \dfrac{12\pi}{12} = \dfrac{13\pi}{12} & \checkmark \\
x = \dfrac{5\pi}{12} + \dfrac{12\pi}{12}= \dfrac{17\pi}{12} & \checkmark
\end{array} \right.$$

$$n=2: \left\{ \begin{array}{lll}
x = \dfrac{\pi}{12} + \dfrac{24\pi}{12} = \dfrac{25\pi}{12} & \times \\
x = \dfrac{5\pi}{12} + \dfrac{24\pi}{12}= \dfrac{29\pi}{12} & \times
\end{array} \right.$$

Therefore the solution of the original equation is $x= \dfrac{\pi}{12}, \dfrac{5\pi}{12}, \dfrac{13\pi}{12}, \dfrac{17\pi}{12}$. 
\end{parts} 

\end{questions}
\end{document} 