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\begin{coverpages}
\begin{center}{\huge MATH 1540 - EXAM 3 FALL 2019 \\
\vspace*{0.4in}
SOLUTION
%\large Name: \makebox[3in]{\hrulefill}\\
}
\end{center}

\begin{flushleft}\vspace{0.5in}
Friday, 22 November \\
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}

\begin{center}\underline{Formulas}\end{center}
\begin{itemize} 
\item $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$
\item $\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)$
\item $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
\item $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$
\item $\sin(2\theta)=2\sin(\theta)\cos(\theta)$
\item $\begin{array}{ll} 
\cos(2\theta) &= \cos^2(\theta)-\sin^2(\theta) \\
&=1-2\sin^2(\theta) \\
&=2\cos^2(\theta)-1
\end{array}$
\item $\sin \left( \dfrac{\alpha}{2} \right) = \pm \sqrt{\dfrac{1-\cos(\alpha)}{2}}$
\item $\cos \left( \dfrac{\alpha}{2} \right) = \pm \sqrt{\dfrac{1+\cos(\alpha)}{2}}$
\end{itemize}
\end{coverpages}

\begin{questions}
\question[10] Prove the identity.
\begin{parts}
\part[5] $\cos(x)-\cos^3(x)=\cos(x)\sin^2(x)$ \\ 
\textit{Solution}: Start with the left:
\[ \begin{array}{ll}
\cos(x) - \cos^3(x) &\stackrel{\text{factor}}{=} \cos(x)\Big( 1-\cos^2(x) \Big) \\
&\hspace{-22pt}\stackrel{\underset{\cos^2(x)+\sin^2(x)=1}{1-\cos^2(x)=\sin^2(x)}}{=} \cos(x)\sin^2(x).
\end{array}\]
\part[5] $\dfrac{\sin(x+h)-\sin(x)}{h} = \sin(x) \left( \dfrac{\cos(h)-1}{h} \right) + \cos(x) \left( \dfrac{\sin(h)}{h} \right)$ \\
\textit{Solution}: Using the sum identity for sine, start with the left and compute
\[ \begin{array}{ll}
\dfrac{\sin(x+h)-\sin(x)}{h} &= \dfrac{\Big( \sin(x)\cos(h)+\cos(x)\sin(h)\Big)-\sin(x)}{h} \\
&= \dfrac{\sin(x)\cos(h)-\sin(x)}{h} + \dfrac{\cos(x)\sin(h)}{h} \\
&= \sin(x) \left( \dfrac{\cos(h)-1}{h} \right) + \cos(x) \left( \dfrac{\sin(h)}{h} \right).
\end{array} \]
\end{parts}

\question[14] Find the exact value.
\begin{parts}
\part[7] $\cos \left( \dfrac{5\pi}{12} \right)$ \\
\textit{Solution}: Note that 
\[ \dfrac{\pi}{4} + \dfrac{\pi}{6} = \dfrac{3\pi}{12} + \dfrac{2\pi}{12} = \dfrac{5\pi}{12}.\]
Therefore we may use the sum rule for cosine to compute
\[ \begin{array}{ll}
\cos\left( \dfrac{5\pi}{12} \right) &= \cos \left( \dfrac{\pi}{4} + \dfrac{\pi}{6} \right)\\
&= \cos \left( \dfrac{\pi}{4} \right) \cos \left( \dfrac{\pi}{6} \right) - \sin \left( \dfrac{\pi}{4} \right) \sin\left( \dfrac{\pi}{6} \right) \\
&=\left( \dfrac{\sqrt{2}}{2} \right) \left( \dfrac{\sqrt{3}}{2} \right) - \left( \dfrac{\sqrt{2}}{2} \right) \left( \dfrac{1}{2} \right) \\
&= \dfrac{\sqrt{6}-\sqrt{2}}{4}.
\end{array} \]
\part[7] $\cos \left( \dfrac{\pi}{8} \right)$ \\ 
\textit{Solution}: Consider
\[ \dfrac{\pi}{8} = \dfrac{\alpha}{2},\]
implying that
\[ \alpha = \dfrac{\pi}{4}.\]
This means that we can use the half-angle identity:
\[ \begin{array}{ll}
\cos \left( \dfrac{\pi}{8} \right) &= \cos \left( \dfrac{\frac{\pi}{4}}{2} \right) \\
&= +\sqrt{\dfrac{1+\cos\left(\frac{\pi}{4}\right)}{2}} \\
&=\sqrt{\dfrac{1+\frac{\sqrt{2}}{2}}{2}} \\
&=\sqrt{\dfrac{2+\sqrt{2}}{4}}.
\end{array} \]
\end{parts}

\question[6] Solve for $\theta$ in $[0,2\pi)$: $2\sin(\theta)=1$. \\
\textit{Solution}: First divide by 2 to get
\[ \sin(\theta) = \dfrac{1}{2}.\]
Now we look at the unit circle for all points whose second coordinate is $\dfrac{1}{2}$. That search reveals that $\theta=\dfrac{\pi}{6}$ and $\theta=\dfrac{5\pi}{6}$ are the solutions.

\question[14] Consider the following triangle:
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (4,4) -- cycle;
\draw (4.5,1.7) node {$2$};
\draw (1.8,2.6) node {$7$};
\draw (2,-0.5) node {$a$};
\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}
Find
\begin{parts}
\part[7] $\cos \left(\dfrac{\alpha}{2} \right)$ \\
\textit{Solution}: First find the missing leg $a$ using the Pythagorean theorem:
\[ a^2+2^2=7^2, \]
hence
\[ a = \sqrt{49-4} = \sqrt{45}.\]
Now, using the half-angle identity,
\[ \begin{array}{ll}
\cos\left(\dfrac{\alpha}{2}\right) &= +\sqrt{\dfrac{1+\cos(\alpha)}{2}} \\
&=\sqrt{\dfrac{1+\frac{\sqrt{45}}{7}}{2}} \\
&=\sqrt{\dfrac{7+\sqrt{45}}{14}}.
\end{array}\]
\part[7] $\tan(2\theta)$ \\ 
\textit{Solution}: Using the double angle formulas, we get
\[ \begin{array}{ll}
\tan(2\theta) &= \dfrac{\sin(2\theta)}{\cos(2\theta)} \\
&= \dfrac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta)-\sin^2(\theta)} \\
&=\dfrac{2\left(\frac{\sqrt{45}}{7}\right)\left(\frac{2}{7}\right)}{\left( \frac{2}{7} \right)^2 - \left( \frac{\sqrt{45}}{7} \right)^2} \\
&= \dfrac{\frac{4\sqrt{45}}{49}}{\frac{4}{49}-\frac{45}{49}} \\
&=\left( \dfrac{4\sqrt{45}}{49} \right) \left( \dfrac{49}{-41} \right) \\
&= -\dfrac{4\sqrt{45}}{41}.
\end{array}\]
\end{parts}

\question[14] If $\cos(x)=-\dfrac{1}{3}$ and $x$ is in quadrant II, then find
\begin{parts}
\part[7] $\cos(2x)$ \\
\textit{Solution}: First draw a triangle:
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (4,4) -- cycle;
\draw (4.5,1.7) node {$?$};
\draw (1.8,2.6) node {$3$};
\draw (2,-0.5) node {$1$};
\draw (0.6,0.2) node {$x$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}
\end{center}
Using Pythagorean theorem, $1^2+?^2=3^2$, hence $?=\sqrt{9-1}=\sqrt{8}$. Therefore, $\sin(x)=\dfrac{\sqrt{8}}{3}$. Now we may compute
\[ \cos(2x) = \cos^2(x) - \sin^2(x) = \left( -\dfrac{1}{3}\right)^2-\left( \dfrac{\sqrt{8}}{3} \right)^2 = \dfrac{1}{9} - \dfrac{8}{9} = -\dfrac{7}{9}.\]
\part[7] $\sin \left( \dfrac{x}{2} \right)$ \\ 
\textit{Solution}: Using the half-angle identity and the fact that if $x$ is in QII, then $\dfrac{x}{2}$ is in QI,
\[ \sin\left(\dfrac{x}{2}\right) = +\sqrt{\dfrac{1-\cos(x)}{2}} = \sqrt{\dfrac{1-\left(-\frac{1}{3}\right)}{2}} = \sqrt{\dfrac{\frac{4}{3}}{2}} = \sqrt{\dfrac{4}{6}}=\sqrt{\dfrac{2}{3}}.\]
\end{parts}

\question[10] Consider the vectors $\vec{u}=\left\langle 1,2 \right\rangle$ and $\vec{v}=\left\langle -1,3\right\rangle$. 
\begin{parts}
\part[5] Compute $\vec{u}+\vec{v}$. \\
\textit{Solution}: Calculate
\[ \vec{u}+\vec{v}=\langle 1,2 \rangle + \langle -1,3 \rangle = \langle 1+(-1), 2+3 \rangle = \langle 0,5 \rangle.\]
\part[5] Compute $3\vec{u}-2\vec{v}$.  \\
\textit{Solution}: Calculate
\[\begin{array}{ll}
3\vec{u}-2\vec{v} &= 3\langle 1,2 \rangle -2\langle -1,3 \rangle =\langle 3,6 \rangle + \langle 2,-6\rangle =\langle 5,0\rangle.
\end{array}\]
\end{parts}

\question[16] Solve the triangle. Express your answers with two decimal places. \\
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (2,4) -- cycle;
\draw (3.4,0.3) node {$B$};
\draw (2,3.1 ) node {$113^{\circ}$};
\draw (2,-0.5) node {$32$};
\draw (3.35,2.2) node {$a$};
\draw (0.6,2.2) node {$10$};
\draw (0.6,0.3) node {$A$};
\end{tikzpicture}
\end{center}
\textit{Solution}: \underline{\textbf{Find $B$}} \\
By law of sines,
\[ \dfrac{\sin(B)}{10} = \dfrac{\sin(113^{\circ})}{32},\]
so multiplying by $10$ yields
\[ \sin(B) = \dfrac{10\sin(113^{\circ})}{32}. \]
Taking inverse sine of both sides yields
\[ B = \sin^{-1} \left( \dfrac{10\sin(113^{\circ})}{32} \right) \approx 16.72^{\circ}. \]
{\color{blue} (note: we need to check for a second solution. The second possible value of $B$ would be 
\[ B_2 = 180^{\circ} - 16.72^{\circ} =163.28^{\circ},\]
but this value is \textbf{too large}, so we do not have a second solution to this triangle)} \\
\underline{\textbf{Find $A$}} \\
Using the sum of angles of a triangle being $180^{\circ}$, we see
\[ 113^{\circ} + 16.72^{\circ} + A = 180^{\circ},\]
so
\[ A=180^{\circ}-113^{\circ}-16.72^{\circ} = 50.28^{\circ}.\]
\underline{\textbf{Find $a$}} \\
Using law of sines,
\[ \dfrac{\sin(50.28^{\circ})}{a} = \dfrac{\sin(113^{\circ})}{32}.\]
Therefore multiply by $a$ and by $32$ to get
\[ 32\sin(50.28^{\circ}) = a\sin(113^{\circ}),\]
and divide by the number $\sin(113^{\circ})$ to arrive at
\[a=\dfrac{32\sin(50.28^{\circ})}{\sin(113^{\circ})} \approx 26.739.\]
\question[16] Solve the triangle. Express your answers with two decimal places. \\
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (2,4) -- cycle;
\draw (3.4,0.3) node {$B$};
\draw (2,3.4) node {$C$};
\draw (2,-0.5) node {$30$};
\draw (3.35,2.2) node {$42$};
\draw (0.6,2.2) node {$19$};
\draw (0.6,0.3) node {$A$};
\end{tikzpicture}
\end{center}
\textit{Solution}: \textbf{\underline{Find $A$}} \\
Using law of cosines,
\[ 42^2 = 19^2 + 30^2 - 2(19)(30)\cos(A),\]
and algebra yields
\[ \cos(A) = \dfrac{42^2-19^2-30^2}{-2(19)(30)},\]
so taking the inverse cosine gives us
\[ A = \cos^{-1} \left( \dfrac{42^2-19^2-30^2}{-2(19)(30)} \right) \approx 116.2^{\circ}.\]
\textbf{\underline{Find $B$}} \\
Using law of cosines, 
\[ 19^2 = 30^2 + 42^2 - 2(30)(42)\cos(B),\]
and hence
\[ B = \cos^{-1} \left( \dfrac{19^2-30^2-42^2}{-2(30)(42)} \right) \approx 23.95^{\circ}.\]
\underline{\textbf{Find $C$}} \\
Using the sum of the angles of a triangle being $180^{\circ}$, we get
\[ 116.2^{\circ} + 23.95^{\circ}+C = 180^{\circ},\]
and so
\[C = 180^{\circ} - 116.2^{\circ} - 23.95^{\circ} = 39.85^{\circ}.\]
\iffalse
\begin{minipage}{0.49\textwidth} 
\textbf{Law of Sines}: $\dfrac{\sin(A)}{a}=\dfrac{\sin(B)}{b}=\dfrac{\sin(C)}{c}$ 
\end{minipage}%
\begin{minipage}{0.49\textwidth}
\textbf{Law(s) of Cosines}: 
\begin{itemize} 
\item $a^2=b^2+c^2-2bc\cos(A)$ 
\item $b^2=a^2+c^2-2ac\cos(B)$
\item $c^2=a^2+b^2-2ab\cos(C)$
\end{itemize}
\end{minipage}
\fi
\end{questions}
\end{document}

