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\begin{coverpages}
\begin{center}{\huge MATH 1540 - EXAM 3 FALL 2018 \\
SOLUTION
%\vspace*{1in}
%\large Name: \makebox[3in]{\hrulefill}\\
}
\end{center}

\begin{flushleft}\vspace{0.5in}
Friday, 16 November 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}
\begin{center}
\[ \sin(\alpha+\beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \]
\[ \sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) \]
\[ \cos(\alpha+\beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \]
\[ \cos(\alpha-\beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) \]
\[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \]
\[ \begin{array}{ll}
\cos(2\theta) &= \cos^2(\theta)-\sin^2(\theta) \\
&= 1-2\sin^2(\theta) \\
&= 2\cos^2(\theta)-1 
\end{array} \]
\[ \sin \left( \dfrac{\alpha}{2} \right) = \pm \sqrt{ \dfrac{1-\cos(\alpha)}{2}} \]
\[ \cos \left( \dfrac{\alpha}{2} \right) = \pm \sqrt{ \dfrac{1+\cos(\alpha)}{2} } \]
\[ \sin^2(\theta) = \dfrac{1-\cos(2\theta)}{2} \]
\[ \cos^2(\theta) = \frac{1+\cos(2\theta)}{2} \]
\textbf{Law of sines} \\
$\dfrac{\sin(\alpha)}{a} = \dfrac{\sin(\beta)}{b} = \dfrac{\sin(\gamma)}{c}$ \\
\textbf{Law of cosines} \\
$a^2=b^2+c^2-2bc\cos(\alpha)$ \\
$b^2 = a^2+c^2 - 2ac\cos(\beta)$ \\
$c^2 = a^2+b^2 - 2ab\cos(\gamma)$
\end{center}

\end{coverpages}

\begin{questions}
\question[10] Find an exact value for $\cos \left( \dfrac{\pi}{8} \right)$ \\
\textit{Solution}: Since $\dfrac{\pi}{4}$ is in QI,
\[ \cos \left( \dfrac{\pi}{8} \right) = \cos \left( \dfrac{1}{2} \left( \dfrac{\pi}{4} \right) \right) \stackrel{\text{half-angle}}{=} + \sqrt{ \dfrac{1+\cos(\frac{\pi}{4})}{2}} = \sqrt{\dfrac{1+\frac{\sqrt{2}}{2}}{2}} = \sqrt{ \dfrac{2+\sqrt{2}}{4}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}.\]
\question[10] Given that $\cos(x)=-\dfrac{1}{4}$ and $x$ is in quadrant III, compute $\sin \left( 2x \right)$ \\
\textit{Solution}: First draw a triangle corresponding to $\cos(x)=-\dfrac{1}{4}$: \\
\begin{center}\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (4,4) -- cycle;
\draw (4.5,1.7) node {$?$};
\draw (1.8,2.6) node {$4$};
\draw (2,-0.5) node {$1$};
%\draw (3.7,3.3) node {$A$};
\draw (0.6,0.2) node {$x$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}\end{center}
Find the missing side using the Pythagorean theorem: $1^2+?^2=4^2$, hence $?=\sqrt{15}$. Therefore since $x$ is in QIII, $\sin(x) = -\dfrac{\sqrt{15}}{4}$. Now we may compute
\[ \sin(2x) \stackrel{\text{double angle}}{=} 2\sin(x)\cos(x) = 2 \left( -\dfrac{\sqrt{15}}{4} \right) \left( -\dfrac{1}{4} \right) = \dfrac{\sqrt{15}}{8}. \]

\question[15] Prove the identity.
\begin{parts}
\part $\dfrac{\sin(x+h)-\sin(x)}{h} = \dfrac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}$ \\
\textit{Solution}: Start with the left: \\
\[ \begin{array}{ll}
\dfrac{\sin(x+h)-\sin(x)}{h} &\stackrel{\text{sum formula}}{=} \dfrac{\Big(\sin(x)\cos(h)+\cos(x)\sin(h) \Big)-\sin(x)}{h} \\
&= \dfrac{\sin(x)(\cos(h)-1) + \cos(x)\sin(h)}{h},
\end{array} \]
as was to be shown.

\part $\cos(3x)=\cos(x)-4\sin^2(x)\cos(x)$ \\
\textit{Solution}: Notice that $3x=x+2x$. So, start with the left and compute 
\[ \begin{array}{ll}
\cos(3x) &= \cos(x+2x) \\
&\stackrel{\text{sum formula}}{=} \cos(x)\cos(2x) - \sin(x)\sin(2x) \\
&\stackrel{\text{double angle formulas}}{=} \cos(x)\Big( 1-2\sin^2(x) \Big) - \sin(x) \Big( 2\sin(x)\cos(x) \Big) \\
&\stackrel{\text{distribute}}{=} \cos(x) -2\sin^2(x)\cos(x) - 2\sin^2(x)\cos(x) \\
&= \cos(x) -4\sin^2(x)\cos(x),
\end{array}\]
as was to be shown.
\end{parts}
\question[15] Find an exact value for
\[ \cos\left( \sin^{-1}\left(\dfrac{1}{4} \right) + \tan^{-1} \left( \dfrac{1}{7} \right) \right).\]
\textit{Solution}: First use the sum formula to see
\begin{eqnarray*} 
\lefteqn{\cos \left( \sin^{-1} \left( \dfrac{1}{4} \right) + \tan^{-1} \left( \dfrac{1}{7} \right) \right)} \\
&& = \cos \left( \sin^{-1} \left( \dfrac{1}{4} \right) \right) \cos \left( \tan^{-1} \left( \dfrac{1}{7} \right) \right) - \sin \left( \sin^{-1} \left(\dfrac{1}{4} \right) \right) \sin \left( \tan^{-1} \left( \dfrac{1}{7} \right) \right) \\
&& = \cos \left( \sin^{-1} \left( \dfrac{1}{4} \right) \right) \cos \left( \tan^{-1} \left( \dfrac{1}{7} \right) \right) - \dfrac{1}{4} \sin \left( \tan^{-1} \left( \dfrac{1}{7} \right) \right)
\end{eqnarray*} 
\underline{Deal with $\sin^{-1}$} \\
Let $\theta=\sin^{-1} \left( \dfrac{1}{4} \right)$ (so $\theta$ is in QI or QIV), thus $\sin(\theta)=\dfrac{1}{4}$ (so $\theta$ is in QI or QII). We conclude that $\theta$ is in QI. Now draw a triangle: \\
\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 {$4$};
\draw (2,-0.5) node {$?$};
%\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}
Using Pythagorean theorem to find $?$ yields $?+1^2=4^1$, i.e. $?=\sqrt{15}$. Therefore we can conclude that 
\[ \cos \left( \sin^{-1} \left( \dfrac{1}{4} \right) \right) = \cos(\theta) = \dfrac{\sqrt{15}}{4}.\]
\underline{Deal with $\tan^{-1}$} \\
Let $\psi = \tan^{-1} \left( \dfrac{1}{7} \right)$ (so $\psi$ is in QI or QIV), thus $\tan(\psi) = \dfrac{1}{7}$ (so $\psi$ is in QI or QIII). We conclude that $\psi$ is in QIII. Now draw a triangle: \\
\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 {$7$};
%\draw (3.7,3.3) node {$A$};
\draw (0.6,0.2) node {$\psi$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}\end{center}
Using Pythagorean theorem to find $?$ yields $7^1+1^2=?^2$, i.e. $?=\sqrt{50}$. Therefore we can conclude that
\[ \cos \left( \tan^{-1} \left( \dfrac{1}{7} \right) \right) = \cos(\psi) = \dfrac{7}{\sqrt{50}}, \]
and
\[ \sin \left( \tan^{-1} \left( \dfrac{1}{7} \right) \right) = \sin(\psi) = \dfrac{1}{\sqrt{50}}.\]

Finally we now see that
\[ \begin{array}{ll}
\cos \left( \sin^{-1} \left( \dfrac{1}{4} \right) + \tan^{-1} \left( \dfrac{1}{7} \right) \right) &= \cos \left( \theta \right) \cos \left( \psi \right) - \dfrac{1}{4} \sin \left( \psi \right) \\
&= \left( \dfrac{\sqrt{15}}{4} \right) \left( \dfrac{7}{\sqrt{50}} \right) - \left( \dfrac{1}{4} \right) \left( \dfrac{1}{\sqrt{50}} \right).
\end{array} \]

\question[10] Solve the equation for $0 \leq \theta \leq 2\pi$.
\begin{parts}
\part $\cos(\theta)-\dfrac{1}{2}=0$ \\
\textit{Solution}: First solve for $\cos(\theta)$ to get
\[ \cos(\theta) = \dfrac{1}{2}.\]
Now we write down all angles $\theta$ in the unit circle whose cosine is $\dfrac{1}{2}$, i.e. $\theta=\dfrac{\pi}{3}, \dfrac{5\pi}{3}$.
\part $2\sin^2(\theta)-\sin(\theta)-1=0$ \\
\textit{Solution}: This is a quadratic equation of the form $ax^2+bx+c=0$ where ``$x$" is actually ``$\sin(\theta)$" and $a=2, b=-1, c=-1$. Therefore we use the quadratic formula to see 
\[ \sin(\theta) = \dfrac{-(-1) \pm \sqrt{(-1)^2-4(2)(-1)} }{2(2)} = \dfrac{1 \pm \sqrt{9}}{4} = \dfrac{1\pm 3}{4}, \]
meaning we get two solutions: the ``$+$" solution 
\[ \sin(\theta) = \dfrac{1+3}{4} = \dfrac{4}{4}=1 \]
and the ``$-$" solution
\[ \sin(\theta) = \dfrac{1-3}{4} = \dfrac{-2}{4} = -\dfrac{1}{2}. \]
Solving $\sin(\theta)=1$ yields $\theta = \dfrac{\pi}{2}$ and solving $\sin(\theta) = -\dfrac{1}{2}$ yields $\theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$. 
\end{parts}

\question[10] Let $\vec{u}=\langle 1,2 \rangle$ and let $\vec{v}=\langle -1,1\rangle$. Compute $2\vec{u}-3\vec{v}$. \\
\textit{Solution}: Compute 
\[ \begin{array}{ll}
2\vec{u} - 3\vec{v} &= 2 \langle 1,2 \rangle - 3 \langle -1,1 \rangle \\
&= \langle 2,4 \rangle + \langle 3, -3 \rangle \\
&= \langle 2+3, 4+(-3) \rangle \\
&= \langle 5, 1 \rangle.
\end{array} \]
\vfill

\newpage

\question[15] Solve the triangle or explain why it cannot be solved.
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (2,4) -- cycle;
\draw (3.4,0.3) node {$88^{\circ}$};
\draw (2,3.4) node {$\gamma$};
\draw (2,-0.5) node {$5.2$};
\draw (3.35,2.2) node {$4.4$};
\draw (0.6,2.2) node {$b$};
\draw (0.6,0.3) node {$\alpha$};
\end{tikzpicture}
\end{center} 
\textit{Solution}: We cannot use law of sines (all side-angle pairs contain an unknown), so we must use law of cosines. \\
\underline{Find $b$} \\
By law of cosines,
\[ b^2 = 5.2^2 + 4.4^2 - 2(5.2)(4.4) \cos(88^{\circ}) .\]
Therefore
\[ b = \sqrt{5.2^2 + 4.4^2 - 2(5.2)(4.4) \cos(88^{\circ})} \approx \sqrt{44.8} \approx 6.69.\] 
\underline{Find $\gamma$} \\
By law of cosines,
\[ 5.2^2 = 4.4^2 + 6.69^2 - 2(4.4)(6.69)\cos(\gamma) .\]
Solving for $\cos(\gamma)$ yields
\[ \cos(\gamma) = \dfrac{5.2^2-4.4^2-6.69^2}{-2(4.4)(6.69)} \approx 0.6297. \]
Therefore
\[ \gamma = \cos^{-1} \left( 0.6297 \right) \approx 0.88 \text{ rad} = 50.97^{\circ}.\]
\underline{Find $\alpha$} \\
Using the fact that the sum of angles in a triangle is $180^{\circ}$, we see that
\[ 88^{\circ} + 50.97^{\circ} + \alpha = 180^{\circ}, \]
i.e.
\[ \alpha = 180^{\circ} - 88^{\circ} - 50.97^{\circ} = 41.03^{\circ}. \]
\newpage
\question[15] Solve the triangle or explain why it cannot be solved.
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (2,4) -- cycle;
\draw (3.4,0.3) node {$\beta$};
\draw (2,3.4) node {$\gamma$};
\draw (2,-0.5) node {$c$};
\draw (3.35,2.2) node {$15$};
\draw (0.6,2.2) node {$25$};
\draw (0.6,0.3) node {$80^{\circ}$};
\end{tikzpicture}
\end{center} 
\textit{Solution}: Since we know a side-angle pair we can use law of sines. By law of sines,
\[ \dfrac{\sin(80^{\circ})}{15} = \dfrac{\sin(\beta)}{25}.\]
Therefore
\[ \sin(\beta) = \dfrac{25\sin(80^{\circ})}{15} \approx 1.641.\]
We cannot find $\beta$ because to do so would require computing $\sin^{-1}(1.641)$, but this produces an error since $1.641$ is not in the domain of $\sin^{-1}$. Therefore this problem does not define a triangle.
\vfill
\end{questions}
\end{document}
