 \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 3 - SPRING 2018 \\
SOLUTION
\end{center}
\begin{flushleft}\vspace{0.5in}
12 April 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 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[6] $\cot(x)\sin(x)=\cos(x)$ \\
\textit{Solution}: Start with the left:
$$\cot(x)\sin(x) = \dfrac{\cos(x)}{\cancel{\sin(x)}} \cancel{\sin(x)} = \cos(x).$$
\part[9] $\sin(x)-\sin^3(x)=\sin(x)\cos^2(x)$ \\ 
\textit{Solution}: Start with the left:
$$\sin(x)-\sin^3(x) = \sin(x) \left( \underbrace{1-\sin^2(x)}_{=\cos^2(x)} \right)= \sin(x)\cos^2(x).$$
%\part[7] $\cos(2\theta) = \dfrac{1-\tan^2(\theta)}{1+\tan^2(\theta)}$
\part[9] $\dfrac{\sin(x+h)-\sin(x)}{h} = \sin(x) \left( \dfrac{\cos(h)-1}{h} \right) + \cos(x) \left( \dfrac{\sin(h)}{h} \right)$ \\
\textit{Solution}: Recall by the sum identity for sine, 
$$\sin(x+h)=\sin(x)\cos(h)+\cos(x)\sin(h).$$
Using that identity and starting on the left, we get
$$\begin{array}{ll}
\dfrac{\sin(x+h)-\sin(x)}{h} &= \dfrac{ \Big[ \sin(x)\cos(h)+\cos(x)\sin(h) \Big] - \sin(x)}{h} \\
&= \dfrac{\Big[ \sin(x)\cos(h) - \sin(x) \Big] + \cos(x)\sin(h)}{h} \\
&= \dfrac{\sin(x)\cos(h) - \sin(x)}{h} + \dfrac{\cos(x)\sin(h)}{h} \\
&= \sin(x) \left( \dfrac{\cos(h)-1}{h} \right) + \cos(x) \left( \dfrac{\sin(h)}{h} \right).
\end{array}$$
\end{parts}

\question[16] Find the exact value.
\begin{parts}
\part[8] $\sin \left( \dfrac{7\pi}{12} \right)$ \\
\textit{Solution}: Note that $\dfrac{7\pi}{12}=\dfrac{\pi}{4}+\dfrac{\pi}{3}$. Therefore using the sum identity for sine,
$$\begin{array}{ll}
\sin \left( \dfrac{7\pi}{12} \right) &= \sin \left( \dfrac{\pi}{4} + \dfrac{\pi}{3} \right) \\
&=\sin \left( \dfrac{\pi}{4} \right) \cos \left( \dfrac{\pi}{3} \right) + \cos \left( \dfrac{\pi}{4} \right) \sin \left( \dfrac{\pi}{3} \right) \\
&\stackrel{\text{unit circle}}{=}  \dfrac{\sqrt{2}}{2} \left( \dfrac{1}{2} \right) + \dfrac{\sqrt{2}}{2} \left( \dfrac{\sqrt{3}}{2} \right).
\end{array}$$
\part[8] $\sin \left( \dfrac{\pi}{8} \right)$ \\ 
\textit{Solution}: First note that $\dfrac{\pi}{8}$ is in quadrant $1$, so the sine is positive there. Now note that we may write $\dfrac{\pi}{8} = \dfrac{\frac{\pi}{4}}{2}$, so we use the half-angle identity for sine with $\alpha=\dfrac{\pi}{4}$ to arrive at
$$\begin{array}{ll}
\sin \left( \dfrac{\pi}{8} \right) &= \sin \left( \dfrac{\frac{\pi}{4}}{2} \right) \\
&= + \sqrt{ \dfrac{1-\cos(\frac{\pi}{4})}{2} } \\
&= \sqrt{ \dfrac{1-\frac{\sqrt{2}}{2}}{2} } \\
\end{array}$$
\vfill
\end{parts}

\newpage
\question[16] 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}
Now find
\begin{parts}
\part[8] $\cos \left(\dfrac{\alpha}{2} \right)$ \\
\textit{Solution}: First find side length $a$ by using the Pythagorean theroem: $a^2+2^2=7^2$, so $a=\sqrt{45}$.  Thus $\cos(\alpha) = \dfrac{\sqrt{45}}{7}$. Note that $\alpha$ is in quadrant I and so is $\dfrac{\alpha}{2}$, so the cosine will be positive here. Using the half-angle identity for cosine, we write
$$\cos \left( \dfrac{\alpha}{2} \right) = + \sqrt{ \dfrac{1+\cos(\alpha)}{2} } = \sqrt{\dfrac{1+\frac{\sqrt{45}}{7}}{2}} $$
\part[8] $\tan(2\theta)$ \\ 
\textit{Solution}: Note that $\sin(\theta)=\dfrac{\sqrt{45}}{7}$ and $\cos(\theta) = \dfrac{2}{7}$. Now using the double angle identitities for sine and cosine, we arrive at
$$\tan \left( 2 \theta \right) = \dfrac{\sin(2\theta)}{\cos(2\theta)}=\dfrac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta)-\sin^2(\theta)}= \dfrac{2\frac{\sqrt{45}}{7} \frac{2}{7}}{\left( \frac{2}{7} \right)^2 - \left( \frac{\sqrt{45}}{7} \right)^2}$$
(note: since there are 3 different ways to write the double angle identity for cosine, it was ok to use any of those other ones as well!)
\end{parts}

\question[16] If $\tan(x)=-\dfrac{1}{5}$ and $x$ is in quadrant II, find
\begin{parts}
\part[8] $\sin(2x)$ \\
\textit{Solution}: From the statement that $\tan(x) = \dfrac{1}{5}$, we 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 {$5$};
\draw (0.6,0.2) node {$x$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}
\end{center}
Find the hypotenuse (labelled $?$) by using the Pythagorean theorem: $5^2+1^2=?^2$ and hence $?=\sqrt{26}$. Also note that if $x$ is in Quadrant II, then $2x$ is in either quadrant III or quadrant IV. In either case, $\sin(2x)$ is negative. 

Therefore use the double angle identity to compute
$$\sin(2x) = 2\sin(x)\cos(x) = -2 \left( \dfrac{1}{\sqrt{26}} \right) \left( \dfrac{5}{\sqrt{26}} \right)$$
\part[8] $\cos \left( \dfrac{x}{2} \right)$ \\ 
\textit{Solution}: Using the same triangle in (a) and the fact that if $x$ is in Quadrant II, then $\dfrac{x}{2}$ is in Quadrant I and hence the cosine is positive there. So, use the half-angle idnentity to compute
$$\cos \left( \dfrac{x}{2} \right) = + \sqrt{\dfrac{1+\cos(x)}{2}} = \sqrt{\dfrac{1+\frac{5}{\sqrt{26}}}{2}}.$$ 
\end{parts}

\question[12] Find the exact value of 
$$\cos \left( \sin^{-1} \left( \dfrac{1}{5} \right) + \tan^{-1} \left( -\dfrac{2}{5} \right) \right).$$
\textit{Solution}: First use the sum identity for cosine to get
\begin{multline*}
(*) \hspace{35pt} \cos \left( \sin^{-1} \left(\dfrac{1}{5} \right) + \tan^{-1} \left(-\dfrac{2}{5} \right) \right) \\
= \cos \left( \sin^{-1} \left( \dfrac{1}{5} \right)\right) \cos \left( \tan^{-1} \left( - \dfrac{2}{5} \right) \right) - \sin \left( \sin^{-1} \left( \dfrac{1}{5} \right) \right) \sin \left( \tan^{-1} \left( -\dfrac{2}{5} \right) \right) 
\end{multline*}
Now we must determine how to handle each of the factors in each term. Firstly, it is easy to see from how inverse functions work that
$$\sin \left( \sin^{-1} \left( \dfrac{1}{5} \right) \right) = \dfrac{1}{5}.$$
Now, let $\theta=\sin^{-1} \left( \dfrac{1}{5} \right)$ (and hence $\theta$ must lie in Quadrant I or Quadrant IV). Then $\sin(\theta)=\dfrac{1}{5}$ (and hence $\theta$ must lie in Quadrant I or Quadrant II). Therefore $\theta$ lies in Quadrant I. Draw a triangle for this situation:
\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 {$5$};
\draw (2,-0.5) node {$?$};
\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 see that $?^2+1^2=5^2$ and hence the unknown side is $?=\sqrt{24}$. From this we may conclude that
$$\cos \left( \sin^{-1} \left( \dfrac{1}{5} \right) \right) = \cos(\theta) = + \dfrac{\sqrt{24}}{5}$$
For the other two, let $\psi=\tan^{-1} \left( - \dfrac{2}{5} \right)$ (hence $\psi$ lies in Quadrant I or Quadrant IV). Then $\tan(\psi)=-\dfrac{2}{5}$ (hence $\psi$ lies in Quadrant II or Quadrant IV). Therefore $\psi$ lies in Quadrant IV. Draw a triangle for this situation:
\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 {?`};
\draw (2,-0.5) node {$5$};
\draw (0.6,0.2) node {$\psi$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}
\end{center}
To find the missing side, use the Pythagorean theorem: $5^2+2^2=\text{?`}^2$ and conclude $\text{?`}=\sqrt{29}$. From this we may conclude that
$$\cos \left( \tan^{-1} \left( - \dfrac{2}{5} \right) \right) = \cos(\psi) = +\dfrac{5}{\sqrt{29}},$$
and
$$\sin \left( \tan^{-1} \left( - \dfrac{2}{5} \right) \right) = \sin(\psi) = - \dfrac{2}{\sqrt{29}}.$$
Therefore we may finally complete the problem by going back to $(*)$:
\[ \cos \left( \sin^{-1} \left(\dfrac{1}{5} \right) + \tan^{-1} \left(-\dfrac{2}{5} \right) \right) = \left( \dfrac{\sqrt{24}}{5} \right) \left( \dfrac{5}{\sqrt{29}} \right) - \left( \dfrac{1}{5} \right) \left( - \dfrac{2}{\sqrt{29}} \right).  \]

\question[16] The Chebyshev polynomials (of the first kind), $T_n(x)$, are defined by
$$T_n(x)=\cos(n\cos^{-1}(x)).$$
\begin{parts}
\part[5] Compute $T_1(x)$.\\
\textit{Solution}: Substituting $n=1$ into the definition of $T_n(x)$ yields 
$$T_1(x)=\cos(\cos^{-1}(x))=x.$$
\part[11] Compute $T_2(x)$. \\
\textit{Solution}: Substituting $n=2$ into the definition of $T_n(x)$ yields
$$T_2(x)=\cos(2\cos^{-1}(x)).$$
Using the double angle identity for cosine (any of them will be ok, but I will only use one of them here), we get
$$\begin{array}{ll}
(*) \hspace{35pt} T_2(x) &= \cos(2\cos^{-1}(x)) \\
&= \cos^2(\cos^{-1}(x)) - \sin^2(\cos^{-1}(x)) \\
&= \Big( \cos(\cos^{-1}(x)) \Big)^2 - \Big( \sin(\cos^{-1}(x)) \Big)^2 \\
&= x^2 - \Big( \sin(\cos^{-1}(x)) \Big)^2
\end{array}$$
To complete the problem, we must find $\sin(\cos^{-1}(x))$. To do this, let $\theta=\cos^{-1}(x)$, so that $\cos(\theta)=x$. 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 {$1$};
\draw (2,-0.5) node {$x$};
\draw (0.6,0.2) node {$\theta$};
\draw (3.8,0) -- (3.8,0.2) -- (4,0.2);
\end{tikzpicture}
\end{center}
To find the unknown side, use the Pythagorean theorem: $x^2+?^2=1^2$, hence $?=\sqrt{1-x^2}$. Now we can finish the calculation above in $(*)$:
$$T_2(x) = x^2- \left( \sin (\cos^{-1}(x)) \right)^2 = x^2- \left( \sqrt{1-x^2} \right)^2=2x^2-1.$$
\end{parts}
\end{questions}
\end{document} 