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\newcommand\ds{\displaystyle}
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QUESTION

\begin{description}

\item[(a)]
Write $\cos(2x)-\cos(4x)$ as a product of trigonometric functions,
and hence deduce ALL solutions of the equation

$$\cos(2x)=\cos(4x).$$

\item[(b)]
Differentiate with respect to $x$ the following functions

${\rm (i)}\ \ds\frac{x}{\sqrt{1+x^2}},\hspace{1cm}{\rm (ii)}\
\exp(x^2\sinh x),\hspace{1cm}{\rm (iii)}\ \tan^3\{(1-x^2)^2\}.$

\end{description}

\bigskip

ANSWER
\begin{description}

\item[(a)]

\begin{eqnarray*}
\cos(2x)-\cos(4x)&=&-2\sin\left(\frac{2x+4x}{2}\right)\sin\left(\frac{2x-4x}{2}\right)\\
&=&-2\sin(3x)\sin(-x)\\ &=&2\sin(3x)\sin(x)\\
\cos(2x)=\cos(4x)&\Rightarrow&\cos(2x)-\cos(4x)=0\\
&\textrm{i.e.}&2\sin(3x)\sin(x)=0\\ \textrm{ therefore }\sin x=0
&\textrm{or}&\sin3x=0
\end{eqnarray*}

$\sin x=0\Rightarrow x=n\pi;$ i.e. $x=0,\pm\pi,\pm2\pi\ldots\\
\sin(3x)=0,\Rightarrow3x=n\pi,$ therefore $\ds x=\frac{n\pi}{3}$
where $n$ is any integer. \\Both conditions are satisfied by $\ds
x=\frac{n\pi}{3}$ where $n$ is any integer.

\item[(b)]

\begin{description}

\item[(i)]
\begin{eqnarray*}
\frac{d}{dx}\left\{\frac{x}{\sqrt{1+x^2}}\right\}&=&
\frac{\sqrt{(1+x^2)}.1-x\left\{\frac{1}{2}(1+x^2)^{-\frac{1}{2}}(2x)\right\}}{\left\{\sqrt{(1+x^2)}\right\}^2}\\
&=&\frac{\sqrt{(1+x^2)}-\frac{x^2}{\sqrt{(1+x^2)}}}{1+x^2}\\
&=&\frac{\frac{(1+x^2)-x^2}{\sqrt{(1+x^2)}}}{1+x^2}\\
&=&\frac{1}{(1+x^2)^\frac{3}{2}}
\end{eqnarray*}

\item[(ii)]
\begin{eqnarray*}
\frac{d}{dx}\left\{e^{x^2\sinh x}\right\}&=&e^{x^2\sinh
x}\left\{x^2\cosh x+2x\sinh x\right\}\\ &=&x(x\cosh x+2 \sinh
x)e^{x^2\sinh x}
\end{eqnarray*}

\item[(iii)]
\begin{eqnarray*}
&&\frac{d}{dx}\left\{\tan^3((1-x^2))\right\}\\&=&3\tan^2((1-x^2)^2)\sec^2((1-x^2)^2)2(1-x^2)(-2x)\\
&=&-12x(1-x^2)\tan^2((1-x^2)^2)\sec^2((1-x^2)^2) \end{eqnarray*}

\end{description}

\end{description}



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