\documentclass[a4paper,12pt]{article}
\newcommand\ds{\displaystyle}
\begin{document}

\parindent=0pt

QUESTION
\begin{description}

\item[(a)]
Obtain $\sin(2x)\cos(4x)$ as a sum of trigonometric functions, and
hence evaluate

$$\int_0^\frac{\pi}{2} x\sin(2x)\cos(4x)\,dx.$$

\item[(b)]

\begin{description}

\item[(i)]
Show that

$$\frac{d}{dx}(\tanh x)=\frac{1}{\cosh^2x}.$$

\item[(ii)]
Given that $x=2$ is an approximate solution of the equation
$2\tanh x=x$ use the Newton Raphson formula TWICE, and the result
in part (i), to obtain a better approximation (correct to four
decimal places).

\end{description}

\end{description}



ANSWER

\begin{description}

\item[(a)]
$\sin(2x)\cos(4x)$

From formula sheet, $\sin A+\sin
B=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$\\ Therefore we require
$\frac{A+B}{2}=2x,\ \frac{A-B}{2}=4x$ i.e. \\
$\left.\begin{array}{c}A+B=4x\\A-B=8x\end{array}\right\}$\ add:\
$2A=12x,A=6x\Rightarrow B=-2x.$\\ Therefore \\ $\sin2x\cos
4x=\frac{1}{2}\{\sin(6x)+\sin(-2x)\}=\frac{1}{2}\{\sin(6x)-\sin(2x)\}$\\
Hence
\begin{eqnarray*}
&&\int_0^\frac{\pi}{2}x\sin2x\cos4x\,dx\\
&=&\int_0^\frac{\pi}{2}\frac{x}{2}\{\sin6x-\sin2x\}\,dx\\
&=&\frac{1}{2}\left\{\left[x\left(-\frac{\cos6x}{6}+\frac{\cos2x}{2}\right)\right]_0^\frac{\pi}{2}-\int_0^\frac{\pi}{2}\left(-\frac{\cos6x}{6}+\frac{\cos2x}{2}\right)\,dx\right\}\\
&=&\frac{1}{2}\left\{\left[\frac{\pi}{2}\left(-\frac{\cos3\pi}{6}+\frac{\cos\pi}{2}\right)-0\right]+\left[\frac{\sin6x}{36}-\frac{\sin2x}{4}\right]_0^\frac{\pi}{2}\right\}\\
&=&\frac{1}{2}\left\{\frac{\pi}{2}\left(\frac{1}{6}-\frac{1}{2}\right)+\frac{\sin
3\pi}{36}-\frac{\sin\pi}{4}-0\right\}\\
&=&\frac{\pi}{4}\left(-\frac{2}{6}\right)=-\frac{\pi}{12}
\end{eqnarray*}


\item[(b)]

\begin{description}

\item[(i)]
Now $\ds u=\tanh x=\frac{\sinh x}{\cosh x},\
\frac{du}{dx}=\frac{\cosh x\cosh x-\sinh x\sinh x}{(\cosh
x)^2}=\frac{1}{\cosh^2x}$

\item[(ii)]
$f(x)=2\tanh x-x,\ f'(x)=2\textrm{sech}^2x-1$\\ given $\ds
x_0=2,\\
x_1=2-\frac{f(2)}{f'(2)}=2-\frac{(-0.071945)}{(-0.85870)}=1.91622\\
x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1.91622-\frac{(-0.00101)}{(-0.83401)}=1.91501$

Second approximation $\Rightarrow x=1.9150$ (correct to 4 decimal
places.)

\end{description}

\end{description}




\end{document}
