Summer 2017 – Q.4

Q.4 a) Obtain tan5θ in terms of tanθ and show that $latex 1-10\;\tan^2\frac\pi{10}+5\;\tan^4\frac\pi{10}=0$ [6M]

Get the value of tan5θ in terms of tanθ

T.P.T =$latex 1-10\tan^2\frac\pi{10}+5\tan^4\frac\pi{10}=0$

We know

$latex \tan n\theta=\frac{{}^nc_1\tan\theta-{}^nc_3\tan^3\theta+{}^nc_5\tan^5\theta…..}{1-{}^nc_2\tan^2\theta+{}^nc_4+\tan^4\theta-{}^nc_6\tan^6\theta…….}$

$latex \tan5\theta=\frac{{}^5c_1\tan\theta-{}^5c_3\tan^3\theta+{}^5c_5\tan^5\theta}{1-{}^5c_2\tan^2\theta+{}^5c_4\tan^4\theta.}$

$latex {}^nc_r=\frac{\left.n\right|}{r\left|\left(n-r\right)\right|}$

$latex \tan5\theta=\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\;\tan^2\theta+5\tan^4\theta}$

Put $latex \theta=\frac\pi{10},$ we get

$latex \infty=\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\;\tan^2\frac\pi{10}+5\tan^4\frac\pi{10}}$

Since,

$latex \begin{array}{l}\tan\;5\theta=\tan\;90^\circ\\\\\;\;\;\;\;\;\;\;\;\;\;=\infty\end{array}$

We can say the dinominator well be 0 so, we get

$latex 1-10\tan^2\frac\pi{10}+5\tan\frac\pi{10}=0$

Hence Proved.

b) If $latex y=e^{\tan^{-1}x},$ prove that [6M]

If $latex y=e^{\tan^{-1}x},$

T.P.T.-$latex \left(1+x^2\right)y_{n+2}+\left[2\left(n+1\right)x-1\right]y_{n+2}+n\left(n+1\right)y_n=0$

$latex y=e^{\tan^{-1}x}$

Differentiate both sides with respect to x

$latex \begin{array}{l}y_1=\frac{e^{\tan^{-1}x}}{1+x^2}\\\\\;\;\;\;=\frac y{1+x^2}\\\\\left(1+x^2\right)y_1=y\end{array}$

Differentiate again with respect to x

$latex \left(1+x^2\right)y_2+2xy_1=y_1—–\left(i\right)$

Differentiate (iii) by Leibnitz’s theorem,

$latex \begin{array}{l}\left(1+x^2\right)y_{n+2}+{}^nc_1\left(2x\right)y_{n+1}+{}^nc_2\left(2\right)y_n+{}^nc_1\left(2\right)y_n=y_{n+1}\\\\\left(1+x^2\right)y_{n+2}+2nxy_{n+1}+\frac{\left(n-1\right)n}2y_n+2xy_{n+1}+2ny_n=y_{n+1}\\\\\left(1+x^2\right)y_{n+2}+2\left(n+1\right)xy_{n+1}-y_{n+1}+\left[n\left(n-1\right)+2n\right]y_n=0\\\\\left(1+x^2\right)y_{n+2}+\left[2\left(n+1\right)x-1\right]y_{n+1}+n\left(n+1\right)y_n=0\\\\\end{array}$

Hence proved.

c) i. Express $latex \left(2x^3+3x^2-8x+7\right)$ in terms of (x – 2) using Taylor’s theorem. [4M]

ii. Prove that $latex \tan^{-1}x=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+—–\;$ [4M]

i. $latex \left(2x^3+3x^2-8x+7\right)$

By Taylors theorem,

$latex f\left(x\right)=f\left(a\right)+\left(x-a\right)f’\left(a\right)+\frac{\left(x-a\right)^2}{2!}f”\left(a\right)\frac{f\left(x-a\right)^3}{3!}f”’\left(a\right)+….$

put, a=2

$latex f\left(x\right)=f\left(2\right)+\left(x-2\right)f’\left(2\right)+\frac{\left(x-2\right)^2}{2!}f”\left(2\right)\frac{f\left(x-2\right)^3}{3!}f”’\left(2\right)+….\left(i\right)$

$latex \begin{array}{l}f\left(x\right)=2x^3+3x^2-8x+7\\\\f\left(2\right)=16+12+7-16+2\\\\\;\;\;\;\;\;=19\end{array}$

$latex \begin{array}{l}f’\left(x\right)=6x^2+6x-8\\\\f’\left(2\right)=24+12-8\\\\\;\;\;\;\;\;=28\end{array}$

$latex \begin{array}{l}f”\left(x\right)=12x+6\\\\f”\left(2\right)=24+6\\\\\;\;\;\;\;\;=30\end{array}$

$latex \begin{array}{l}f”’\left(x\right)=12\\\\f”\left(2\right)=12\\\\\;\;\;\;\;\;=12\end{array}$

Substituting the values in equation (i) we get

$latex \begin{array}{l}f\left(x\right)=19+\left(x-2\right)28+\frac{\left(x-2\right)^2\times30}{2!}+\frac{\left(x-2\right)^3\times12}{3!}\\\\f\left(x\right)=19+28\left(x-2\right)+15\left(x-2\right)^2+2\left(x-2\right)^3\end{array}$

ii. T.P.T

$latex \tan^{-1}x=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+…..$

Let

$latex y=\tan^{-1}x$

Differentiate with respect to x

$latex \begin{array}{l}\frac{dy}{dx}=\frac1{1+x^2}\\\\=\left(1+x^2\right)^{-1}\\\\=1-x^2+x^4-x^6+x^8…..\end{array}$

(Binomial Theorem)

Integrating both sides 1/w 0 and x

$latex \begin{array}{l}y=\left[\tan^{-1}x\right]_0^x\\\\\;\;\;=\left[x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+…..\right]_0^x\\\\\tan^{-1}x=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+…..\end{array}$

Hence proved.