Question

solve using C+IS method​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\rm{1+\dfrac{{x}^{2}\,cos(2\theta)}{2!}+\dfrac{{x}^{4}\,cos(4\theta)}{4!}+\cdots}$

$\displaystyle\rm{=\sum\limits^{\infty}_{k=0}\dfrac{{x}^{2k}\,cos(2k\theta)}{(2k)!}}$

We know,

$\boxed{\bf{cos(\theta)=\dfrac{{e}^{i\theta}+{e}^{-i\theta}}{2}}}$

$\displaystyle\rm{=\sum\limits^{\infty}_{k=0}\dfrac{{x}^{2k}\cdot \dfrac{{e}^{i\cdot2k\theta}+{e}^{-i\cdot2k\theta}}{2}}{(2k)!}}$

$\displaystyle\rm{=\dfrac{1}{2}\sum\limits^{\infty}_{k=0}\dfrac{{x}^{2k}\cdot \left({e}^{i\cdot2k\theta}+{e}^{-i\cdot2k\theta}\right)}{(2k)!}}$

$\displaystyle\rm{=\dfrac{1}{2}\sum\limits^{\infty}_{k=0}\dfrac{{x}^{2k}\,{e}^{i\cdot2k\theta}}{(2k)!}+\dfrac{1}{2}\sum\limits^{\infty}_{k=0}\dfrac{{x}^{2k}\,{e}^{-i\cdot2k\theta}}{(2k)!}}$

$\displaystyle\rm{=\dfrac{1}{2}\left\{1+\dfrac{{x}^{2}\,{e}^{i\cdot2\theta}}{2!}+\dfrac{{x}^{4}\,{e}^{i\cdot4\theta}}{4!}+\dfrac{{x}^{6}\,{e}^{i\cdot6\theta}}{6!}+\cdots\right\}+\dfrac{1}{2}\left\{1+\dfrac{{x}^{2}\,{e}^{-i\cdot2\theta}}{2!}+\dfrac{{x}^{4}\,{e}^{-i\cdot4\theta}}{4!}+\dfrac{{x}^{6}\,{e}^{-i\cdot6\theta}}{6!}+\cdots\right\}}$

$\displaystyle\rm{=\dfrac{1}{2}\left\{1+\dfrac{\left(x\,{e}^{i\theta}\right)^2}{2!}+\dfrac{\left(x\,{e}^{i\theta}\right)^4}{4!}+\cdots\right\}+\dfrac{1}{2}\left\{1+\dfrac{\left(x\,{e}^{-i\theta}\right)^2}{2!}+\dfrac{\left(x\,{e}^{-i\theta}\right)^4}{4!}+\cdots\right\}}$

We also know,

$\boxed{\bf{cosh(x)=1+\dfrac{{x}^{2}}{2!}+\dfrac{{x}^{4}}{4!}+\cdots}}$

So,

$\displaystyle\rm{=\dfrac{1}{2}\cdot\,cosh\left(x\,{e}^{i\theta}\right)+\dfrac{1}{2}\cdot\,cosh\left(x\,{e}^{-i\theta}\right)}$

$\displaystyle\rm{=\dfrac{1}{2}\left\{cosh\left(x\,{e}^{i\theta}\right)+cosh\left(x\,{e}^{-i\theta}\right)\right\}}$

We know,

$\boxed{\bf{cosh(x)=\dfrac{e^{x}+e^{-x}}{2}}}$

So,

$\displaystyle\rm{=\dfrac{1}{2}\left\{\dfrac{e^{\displaystyle\,x\,{e}^{i\theta}}+e^{\displaystyle\,-x\,{e}^{i\theta}}}{2}+\dfrac{e^{\displaystyle\,x\,{e}^{-i\theta}}+e^{\displaystyle\,-x\,{e}^{-i\theta}}}{2}\right\}}$

$\displaystyle\rm{=\dfrac{1}{4}\left\{e^{\displaystyle\,x\,{e}^{i\theta}}+e^{\displaystyle\,-x\,{e}^{i\theta}}+e^{\displaystyle\,x\,{e}^{-i\theta}}+e^{\displaystyle\,-x\,{e}^{-i\theta}}\right\}}$

$\displaystyle\rm{=\dfrac{1}{4}\bigg\{e^{\displaystyle\,x\left(cos(\theta)+i\,sin(\theta)\right)}+e^{\displaystyle\,-x\left(cos(\theta)+i\,sin(\theta)\right)}+e^{\displaystyle\,x\left(cos(\theta)-i\,sin(\theta)\right)}}\\\\\rm{+e^{\rm\displaystyle\,-x\left(cos(\theta)-i\,sin(\theta)\right)}\bigg\}}$

$\displaystyle\rm{=\dfrac{1}{4}\bigg\{e^{\displaystyle\,x\,cos(\theta)}\cdot\,e^{\displaystyle\,i(x\,sin(\theta))}+e^{\displaystyle\,-x\,cos(\theta)}\cdot\,e^{\displaystyle\,-i(x\,sin(\theta))}+e^{\displaystyle\,x\,cos(\theta)}\cdot\,e^{\displaystyle\,-i(x\,sin(\theta))}}\\\\\rm{+e^{\displaystyle\,-x\,cos(\theta)}\cdot\,e^{\displaystyle\,i(x\,sin(\theta))}\bigg\}}$

$\displaystyle\rm{=\dfrac{1}{4}\bigg\{e^{\displaystyle\,x\,cos(\theta)}\left\{e^{\displaystyle\,i(x\,sin(\theta))}+e^{\displaystyle\,-i(x\,sin(\theta))}\right\}+e^{\displaystyle\,-x\,cos(\theta)}\left\{e^{\displaystyle\,-i(x\,sin(\theta))}+e^{\displaystyle\,i(x\,sin(\theta))}\right\}\bigg\}}$$\displaystyle\rm{=\dfrac{1}{4}\bigg\{\left\{e^{\displaystyle\,x\,cos(\theta)}+e^{\displaystyle\,-x\,cos(\theta)}\right\}\left\{e^{\displaystyle\,-i(x\,sin(\theta))}+e^{\displaystyle\,i(x\,sin(\theta))}\right\}\bigg\}}$

$\displaystyle\rm{=\left\{\dfrac{e^{\displaystyle\,x\,cos(\theta)}+e^{\displaystyle\,-x\,cos(\theta)}}{2}\right\}\left\{\dfrac{e^{\displaystyle\,-i(x\,sin(\theta))}+e^{\displaystyle\,i(x\,sin(\theta))}}{2}\right\}}$

$\displaystyle\rm{=cosh\left(x\,cos(\theta)\right)\cdot cos\left(x\,sin(\theta)\right)}$