Question Number 1838 by 112358 last updated on 11/Oct/15
![Show that (1) (1/(sinz))=(1/z)+Σ_(n=1) ^∞ (−1)^n ((1/(z+nπ))+(1/(z−nπ))) (2) cotz=(1/z)+Σ_(n=1) ^∞ ((1/(z+nπ))+(1/(z−nπ))) where z≠mπ, m∈Z , given the fact that cosax=((2sinaπ)/π)[(1/(2a))+Σ_(n=1) ^∞ ((((−1)^n acosnx)/(a^2 −n^2 )))] where a∉Z, −π≤x≤π.](Q1838.png)
$${Show}\:{that} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{sinz}}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{{z}+{n}\pi}+\frac{\mathrm{1}}{{z}−{n}\pi}\right) \\ $$$$\left(\mathrm{2}\right)\:{cotz}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{z}+{n}\pi}+\frac{\mathrm{1}}{{z}−{n}\pi}\right) \\ $$$${where}\:{z}\neq{m}\pi,\:{m}\in\mathbb{Z}\:,\:{given}\:{the}\:{fact}\:{that} \\ $$$${cosax}=\frac{\mathrm{2}{sina}\pi}{\pi}\left[\frac{\mathrm{1}}{\mathrm{2}{a}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} {acosnx}}{{a}^{\mathrm{2}} −{n}^{\mathrm{2}} }\right)\right] \\ $$$${where}\:{a}\notin\mathbb{Z},\:−\pi\leqslant{x}\leqslant\pi.\: \\ $$
Answered by 112358 last updated on 14/Oct/15
![We are given that cosax=((2sinaπ)/π)[(1/(2a))+Σ_(n=1) ^∞ ((((−1)^n acosnx)/(a^2 −n^2 )))] where a∉Z and x∈[−π,π]. (1) Let x=0. The equation then becomes cos(0)=((2sinaπ)/π)[(1/(2a))+Σ_(n=1) ^∞ ((((−1)^n acos(0))/(a^2 −n^2 )))] 1=((2sinaπ)/π)[(1/(2a))+Σ_(n=1) ^∞ ((((−1)^n a)/(a^2 −n^2 )))] z is not an integer multiple of π and a is not an integer so that we can write z=aπ since aπ appears under the sine function. ∴ a=(z/π). ⇒1=((2sinz)/π)[(π/(2z))+Σ_(n=1) ^∞ ((((−1)^n (z/π))/((z^2 /π^2 )−n^2 )))] ⇒(1/(sinz))=(2/π)[(π/(2z))+Σ_(n=1) ^∞ ((((−1)^n πz)/((z−πn)(z+πn))))] (1/(sinz))=(1/z)+Σ_(n=1) ^∞ ((((−1)^n 2z)/((z−πn)(z+πn)))) (1/(sinz))=(1/z)+Σ_(n=1) ^∞ (−1)^n (((z−πn+z+πn)/((z−πn)(z+πn)))) (1/(sinz))=(1/z)+Σ_(n=1) ^∞ (−1)^n ((1/(z+nπ))+(1/(z−nπ))) (2) If we let x=π and z=aπ we can obtain the second equation since (−1)^n cosnπ=(−1)^n (−1)^n =(−1)^(2n) =1.](Q1855.png)
$${We}\:{are}\:{given}\:{that} \\ $$$${cosax}=\frac{\mathrm{2}{sina}\pi}{\pi}\left[\frac{\mathrm{1}}{\mathrm{2}{a}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} {acosnx}}{{a}^{\mathrm{2}} −{n}^{\mathrm{2}} }\right)\right] \\ $$$${where}\:{a}\notin\mathbb{Z}\:{and}\:{x}\in\left[−\pi,\pi\right].\: \\ $$$$\left(\mathrm{1}\right)\:{Let}\:{x}=\mathrm{0}.\:{The}\:{equation}\:{then} \\ $$$${becomes}\: \\ $$$${cos}\left(\mathrm{0}\right)=\frac{\mathrm{2}{sina}\pi}{\pi}\left[\frac{\mathrm{1}}{\mathrm{2}{a}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} {acos}\left(\mathrm{0}\right)}{{a}^{\mathrm{2}} −{n}^{\mathrm{2}} }\right)\right] \\ $$$$\mathrm{1}=\frac{\mathrm{2}{sina}\pi}{\pi}\left[\frac{\mathrm{1}}{\mathrm{2}{a}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} {a}}{{a}^{\mathrm{2}} −{n}^{\mathrm{2}} }\right)\right] \\ $$$${z}\:{is}\:{not}\:{an}\:{integer}\:{multiple}\:{of}\:\pi \\ $$$${and}\:{a}\:{is}\:{not}\:{an}\:{integer}\:{so}\:{that} \\ $$$${we}\:{can}\:{write}\:{z}={a}\pi\:{since}\:{a}\pi\:{appears} \\ $$$${under}\:{the}\:{sine}\:{function}.\:\therefore\:{a}=\frac{{z}}{\pi}. \\ $$$$\Rightarrow\mathrm{1}=\frac{\mathrm{2}{sinz}}{\pi}\left[\frac{\pi}{\mathrm{2}{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} \frac{{z}}{\pi}}{\frac{{z}^{\mathrm{2}} }{\pi^{\mathrm{2}} }−{n}^{\mathrm{2}} }\right)\right] \\ $$$$\Rightarrow\frac{\mathrm{1}}{{sinz}}=\frac{\mathrm{2}}{\pi}\left[\frac{\pi}{\mathrm{2}{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} \pi{z}}{\left({z}−\pi{n}\right)\left({z}+\pi{n}\right)}\right)\right] \\ $$$$\frac{\mathrm{1}}{{sinz}}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} \mathrm{2}{z}}{\left({z}−\pi{n}\right)\left({z}+\pi{n}\right)}\right) \\ $$$$\frac{\mathrm{1}}{{sinz}}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{{z}−\pi{n}+{z}+\pi{n}}{\left({z}−\pi{n}\right)\left({z}+\pi{n}\right)}\right) \\ $$$$\frac{\mathrm{1}}{{sinz}}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{{z}+{n}\pi}+\frac{\mathrm{1}}{{z}−{n}\pi}\right) \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:{If}\:{we}\:{let}\:{x}=\pi\:{and}\:{z}={a}\pi\:{we}\:{can}\:{obtain} \\ $$$${the}\:{second}\:{equation}\:{since} \\ $$$$\left(−\mathrm{1}\right)^{{n}} {cosn}\pi=\left(−\mathrm{1}\right)^{{n}} \left(−\mathrm{1}\right)^{{n}} =\left(−\mathrm{1}\right)^{\mathrm{2}{n}} =\mathrm{1}. \\ $$$$ \\ $$