${t_r} = \frac{1}{{4{r^2} – 1}}$
$ = \frac{1}{{(2r – 1)(2r + 1)}} = \frac{1}{2}\left( {\frac{1}{{2r – 1}} – \frac{1}{{2r + 1}}} \right)$
${S_n} = \sum\limits_{r = 1}^n {\frac{1}{2}\left( {\frac{1}{{2r – 1}} – \frac{1}{{2r + 1}}} \right)} $
${S_n} = \frac{1}{2}\left[ {\left( {\frac{1}{1} – \frac{1}{3}} \right) + \left( {\frac{1}{3} – \frac{1}{5}} \right) + \left( {\frac{1}{5} – \frac{1}{7}} \right) + ……. + \left( {\frac{1}{{2n – 1}} – \frac{1}{{2n + 1}}} \right)} \right]$
${S_n} = \frac{1}{2}\left[ {1 – \frac{1}{{2n + 1}}} \right] = \frac{n}{{2n + 1}}$
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The given expression can be written as,
$\frac{1}{2}\left[ {(1 – \cos {2^\circ }) + (1 – \cos {4^\circ }) + ……………. + (1 – \cos {{360}^\circ })} \right]$
$ = \frac{1}{2}\left[ {(1 + 1 + …….. + 1) – (\cos {2^\circ } + \cos {4^\circ } + ……….. + \cos {{360}^\circ })} \right]$
There are 180 terms in the series.
$ = \frac{1}{2}\left[ {180 – \frac{{\cos \left( {{2^\circ } + \frac{{180 – 1}}{2} \times {2^\circ }} \right)\sin \left( {\frac{{180 \times {2^\circ }}}{2}} \right)}}{{\sin \left( {\frac{{{2^\circ }}}{2}} \right)}}} \right]$
$ = \frac{1}{2}\left( {180 – \frac{{\cos {{181}^\circ }\sin {{180}^\circ }}}{{\sin {1^\circ }}}} \right) = 90$
LHS = ${\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{\{ 1 + (\sin \theta – i\cos \theta )\} (\sin \theta + i\cos \theta )}}} \right]^n}$
$ = {\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{(\sin \theta + i\cos \theta ) + \{ {{\sin }^2}\theta – {{(i\cos \theta )}^2}\} }}} \right]^n}$
$ = {\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{(\sin \theta + i\cos \theta ) + ({{\sin }^2}\theta + {{\cos }^2}\theta )}}} \right]^n}$
$ = {\left[ {\frac{{(1 + \sin \theta + i\cos \theta )(\sin \theta + i\cos \theta )}}{{(1 + \sin \theta + i\cos \theta )}}} \right]^n}$
$ = {(\sin \theta + i\cos \theta )^n}$
$ = {\left\{ {\cos \left( {\frac{\pi }{2} – \theta } \right) + i\sin \left( {\frac{\pi }{2} – \theta } \right)} \right\}^n}$
$ = \cos n\left( {\frac{\pi }{2} – \theta } \right) + i\sin n\left( {\frac{\pi }{2} – \theta } \right)$
= RHS
We have, ${e^{ – 1}} = 1 + ( – 1) + \frac{{{{( – 1)}^2}}}{{2!}} + \frac{{{{( – 1)}^3}}}{{3!}} + \frac{{{{( – 1)}^4}}}{{4!}} + ……….$
The given integral = $\int {{{({e^{ – 1}})}^x}dx} = \int {{e^{ – x}}dx} = – {e^{ – x}} + C$
L.H.S.=$\cos \frac{\pi }{7} – \cos \frac{{2\pi }}{7} + \cos \frac{{3\pi }}{7}$
= $\cos \frac{\pi }{7} + \cos \frac{{3\pi }}{7} – \cos \left( {\pi – \frac{{5\pi }}{7}} \right)$
= $\cos \frac{\pi }{7} + \cos \frac{{3\pi }}{7} + \cos \frac{{5\pi }}{7}$
= $\cos \left( {\frac{\pi }{7} + \frac{{3 – 1}}{2}.\frac{{2\pi }}{7}} \right).\frac{{\sin \left( {3.\frac{{2\pi /7}}{2}} \right)}}{{\sin \left( {\frac{{2\pi /7}}{2}} \right)}}$ Continue reading Prove That,
$\cos \frac{\pi }{7}-\cos \frac{{2\pi }}{7}+\cos \frac{{3\pi}}{7}=\frac{1}{2}$ →
We have, $36{x^2} + 60x + 25 = \frac{{35}}{{3{x^2} + 5x + 2}}$
$ \Rightarrow 12(3{x^2} + 5x) + 25 = \frac{{35}}{{(3{x^2} + 5x) + 2}}$
Let, $3{x^2} + 5x = t$
So, $12t + 25 = \frac{{35}}{{t + 2}}$
$ \Rightarrow 12{t^2} + 49t + 15 = 0$ Continue reading Solve for $x \in \mathbb{R}$
${(6x + 5)^2} = \frac{{35}}{{3{x^2} + 5x + 2}}$ →
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