Question

given 1 + cos alpha + { cos ^2 } alpha +..... Infinity = 2 - √2'then ( 0 < alpha < pi) is =? ​

Answer 1

${ \bf{ \underline{ \blue{ \underline{ \blue{ Given}}}} : - }}$

$\\ \rm \implies 1 + \cos( \alpha ) + { \cos}^{2} ( \alpha ) + ...... \infty = 2 - \sqrt{2}$

${ \bf{ \underline{ \blue{ \underline{ \blue{ To-Find}}}} : - }}$

$\\ \to \rm \alpha(0 < \alpha < \pi) = ?$

${ \bf{ \underline{ \blue{ \underline{ \blue{ Solution }}}} : - }}$

Given that,

$\\ \rm \implies 1 + \cos( \alpha ) + { \cos}^{2} ( \alpha ) + ...... \infty = 2 - \sqrt{2}$

We know that, sum of infinite G.P. series is –

$\\ \dashrightarrow \: { \boxed{ \rm S_{n} = \dfrac{a}{1 - r}}}$

Where,

$\\ \rm \:\:\: { \huge{.}} \:\:\: a = 1$

$\\ \rm \:\:\: { \huge{.}} \:\:\: r = \cos( \alpha )$

Again,

$\\ \rm \implies \dfrac{1}{1 - \cos( \alpha ) } = 2 - \sqrt{2}$

$\\ \rm \implies 1 - \cos( \alpha ) = \dfrac{1}{2 - \sqrt{2}}$

( Rationalization )

$\\ \rm \implies 1 - \cos( \alpha ) = \dfrac{1}{2 - \sqrt{2}} \times \dfrac{2 + \sqrt{2} }{2 + \sqrt{2} }$

$\\ \rm \implies 1 - \cos( \alpha ) = \dfrac{2 + \sqrt{2} }{(2 - \sqrt{2})(2 + \sqrt{2})}$

$\\ \rm \implies 1 - \cos( \alpha ) = \dfrac{2 + \sqrt{2} }{(2)^{2} - (\sqrt{2})^{2}}$

$\\ \rm \implies 1 - \cos( \alpha ) = \dfrac{2 + \sqrt{2} }{4 - 2}$

$\\ \rm \implies 1 - \cos( \alpha ) = \dfrac{2 + \sqrt{2} }{2}$

$\\ \rm \implies 1 - \cos( \alpha ) = 1 + \dfrac{1}{ \sqrt{2} }$

$\\ \rm \implies \cos( \alpha ) = - \dfrac{1}{ \sqrt{2} }$

$\\ \rm \implies \cos( \alpha ) = \cos \left( \dfrac{\pi}{2} + \dfrac{\pi}{4} \right)$

$\\ \rm \implies \cos( \alpha ) = \cos \left( \dfrac{3\pi}{4} \right)$

$\\ \rm \implies \alpha = \dfrac{3\pi}{4} \in \left(0 < \alpha < \pi \right)$

$\small{\underline{\sf{\blue{Hence-}}}}$

It’s the required answer.

Answer 2

Step-by-step explanation:

Using infinite gp series

a/1-r

1/1-cos alpha = 2-2√2

(2-2√2)(1-cos alpha)= 1

2-2cos alpha -2√2 +2√2 cos alpha = 1

-2cos alpha +2√2 cos alpha = 1-2+2√2

-2cos alpha (1-√2) = 1-2+2√2

cos alpha = 1-2+2√2 / -2(1-√2)

Therefore solving this you get alpha= 3 pi/4