Question

If 1 + cos alpha + {cos ^2}alpha + .........infty = 2 - sqrt 2 , then (0 < alpha < pi ) is

Answer 1

GIVEN :–

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

TO FIND :–

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

SOLUTION :–

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

• This is an infinite G.P. series and sum of infinite G.P. series is –

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

• Here –

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

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

• So that –

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

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

• Now rationalization –

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

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

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

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

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

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

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

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

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

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

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

Answer 2

Step-by-step explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

• $\sf1 + \cos \alpha + { \cos }^{2} \alpha + .... + \infty = 2 - \sqrt{2}$

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

• $\alpha(0 < \alpha < \pi)$

${\green{\underline{\underline{\bold{Solution:-}}}}}$

$\sf1 + \cos \alpha + { \cos }^{2} \alpha + .... + \infty = 2 - \sqrt{2}$

The given series is in G.P ( Geometric progession)

• a = 1

• $\sf r = \dfrac{cos \alpha}{1} = cos \alpha$

• $\sf n = \infty$

In G.P sum of n terms in given by:-

$\boxed{ \rm\pink{ \rightarrow S_{n} = \dfrac{a}{1 - r} }}$

Substituting the values:-

$\sf \rightarrow S_{ \infty } = \dfrac{1}{1 - \cos \alpha}$

$\sf \rightarrow \dfrac{1}{1 - \cos \alpha} = 2 - \sqrt{2}$

$\sf \rightarrow {1 - \cos \alpha} = \dfrac{1}{2 - \sqrt{2} }$

$\sf \rightarrow { \cos \alpha} = 1 - \dfrac{1}{2 - \sqrt{2} }$

$\sf \rightarrow { \cos \alpha} = \dfrac{2 - \sqrt{2 } - 1 } {2 - \sqrt{2} }$

$\sf \rightarrow { \cos \alpha} = \dfrac{1- \sqrt{2 } } {2 - \sqrt{2} }$

By rationalising of denominator:-

$\sf \rightarrow { \cos \alpha} = \dfrac{1- \sqrt{2 } } {2 - \sqrt{2} } \times \dfrac{2 + \sqrt{2} }{2 + \sqrt{2} }$

$\sf \rightarrow { \cos \alpha} = \dfrac{2 + \sqrt{2 } - 2 \sqrt{2} - 2} { {2}^{2} - {( \sqrt{2}) }^{2} }$

$\sf \rightarrow { \cos \alpha} = \dfrac{ - \sqrt{2} } { 4- 2 }$

$\sf \rightarrow { \cos \alpha} = \dfrac{ - \sqrt{2} } { 2 }$

$\sf \rightarrow { \cos \alpha} = \dfrac{ - \sqrt{2} } { \sqrt{2} \times \sqrt{2} }$

$\sf \rightarrow { \cos \alpha} = \dfrac{ - 1 } { \sqrt{2} }$

$\sf \rightarrow { \cos \alpha} = \cos \bigg( \dfrac{3\pi}{4} \bigg)$

$\sf \rightarrow { \alpha} = { \cos}^{ - 1} \bigg \{ \cos \bigg( \dfrac{3\pi}{4} \bigg) \bigg \}$

$\sf \rightarrow { \alpha} = \dfrac{3\pi}{4}$

$\boxed {\sf \purple{ \rightarrow \alpha(0 < \alpha < \pi) = \frac{3\pi}{4} }}$