Question

prove the following question:-


$\sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8 \ \alpha } } } = 2cos \alpha$
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Answer 1

ÊLLØ'.....!!

Given:-

$\sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8 \ \alpha } } } = 2cos \alpha$

$lhs = \sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8 \alpha } } }$

$= \sqrt{2 + \sqrt{2 + \sqrt{2(1 + cos8 \alpha )} } }$

$= \sqrt{2 + \sqrt{2 + \sqrt{2 \times 2 {cos}^{2} 4 \alpha } } }$

$= \sqrt{2 + \sqrt{2 + 2cos4 \alpha } }$

$= \sqrt{2 + \sqrt{2(1 + cos4 \alpha )} }$

$= \sqrt{2 + \sqrt{2 \times 2 {cos}^{2} \alpha } }$

$= \sqrt{2 + 2cos2 \alpha }$

$= \sqrt{2(1 + cos2 \alpha )}$

$= \sqrt{2 \times 2 {cos}^{2} \alpha }$

$= 2cos \alpha$

$= rhs$

THÅÑKẞ.....!!

Answer 2

Step-by-step explanation:

Given:

$\sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8 \ \alpha } } } = 2cos \alpha$

$LHS = \sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8 \alpha } } }$

$= \sqrt{2 + \sqrt{2 + \sqrt{2(1 + cos8 \alpha )} } }$

$= \sqrt{2 + \sqrt{2 + \sqrt{2 \times 2 {cos}^{2} 4 \alpha } } }$

$= \sqrt{2 + \sqrt{2 + 2cos4 \alpha } }$

$= \sqrt{2 + \sqrt{2(1 + cos4 \alpha )} }$

$= \sqrt{2 + \sqrt{2 \times 2 {cos}^{2} \alpha } }$

$= \sqrt{2 + 2cos2 \alpha }$

$= \sqrt{2(1 + cos2 \alpha )}$

$= \sqrt{2 \times 2 {cos}^{2} \alpha }$

=2cosα

= RHS