Question

Prove that:

$1+co{s}^{2}2\theta =2(co{s}^4\theta +si{n}^4\theta )$

Answer 1

LHS, $1+co{s}^{2}2\theta$

$=1+(co{s}^2\theta -si{n}^2\theta {)}^2[\because cos2\theta =co{s}^2\theta -si{n}^2\theta ]$

$=1+co{s}^4\theta +si{n}^2\theta -2si{n}^2\theta .co{s}^2\theta =(si{n}^2\theta +co{s}^2\theta {)}^2+co{s}^4\theta +si{n}^4\theta -2si{n}^2\theta .co{s}^2\theta [\because si{n}^2\theta +co{s}^2\theta =1]=si{n}^4\theta +co{s}^4\theta +2si{n}^2\theta co{s}^2\theta +co{s}^4\theta +si{n}^4\theta -2si{n}^2\theta .co{s}^2\theta =2(co{s}^4\theta +si{n}^4\theta )$

= RHS