Question

Prove the following identity:
$1+co{s}^{2}2x=2(co{s}^{4}x+si{n}^{4}x)$

Answer 1

To prove : $1+co{s}^{2}2x=2(co{s}^{4}x+si{n}^{4}x)$
L.H.S. $=1+co{s}^{2}2x$
$=1+(co{s}^{2}x-si{n}^{2}x{)}^2$
$\{\because cos2x=co{s}^{2}x-si{n}^{2}x\}$

$=1+co{s}^{4}x+si{n}^{4}x-2co{s}^{2}xsi{n}^{2}x$
$=(co{s}^{2}x+si{n}^{2}x{)}^2+co{s}^{4}x+si{n}^{4}x$

$-2co{s}^{2}xsi{n}^{2}x[\because co{s}^{2}x+si{n}^{2}x=1]$

$=co{s}^{4}x+si{n}^{4}x+2co{s}^{2}xsi{n}^{2}x+co{s}^{4}x+si{n}^{4}x$ $-2co{s}^{2}xsi{n}^{2}x$

$=2(co{s}^{4}x+si{n}^{4}x)$

= R.H.S.

Hence, proved.