Answer
$sinx$*$sin2x$ $+cosx$$cos2x$ $=cosx$
Work Step by Step
We need to prove the identity $sinx$*$sin2x$ $+cosx$$cos2x$ $=cosx$
$sin2x=2sinxcosx$
and
$cos2x=cos^{2}x-sin^{2}x$
Thus,
$sinx$*$sin2x$ $+cosx$$cos2x$ $=sinx(2sinxcosx)+cosx(cos^{2}x-sin^{2}x)$
$sinx$*$sin2x$ $+cosx$$cos2x$ $=2sin^{2}x cosx+cos^{3}x-cosx\times sin^{2}x$
$sinx$*$sin2x$ $+cosx$$cos2x$ $=sin^{2}x cosx+cos^{3}x$
$sinx$*$sin2x$ $+cosx$$cos2x$ $=cosx (sin^{2}x +cos^{2}x)$
$sinx$*$sin2x$ $+cosx$$cos2x$ $=cosx (1)$
Hence, $sinx$*$sin2x$ $+cosx$$cos2x$ $=cosx$