How do you simplify $(-4cosxsinx+2cos2x)^2+(2cos2x+4sinxcosx)^2$ ?

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Answer 1 · 2016-11-20T17:06:18

$(-4cosxsinx+2cos2x)^2+(2cos2x+4cosxsinx)^2=8$

To solve $(-4cosxsinx+2cos2x)^2+(2cos2x+4cosxsinx)^2$ ,

Let us assume $2sin2x=4cosxsinx=a$ and $2cos2x=b$ , then above can be written as

$(b-a)^2+(b+a)^2$ (note $-2ab$ and $+2ab$ cancel out)

= $2b^2+2a^2$

= $2(2sin2x)^2+2(2cos2x)^2$

= $8sin^2 2x+8cos^2 2x$

= $8$

How do you simplify (-4cosxsinx+2cos2x)^2+(2cos2x+4sinxcosx)^2(-4cosxsinx+2cos2x)^2+(2cos2x+4sinxcosx)^2?