How do you simplify ${(- 4 cos x sin x + 2 cos 2 x)}^{2} + {(2 cos 2 x + 4 sin x cos x)}^{2}$ $(-4cosxsinx+2cos2x)^2+(2cos2x+4sinxcosx)^2$ ?

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Answer 1 · 2016-11-20T17:06:18

${(- 4 cos x sin x + 2 cos 2 x)}^{2} + {(2 cos 2 x + 4 cos x sin x)}^{2} = 8$ $(-4cosxsinx+2cos2x)^2+(2cos2x+4cosxsinx)^2=8$

To solve ${(- 4 cos x sin x + 2 cos 2 x)}^{2} + {(2 cos 2 x + 4 cos x sin x)}^{2}$ $(-4cosxsinx+2cos2x)^2+(2cos2x+4cosxsinx)^2$ ,

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

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

= $2 b^{2} + 2 a^{2}$ $2b^2+2a^2$

= $2 {(2 sin 2 x)}^{2} + 2 {(2 cos 2 x)}^{2}$ $2(2sin2x)^2+2(2cos2x)^2$

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

= $8$ $8$

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