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#