(-4 cos x sin x + 2cos 2x)^2 + (2 cos 2x + 4 sin x cos x)^2

jonboy

Full Member
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Jun 8, 2006
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547
The answer to this problem is 8.

I've tried factoring it out and got 20sin2(2x)8sin2xcos2x+8cos2(2x)\displaystyle 20sin^2(2x) -8sin2xcos2x + 8cos^2(2x)

But that doesn't really help me. Can someone give me a hint as to what to do to get the answer?

I would show all my work but it is way too long to type out.
 
jonboy said:
The answer to this problem is 8.

I've tried factoring it out and got 20sin2(2x)8sin2xcos2x+8cos2(2x)\displaystyle 20sin^2(2x) -8sin2xcos2x + 8cos^2(2x) <<< That does not turn into 8 - what is the original problem?

But that doesn't really help me. Can someone give me a hint as to what to do to get the answer?

I would show all my work but it is way too long to type out.
 
Subhotosh Khan said:
jonboy said:
The answer to this problem is 8.

I've tried factoring it out and got 20sin2(2x)8sin2xcos2x+8cos2(2x)\displaystyle 20sin^2(2x) -8sin2xcos2x + 8cos^2(2x) <<< That does not turn into 8

However the original problem does.

Show us what did you do to get 20sin2(2x)8sin2xcos2x+8cos2(2x)\displaystyle 20sin^2(2x) -8sin2xcos2x + 8cos^2(2x)



But that doesn't really help me. Can someone give me a hint as to what to do to get the answer?

I would show all my work but it is way too long to type out.
 
8 is correct.
Expand both binomials and combine. You will then have
32sin2xcos2x+8cos22x\displaystyle 32\sin^2x\cos^2x + 8\cos^22x

Change the 32sin2(2x)\displaystyle 32\sin^2(2x) to 8(2sinxcosx)2\displaystyle 8(2\sin x \cos x)^2

Substitute 2sinxcosx=sin2x\displaystyle 2\sin x \cos x = \sin 2x

From there you can arrive at

8(sin22x+cos22x)\displaystyle 8(\sin^22x + \cos^22x)

which puts you home free.
 
Thank you Subhotosh Khan, and especially Loren! That was an articulate explanation. :)
 
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