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

[jonboy]

The answer to this problem is 8.

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

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.

 

[Deleted member 4993]

The answer to this problem is 8.

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

 

[Deleted member 4993]

The answer to this problem is 8.

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

However the original problem does.

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



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.

 

[Loren]

8 is correct.
Expand both binomials and combine. You will then have
\(\displaystyle 32\sin^2x\cos^2x + 8\cos^22x\)

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

Substitute \(\displaystyle 2\sin x \cos x = \sin 2x\)

From there you can arrive at

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

which puts you home free.

 

[jonboy]

Thank you Subhotosh Khan, and especially Loren! That was an articulate explanation.