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$f(x) = 8 \cos^4 x + 6 \sin (2x + 3 \pi/4) \sin(2x - \pi/4)$.

How can I simplify into a linear combination of simple sine functions?

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  • $\begingroup$ FullSimplify[8 Cos[x]^4 + 6 Sin[2 x + 3 Pi/4] Sin[2 x - Pi/4]] returns a simple expression (4 Cos[2 x] + Cos[4 x] + 3 Sin[4 x]), but it's a combination of $\sin$ and $\cos$. Would that work? $\endgroup$ – MarcoB Sep 29 '17 at 22:27
  • $\begingroup$ how did you get it to 3sin(4x)? $\endgroup$ – SandWu Sep 29 '17 at 22:28
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    $\begingroup$ Is this a question about the computing software Mathematica or is it just a Mathematics question? $\endgroup$ – march Sep 29 '17 at 22:29
  • $\begingroup$ Or TrigReduce[]. $\endgroup$ – David G. Stork Sep 30 '17 at 0:20
  • $\begingroup$ The problem seems to be that Mathematica automatically simplifies Sin[y + Pi/2] to Cos[y], as you can see with Hold[Sin[y + Pi/2]] ReleaseHold[%]. $\endgroup$ – aardvark2012 Sep 30 '17 at 0:33
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Maybe this?:

FourierTrigSeries[8 Cos[x]^4 + 6 Sin[2 x + 3 Pi/4] Sin[2 x - Pi/4], x, 4] /.
 Cos[t_] :> HoldForm[Sin][Pi/2 - t]

Mathematica graphics

I'm assuming it's primarily about formatting the output in terms of sines.

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I figured that if we take the Fourier transform of it to get the discrete frequencies and then invert the transform, that the expression would be simpler. It comes back in exponential form, and the leading coefficient needs to be distributed. From there, you can use Euler's equation to transform it back to trig. The remaining required transforms are trivial. I have no idea if this is how you were supposed to solve it.

FourierTransform[8 Cos[x]^4 + 6Sin[2x + 3Pi/4]Sin[2x - Pi/4],x,w];
InverseFourierTransform[%,w,t];
Distribute@%;
ExpToTrig@%
(*4 Cos[2 t]+Cos[4 t]+3 Sin[4 t]*)
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  • $\begingroup$ is it possible to change 4 cos (2t) + cos (4t) into sine function? ie : A1Sin(2pix + theta) + A2sin(2pix + theta) + .. $\endgroup$ – SandWu Sep 29 '17 at 23:15
  • $\begingroup$ @SandWu: Can't you simply add $\pi/2$ to the arguments of Cosines to get sines? $\endgroup$ – David G. Stork Sep 30 '17 at 1:21
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$4 \cos (2 t) + \cos (4 t) + 3 \sin (4 t) = 4 \sin (2 t + \pi/2) + \sin (4 t + \pi/2) + 3 \sin (4 t)$

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