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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, 2017 at 22:27
  • $\begingroup$ how did you get it to 3sin(4x)? $\endgroup$
    – SandWu
    Sep 29, 2017 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, 2017 at 22:29
  • $\begingroup$ Or TrigReduce[]. $\endgroup$ Sep 30, 2017 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[%]. $\endgroup$ Sep 30, 2017 at 0:33

3 Answers 3

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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, 2017 at 23:15
  • $\begingroup$ @SandWu: Can't you simply add $\pi/2$ to the arguments of Cosines to get sines? $\endgroup$ Sep 30, 2017 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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