$f(x) = 8 \cos^4 x + 6 \sin (2x + 3 \pi/4) \sin(2x - \pi/4)$.
How can I simplify into a linear combination of simple sine functions?
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3 Answers
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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]
I'm assuming it's primarily about formatting the output in terms of sines.
answered Sep 30, 2017 at 0:49
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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) + .. $\endgroup$– SandWuCommented Sep 29, 2017 at 23:15
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$\begingroup$ @SandWu: Can't you simply add $\pi/2$ to the arguments of Cosines to get sines? $\endgroup$ Commented Sep 30, 2017 at 1:21
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$4 \cos (2 t) + \cos (4 t) + 3 \sin (4 t) = 4 \sin (2 t + \pi/2) + \sin (4 t + \pi/2) + 3 \sin (4 t)$
answered Sep 30, 2017 at 1:23
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? $\endgroup$TrigReduce[]. $\endgroup$Sin[y + Pi/2]toCos[y], as you can see withHold[Sin[y + Pi/2]]ReleaseHold[%]. $\endgroup$