Convert this function using trig identities into sine functions - Mathematica Stack Exchange most recent 30 from mathematica.stackexchange.com 2019-09-23T10:35:10Z https://mathematica.stackexchange.com/feeds/question/156838 https://creativecommons.org/licenses/by-sa/4.0/rdf https://mathematica.stackexchange.com/q/156838 1 Convert this function using trig identities into sine functions SandWu https://mathematica.stackexchange.com/users/52563 2017-09-29T22:15:45Z 2017-09-30T01:23:35Z <p>$f(x) = 8 \cos^4 x + 6 \sin (2x + 3 \pi/4) \sin(2x - \pi/4)$.</p> <p>How can I simplify into a linear combination of simple sine functions?</p> https://mathematica.stackexchange.com/questions/156838/-/156842#156842 0 Answer by Chris Chiasson for Convert this function using trig identities into sine functions Chris Chiasson https://mathematica.stackexchange.com/users/14489 2017-09-29T23:13:52Z 2017-09-29T23:13:52Z <p>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.</p> <pre><code>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]*) </code></pre> https://mathematica.stackexchange.com/questions/156838/-/156844#156844 4 Answer by Michael E2 for Convert this function using trig identities into sine functions Michael E2 https://mathematica.stackexchange.com/users/4999 2017-09-30T00:49:03Z 2017-09-30T00:49:03Z <p>Maybe this?:</p> <pre><code>FourierTrigSeries[8 Cos[x]^4 + 6 Sin[2 x + 3 Pi/4] Sin[2 x - Pi/4], x, 4] /. Cos[t_] :&gt; HoldForm[Sin][Pi/2 - t] </code></pre> <p><img src="https://i.stack.imgur.com/KE41u.png" alt="Mathematica graphics"></p> <p>I'm assuming it's primarily about formatting the output in terms of sines.</p> https://mathematica.stackexchange.com/questions/156838/-/156846#156846 -1 Answer by David G. Stork for Convert this function using trig identities into sine functions David G. Stork https://mathematica.stackexchange.com/users/9735 2017-09-30T01:23:35Z 2017-09-30T01:23:35Z <p>$4 \cos (2 t) + \cos (4 t) + 3 \sin (4 t) = 4 \sin (2 t + \pi/2) + \sin (4 t + \pi/2) + 3 \sin (4 t)$</p>