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This is a very straight-forward question:

I'm trying to simplify [sin(2pi*t +pi/4) + sin(2pi*t -pi/4)], and failing at it:

[sin(2pi*t +pi/4) + sin(2pi*t -pi/4)]
[2sin(2pi*t)cos(pi/2)] by sum->product

sqrt(2)[sin(2pi*t +pi/4) + sin(2pi*t -pi/4)]
sqrt(2)[2sin(2pi*t)cos(pi/4)] by backwards product->sum
-20t (It took me waaay to long to realize the redundancy of this 2nd method... -_-)

Wolfram alpha gives sqrt(2)sin(2pi*t), but gives no explanation as to how. What am I doing wrong?

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closed as off topic by Szabolcs, ssch, Verbeia Feb 27 '13 at 5:30

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Does this question relate to the stand-alone product Mathematica in any way? –  Mr.Wizard Feb 27 '13 at 4:15

1 Answer 1

Is this what you're trying to do:

enter image description here


It's possible, as I was overly focused on the slightly rambling free-form input stylized as Mathematica code, that I neglected to read the actual point of the question - namely, I guess that you're expecting an explanation from WolframAlpha as to why the identity is true. I think that the short answer is that trig expansion is just not one of the things that WolframAlpha has a "Show steps" button for like, say, integrating functions has such a button.

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