# Wolfram Alpha's Mysterious Trig Abilities [closed]

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
-20t

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?

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

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

Questions on Mathematica Stack Exchange are expected to relate to Mathematica within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

Is this what you're trying to do:

Edit

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.

-