Re: [math-fun] Trig identity
Interesting! Thanks! I don't see any "trigcontract" in Maxima (as opposed to Macsyma). I guess this feature never made it into Maxima. At 06:42 PM 1/21/2018, Bill Gosper wrote:
Macsyma: (c1) TRIGCONTRACT(SIN(X+2*THETA)-SIN(X));
(d1) 2 sin(theta) cos(x + theta)
(c2) BLOCK([FANCY_DISPLAY : FALSE],PLAYBACK([1,2]));
Mathematica: In[1]:= TrigFactor[Sin[X + 2*THETA] - Sin[X]]
Out[1]= 2 Cos[THETA + X] Sin[THETA]
On 2018-01-21 15:18, Henry Baker wrote:
I can use Maxima to show that the rhs simplifies to the lhs, but I can't go the other way (except by hand).
Any ideas how to force Maxima to derive the rhs from the lhs?
Can Mathematica do this?
(This is a generalization of sin(2*theta)=2*sin(theta)*cos(theta).)
(%i1) sin(x+2*theta)-sin(x) = 2*sin(theta)*cos(x+theta); (%o1) sin(x + 2 theta) - sin(x) = 2 sin(theta) cos(x + theta) (%i2) %,trigexpand,expand; 2 2 (%o2) - sin (theta) sin(x) + cos (theta) sin(x) - sin(x) + 2 cos(theta) sin(theta) cos(x) = 2 cos(theta) sin(theta) cos(x) 2 - 2 sin (theta) sin(x) (%i3) %,trigreduce,expand; (%o3) sin(x + 2 theta) - sin(x) = sin(x + 2 theta) - sin(x)
--rwg
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Henry Baker