Re: Re: Trigonometric simplification

*To*: mathgroup at smc.vnet.net*Subject*: [mg68917] Re: [mg68881] Re: Trigonometric simplification*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Wed, 23 Aug 2006 07:15:43 -0400 (EDT)*References*: <ecbnnc$r29$1@smc.vnet.net><ecc2pn$ajl$1@smc.vnet.net> <200608220920.FAA26920@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

carlos at colorado.edu wrote: >>Hi Carlos, >> >>Using TrigReduce before Simplify will do it: >> >>r = Tan[a]^2/(Sec[a]^2)^(3/2); >>Simplify[TrigReduce[r], Assumptions -> {a > 0, Sec[a] > 0}] >> >>--> Cos[a]*Sin[a]^2 >> >>Best regards, >>Jean-Marc > > > Thanks, that works perfectly. Actually Sec[a]>0 as assumption > is sufficient. This is correct from the problem source, since > the angle is in the range (-Pi/2,Pi/2) One wonders why that important bit of information was not in the original message. Or how you passed it to Simplify: In[8]:= Simplify[Tan[a]^2/(Sec[a]^2)^(3/2), Assumptions->-Pi/2<a<Pi/2] // InputForm Out[8]//InputForm= Cos[a]*Sin[a]^2 > Here is a related question. How can I get Mathematica to pass from > > d = 2 + 3*Cos[a] + Cos[3*a] (* leaf count 10 *) > > to > > 1 + 2*Cos[a]^3 (* leaf count 8 *) > > TrigExpand[d] gives > > 2 + 3*Cos[a] + Cos[a]^3 - 3*Cos[a]*Sin[a]^2 > > Applying Simplify to that yields 2 + 3*Cos[a] + Cos[3*a] so we are > back to the beggining. Thankfully, you can't. In[1]:= d = 2 + 3*Cos[a] + Cos[3*a]; In[2]:= d2 = 1 + 2*Cos[a]^3; In[3]:= Developer`ZeroQ[d-d2] Out[3]= False In[4]:= (d-d2) /. a->Pi/3 // N Out[4]= 1.25 Offhand I do not know if there is a correct equivalent form that is simpler than the original. Daniel Lichtblau Wolfram Research

**References**:**Re: Trigonometric simplification***From:*carlos@colorado.edu