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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
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