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Re: Trigonometric simplification
*To*: mathgroup at smc.vnet.net
*Subject*: [mg68921] Re: Trigonometric simplification
*From*: carlos at colorado.edu
*Date*: Wed, 23 Aug 2006 07:15:53 -0400 (EDT)
*References*: <ecbnnc$r29$1@smc.vnet.net><ecek3f$qpu$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Here is a correction to my second post (Thanks to Paul Abbott for
noticing the factor of 2). The reason I need to get 2 + 4*Cos(a)^3
is to match a published solution in a homework solutions manual.
=====================================================
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)
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
2 + 4*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.
======================================================
Paul Abbott's solution, sent directly by email, is elaborate:
f[d_] := 10*Plus @@ Cases[{d}, Cos[n_ x_] | Sin[n_ x_] -> n, Infinity]
+LeafCount[d]
FullSimplify[d, ComplexityFunction -> f]
4*Cos[a]^3 + 2
Is there a simpler way?
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