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Re: Re: Wrong Simplify[] Answer for

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104543] Re: [mg104410] Re: [mg104400] Wrong Simplify[] Answer for
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Tue, 3 Nov 2009 02:56:50 -0500 (EST)
  • References: <200910300719.CAA27787@smc.vnet.net>
  • Reply-to: drmajorbob at yahoo.com

Sorry... I made a total hash of the double-angle proof (which everyone  
knows anyway, I suppose).

Here's a correct one with Mathematica's help. (By hand is actually easier.)

p1 = ComplexExpand@Exp[x I];
p2 = p1 p1 // Expand;
p3 = ComplexExpand@Through[{Re, Im}@p2]

{Cos[x]^2 - Sin[x]^2, 2 Cos[x] Sin[x]}

d1 = ComplexExpand@Exp[2 x I];
d2 = Through[{Re, Im}@d1] // ComplexExpand

{Cos[2 x], Sin[2 x]}

Thread[d2 == p3]

{Cos[2 x] == Cos[x]^2 - Sin[x]^2, Sin[2 x] == 2 Cos[x] Sin[x]}

Bobby

On Sat, 31 Oct 2009 15:06:47 -0500, DrMajorBob <btreat1 at austin.rr.com>  
wrote:

> In other words,
>
> one = Cos[x]^4 - Sin[x]^4 // Factor
>
> (Cos[x] - Sin[x]) (Cos[x] + Sin[x]) (Cos[x]^2 + Sin[x]^2)
>
> two = MapAt[Simplify, one, {-1}]
>
> (Cos[x] - Sin[x]) (Cos[x] + Sin[x])
>
> three = two // Expand
>
> Cos[x]^2 - Sin[x]^2
>
> four = three // Simplify
>
> Cos[2 x]
>
> The last step is the double-angle formula, which is easily proven from  
> Euler's formula:
>
> Exp[x I] Exp[y I] // ComplexExpand
> % /. y -> x
> Exp[2 x I] // ComplexExpand
>
> Cos[x + y] + I Sin[x + y]
>
> Cos[2 x] + I Sin[2 x]
>
> Cos[2 x] + I Sin[2 x]
>
> (equating real and imaginary parts in the first and last result)
>
> Bobby
>
> On Sat, 31 Oct 2009 01:49:11 -0500, Pratip Chakraborty  
> <pratip.chakraborty at gmail.com> wrote:
>
>> Hi, Please remember the basic identity Cos[x]^2+Sin[x]^2=1 (* We  
>> multiply
>> both sides of the equation with (Cos[x]^2-Sin[x]^2) *)
>> =>(Cos[x]^2+Sin[x]^2)*(Cos[x]^2-Sin[x]^2)=1*(Cos[x]^2-Sin[x]^2) (*  
>> remember
>> (a+b)(a-b)=a^2-b^2 *) =>(Cos[x]^4-Sin[x]^4)=Cos[2x] Also for this type  
>> of
>> doubt one can take help of the Plot function in Mathematica.
>> Plot[Evaluate[{Cos[x]^4 - Sin[x]^4, Cos[2 x], Cos[x]^2 - Sin[x]^2}],  
>> {x, -2
>> Pi, 2 Pi}, PlotStyle -> {{Red}, {Blue, Dashed}, {Cyan}}] You will see  
>> all
>> the three functions that we are plotting will coincide. Hope this helps  
>> you.
>> Regards, Pratip
>>
>> On Fri, Oct 30, 2009 at 8:19 AM, Lawrence Teo <lawrenceteo at yahoo.com>  
>> wrote:
>>
>>> We know that Simplify[Cos[x]^2-Sin[x]^2] -> Cos[2 x]
>>> But why Simplify[Cos[x]^4-Sin[x]^4] -> Cos[2 x] too?
>>>
>>> Doing subtraction between the two expressions will give small delta.
>>> This is enough to prove that the two expression shouldn't be the same.
>>>
>>> Can anyone give me any insight? Thanks.
>>>
>>>
>>
>
>


-- 
DrMajorBob at yahoo.com


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