Trigsimp should do it.

>>> trigsimp(sin(x)*cos(x))
sin(2*x)/2

On 11 July 2015 at 09:11, Pavel Grinfeld <[email protected]> wrote:

> That's almost it! Thank you.
> Could I have the answer as a trig polynomial, meaning sin(2a) instead of
> sin(a)cos(a), etc.
>
> Thanks again.
>
> On Fri, Jul 10, 2015 at 11:36 PM, Sartaj Singh <[email protected]>
> wrote:
>
>> You should do this.
>>
>> >>> ((2+2*I)*exp(I*x) + (2-2*I)*exp(-I*x)).rewrite(sin).expand(sin,trig=
>> True)
>> -4*sin(x) +4*cos(x)
>>
>>
>>
>> On Saturday, 11 July 2015 08:50:00 UTC+5:30, PG wrote:
>>>
>>> I do not believe that is the case:
>>>
>>> >>> expand((2+3*I)*exp(2*I) + (2-3*I)*exp(-2*I), sin, trig=True)
>>> 2*exp(-2*I) + 3*I*exp(2*I) - 3*I*exp(-2*I) + 2*exp(2*I)
>>>
>>> On Fri, Jul 10, 2015 at 6:51 PM, Aaron Meurer <[email protected]> wrote:
>>>
>>>> If you use expand(trig=True) on the expression it will expand the
>>>> trigonometric functions.
>>>>
>>>> Aaron Meurer
>>>>
>>>> On Fri, Jul 10, 2015 at 4:02 PM, Pavel Grinfeld <[email protected]>
>>>> wrote:
>>>> > Yes, you are right, I think I mistyped the original question. So let
>>>> me try
>>>> > to save it from here: is there a way to get the answer as a linear
>>>> > combination of sin and cos?
>>>> >
>>>> >
>>>> >
>>>> > On Fri, Jul 10, 2015 at 3:50 PM, Gaurav Dhingra <[email protected]>
>>>> wrote:
>>>> >>
>>>> >> Since the real part is zero in this case. Hence only imaginary part
>>>> is
>>>> >> there, which I guess is given correctly. In the answer.
>>>> >> Isn't it ?
>>>> >>
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-- 
Regards
*Sartaj Singh*

*Mathematics and Computing*,
Indian Institute of Technology,
Varanasi - 221 005 INDIA

E-mail: [email protected], *[email protected]
<[email protected]>*

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