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 <asme...@gmail.com 
> <javascript:>> 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 <pgei...@gmail.com 
>> <javascript:>> 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 <axyd...@gmail.com 
>> <javascript:>> 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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