[PD] get sinusoid from a sine and a cosine oscillator
Charles Z Henry
czhenry at gmail.com
Thu Jan 30 18:45:17 CET 2014
On Thu, Jan 30, 2014 at 10:58 AM, Alexandros Drymonitis <adrcki at gmail.com>wrote:
>
> On Thu, Jan 30, 2014 at 6:36 PM, Charles Z Henry <czhenry at gmail.com>wrote:
>
>> If you want to use a contribution from both of your axes, you can just
>> sum them together. (x+y)*sqrt(2)/2 is just a projection along the line
>> x-y=0
>>
>
Let me correct myself: the line is x+y=0
> Can't really try it right now, but just to be sure, the last equation is
> to be interpreted like this: (x+y)*(sqrt(2)/2) or like this:
> ((x+y)*sqrt(2))/2?
>
> It's the same either way. That's only needed if you want to be exact in
the projection value.
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