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- mathsyperson
- 2005-11-15 04:56:39
No, it's right. It's just that I wouldn't have thought that such a complicated expression could be simplified into such a simple one like that.
- danny8976
- 2005-11-15 04:52:00
what do you mean? is it not right
- mathsyperson
- 2005-11-15 03:49:27
How very elegant and unexpected.
- ganesh
- 2005-11-14 21:51:26
Sinx*Sin2x + Cosx*Cos2x
= Sinx[Sin(x+x)] + Cosx[(Cos(x+x)]
= Sinx[2SinxCosx]+ Cosx[Cos²x - Sin²x]
Rearranging the terms,
= Cosx[2Sin²x + Cos²x - Sin²x]
= Cosx[Sin²x + Cos²x]
=Cosx (since Sin²x + Cos²x=1).
- Danny8956
- 2005-11-14 21:35:22
Hi,
How would I solve this question:
express sinx sin(2x) + cosx cos(2x) in terms of cosx
I am really stuck in this.
Danny