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#1 2005-11-13 22:35:22
- Danny8956
- Guest
A real tough trig question help will be appreciated
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
#2 2005-11-13 22:51:26
- Jai Ganesh
- Administrator
- Registered: 2005-06-28
- Posts: 46,230
Re: A real tough trig question help will be appreciated
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).
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
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#3 2005-11-14 04:49:27
- mathsyperson
- Moderator
- Registered: 2005-06-22
- Posts: 4,900
Re: A real tough trig question help will be appreciated
How very elegant and unexpected.
Why did the vector cross the road?
It wanted to be normal.
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#4 2005-11-14 05:52:00
- danny8976
- Guest
Re: A real tough trig question help will be appreciated
what do you mean? is it not right
#5 2005-11-14 05:56:39
- mathsyperson
- Moderator
- Registered: 2005-06-22
- Posts: 4,900
Re: A real tough trig question help will be appreciated
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.
Why did the vector cross the road?
It wanted to be normal.
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