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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

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,812

### 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 is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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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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