Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

• Index
•  » Help Me !
•  » A real tough trig question help will be appreciated

## Post a reply

Write your message and submit
|
Options

## Topic review (newest first)

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

## Board footer

Powered by FluxBB