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

You are not logged in.

- Topics: Active | Unanswered

**Danny8956****Guest**

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

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,428

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).

Character is who you are when no one is looking.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

How very elegant and unexpected.

Why did the vector cross the road?

It wanted to be normal.

Offline

**danny8976****Guest**

what do you mean? is it not right

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

Offline