Zmurf,
This formula is easier remembered as

a² = b² + c² - 2bc cosθ
On simplification of the equation given by you, this is what is obtained.
a,b, and c are the three sides of a  triangle.
If the triangle is rightangled and θ is 90 degrees or pi/2 radians,
Cos θ = 0,
which gives us
a² = b² + c²,
the Pythogoras Theorem.

Can I have a go at simplifying?

Start      : a² = (b sin Θ)² + (c – b cos Θ)²
Expand  : a² = b² (sin Θ)² + c² – 2bc (cos Θ) + b² (cos Θ)²
Combine: a² = b² [(sin Θ)²+(cos Θ)²] + c² – 2bc (cos Θ)

Now (sin Θ)²+(cos Θ)² = 1 (a Pythagorean Identity), so:

Finally: a² = b² + c² – 2bc (cos Θ)

(Well done recognising that formula, ganesh)

Zmurf wrote:

Oh ok. But what formula would I use if i want to find the angle oposite to side 'b' if i'm only given the lengths of sides 'a' and 'b' and the angle between them?

With the help of the formula
c² = a² + b² - 2ab Cosθ,
you can find the value of c.
Now, use the formula
b² = a² + c² - 2ac Cosθ.
You know the value of a,b, and c.
You would get
Cosθ = x (some value)
The angle opposite to side 'b' would then be
θ = Cos-¹x.
Is that clear?

Sometimes, I really, really want to be older.

Well, just by expanding.

If you have (x-y)² you can do this:

(x-y)² = (x-y)(x-y) = x(x-y) - y(x-y) = (x²-xy) - (yx-y²) = x²-xy - yx+y² = x² - 2xy + y²     (follow each step carefully)

So, likewise I did:

(c – b cos Θ)² = (c – b cos Θ)(c – b cos Θ) = c(c – b cos Θ) - (b cos Θ)(c – b cos Θ)
... =  c² - bc (cos θ) - bc (cos θ) + b² (cos θ)² = c² - 2bc (cos θ) + b² (cos θ)²

I hope!

Zmurf wrote:

How did you get to the c² - 2bc (cos θ) + b² (cos θ)² ?

You can expand the terms as Mathsisfun said.
Or, you may remember the following formulae:-

1. (a+b)² = a² + 2ab + b²
2. (a-b)² = a² - 2ab + b²
3. (a+b)(a-b) = a² - b²
4. (a+b)³ = a³ + 3a²b+3ab²+ b³
5. (a-b)³ = a³ -3a²b +3ab² - b³
6. a³ + b³ = (a+b)(a² -ab +b²)
7. a³ - b³ = (a-b)(a² +ab + b²)