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)

Sometimes, I really, really want to be older.

You are just on the edge of learning this, I think. For the moment don't be "overawed" by it. In a few years you will say "ah, obviously!"

It is just one or two levels further into "The Game".

Please ask questions yourself!

]]>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!

]]>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)

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

]]>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?

/\

/ \

c / \ b

/ \

B ------------- C

a

If you know two sides and the angle between them, then in this triangle that would be angle A and sides b and c.

You can't find angle B directly from this, but you can use a² = b² + c² 2bc (cos A) to find side a.

Knowing this, you can use sin A/a=sin B/b to find angle B.

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)

]]>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.

a^2 = (b sin Θ)^2 + (c b cos Θ)^2

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