Trig question

Hint: 2sin(x)cos(x)=sin(2x)

In my book (metaphorically speaking) "double angle formulae" could mean you're just doing "sin(a+b)" rather than sin(2a).

They may want you to do the question another way but you can do it by substituting an expression for a in terms of b and c (a, b, c are angles of a triangle...) then using addition/subtraction formulae to get the answer.

The way to do it using "double angle formulae" strictly is to derive the "half angle formulae" i.e. in for example: sin(2x) = 2sin(x)cos(x), substitute x = y/2 and the same for cos(2x). You can derive them or look them up. That would be another way of getting the answer.

The first method above is much faster though

Is $\cos \frac{B - C}{2} = \cos \frac{B}{2} \cos \frac{c}{2} + \sin \frac{B}{2} \sin \frac{C}{2}$?

Or can I not use the regular double angle formula for cos if we half them?

Is $\cos \frac{B - C}{2} = \cos \frac{B}{2} \cos \frac{c}{2} + \sin \frac{B}{2} \sin \frac{C}{2}$?

Or can I not use the regular double angle formula for cos if we half them?

Your equation for cosine subtraction formula is correct. You can use this for cos(x+y) whatever x and y happen to be... Now if you eliminate A and use sin subtraction formula you'll see the answer drop out.

Ok and $\sin \frac{A}{2} = \sqrt\frac{1-\cos A}{2}$, so now there is a square root in our answer?

Ok and $\sin \frac{A}{2} = \sqrt\frac{1-\cos A}{2}$, so now there is a square root in our answer?

See above: I gave the hint that A, B and C are angles of a triangle. So try to eliminate A since it doesn't appear on the right hand side of the equation...

Once you have an expression for A in terms of B and C your answer will fall out

If A, B and C are angles of a triangle, prove that:

$\cos(\frac{B-C}{2})-\sin(\frac{A}{2}) = 2\sin(\frac{B}{2})\sin(\frac{C}{2})$

Literally no idea where to start, it says this comes under the "double angle formulae" but all the stuff is being divided by 2