Help needed with TRIG EXPANSIONS (1 Viewer)

[blackops23]

Hi guys, really need help on this one:

Here's the link to the question:

http://img864.imageshack.us/i/57553476lch.jpg


Q. (a) By considering expansions of sin(X+Y) - sin(X-Y)

Prove that sinA - sin B = 2cos[(A+B)/2]sin[(A-B)/2]


(b) Also given that cos A - cos B = 2sin[(A+B)/2]sin[(B-A)/2]

prove: (sin A - sin B)/(cos A - cos B) = -cot[(A+B)/2

Really appreciate the help, thanks guys
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[Drongoski]

That's easy.

They all come from a rehashing of the 4 well-known trig identities:

sin(x+y) = sinxcosy + cosxsiny . . . . . . . (1)

sin(x-y) = sinxcosy - cosxsiny . . . . . . . . (2)

cos(x+y) = cosxcosy - sinxsiny . . . . . . . . (3)

cos(x-y) = cosxcosy + sinxsiny . . . . . . . . (4)

By (1) - (2) we get:

sin(x+y) - sin(x-y) = 2cosxsiny . . . . . . . . (5)

Now if we call A = x+y and B = x-y we get x = (A+B)/2 and y = (A-B)/2

Rewriting (5):

sin A - sin B = 2 cos(A+B)/2 * sin(A-B)/2

You can now do the 2nd one the same way.

(I've done above virtually off my head - but then I've been doing such things for years)