You can
put this solution on YOUR website!Hi
First thing to do is to tidy up the thing we're trying to prove. Reason being, we're going to have to a bit of messy algebra, and the less we have to do, the better. I'm going to add sin^2(x) to both sides. This leaves the thing we have to prove at:
 + \sin^2(x)=1)
Well, from here, all we do is substitute the definitions of sine and cosine in, then see what happens. To make it a bit simpler, I'm going to let
Using this, it's easy to see that
.
So using my new definitions,
 = \frac{p-q}{2j})
and
 =\frac{p+q}{2})
Substitute these into the thing we are trying to prove, and you should get:
^2}{4} + \frac{(p-q)^2}{-4} = 1)
Multiply the whole lot by 4(we don't like denominators when we don't need them)
^2 - (p-q)^2 = 4)
That just leaves exapnding the brackets, and tidying up
 = 4)
(Remember pq=1)
 + (q^2-q^2) + (2pq - -2pq) = 4)
,
, so
. Which is always true, so what we started with must be true - proved.
Now try a different one yourself, see if only using your definitions of sine and cosine you can prove that
=2\sin(x)\cos(x))
Kev
(