Question 90171: I've been given this problem and I keep getting stuck half was down it and going back on myself, wondered i f anyone could help out with it.
Given that
cos x = (e^jx + e^-jx)/2 and
sin x = (e^jx - e^-jx)/2j
and using only this information, prove that
cos^2 x - sin^2 x = 1 - 2sin^2 x
Answer by kev82(151)   (Show Source):
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
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