Question 90171
Hi<br />

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:<br />

*[tex \cos^2(x) + \sin^2(x)=1]<br />

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 *[tex p=e^{jx}\;q=e^{-jx}] Using this, it's easy to see that *[tex pq=1].<br />

So using my new definitions, *[tex \sin(x) = \frac{p-q}{2j}] and *[tex \cos(x) =\frac{p+q}{2}]<br />

Substitute these into the thing we are trying to prove, and you should get:<br />

*[tex \frac{(p+q)^2}{4} + \frac{(p-q)^2}{-4} = 1]<br />

Multiply the whole lot by 4(we don't like denominators when we don't need them)<br />

*[tex (p+q)^2 - (p-q)^2 = 4]<br />

That just leaves exapnding the brackets, and tidying up<br />

*[tex p^2 + 2pq + q^2 - (p^2 -2pq +q^2) = 4]<br />

(Remember pq=1)<br />

*[tex (p^2-p^2) + (q^2-q^2) + (2pq - -2pq) = 4], *[tex 4pq=4], so *[tex 4=4]. Which is always true, so what we started with must be true - proved.<br />

Now try a different one yourself, see if only using your definitions of sine and cosine you can prove that *[tex \sin(2x)=2\sin(x)\cos(x)]<br />

Kev