Using Trigonometric Identities - Problem 1
We're about to prove a trigonometric identity, but first let's review the Pythagorean identity; cosine squared theta plus sine squared theta equals 1. This identity has two other useful forms. One of them is cosine squared theta equals 1 minus sine squared theta and the other is sine squared theta equals 1 minus cosine squared theta so one of these may come up in the proof we are about to do.
Let's take a look. Prove the identity oops I forgot theta cosine theta over 1 minus sine theta equals 1 plus sine theta over cosine theta and I will write proof here. When you're proving an identity, my feeling is that the best way to do is to pick one side and work on it until you get the other side, work on it algebraically, but don't work on both sides simultaneously some teachers won't like that form of proof.
So let's start with the left side, cosine theta over 1 minus sine theta. Now the first thing I want to do is kind of a trick and if you've studied complex numbers, you will remember this trick as multiplying by the conjugate. See I want to get this into the form of a difference of squares and the way to do that is to multiply by 1 plus sine theta, but I can only do that if I also multiply the top by 1 plus sine theta.
Now I'm thinking ahead here the reason I want to do this is I'll get a 1 minus sine squared theta and I can use one of my identities. So I get cosine theta and I don't want to be in too much of a hurry to distribute this, I'm going to hold off on that over 1 minus sine squared theta.
Now 1 minus sine squared theta in fact by one of our identities is cosine squared theta, so on top I have cosine theta 1 plus sine theta over cosine squared theta and it's good that I didn't distribute this through because these guys cancel I lose one of cosines an I just have one cosine remaining in the denominator and so my final answer not really answer, but I arrive at the right hand side and I'm actually done this is what I wanted to show.
I wanted to show through Algebra and manipulation that cosine theta over 1 minus theta equal to 1 plus sine theta over cosine theta and now that's the end of the proof I can put a little box I'm done.