Using Trigonometric Identities - Problem 3

I want to talk about one more Pythagorean identity and it comes from the original Pythagorean identity cosine squared plus sine squared equals 1. To get it, I just divide both sides of this equation by sine squared theta, and I get cotangent squared theta plus 1 equals co-secant squared theta and we'll have a chance to use this in the next proof.

The problem says prove the identity; cotangent squared minus tangent squared over cotangent plus tangent the quantity squared equals 1 minus 2 sine squared theta and the significance of this expression will come up later when we study the double angle identity, so this is the double angle identity for cosine.

Anyway when you're proving a trig identity it's good to start on one side and do algebraic manipulation and use all the trig identities to get it to the other side, so that's what I'm going to do, I'll start with the lest hand side.

Now this numerator is the difference of squares so I'm just going to factor that and I get a cotangent theta minus tangent theta times cotangent theta plus tangent theta and that's what I have in the denominator cotangent theta plus tangent theta squared so I'll get some nice cancellation and let me do that. Let me cancel one of these with this and I have cotangent theta minus tangent theta over cotangent theta plus tangent theta.

Now where do we go from here? Well the trick's obvious, but if I multiply through by cotangent theta over cotangent theta, remember that cotangent is the reciprocal of tangent, so when I multiply these two I'm going to get one and of course when I multiply cotangent and cotangent I get cotangent squared so I'm going to be able to use my new identity. Cotangent squared theta minus 1 and cotangent squared theta plus 1.

Now cotangent squared theta plus 1 is cosecant squared theta so let me make that substitution. Cotangent squared theta minus 1 over cosecant squared theta and at this point I want to use the trick of switching to sines and cosines.

This equals cosine squared theta over sine squared theta minus 1 over and this is 1 over squared theta. Now I've got a complex fraction that doesn't actually look that simple, but I could easily simplify it by using the trick of fraction bursting which is multiplying by the least common denominator of the smaller fractions these little fractions here and here and that's sine squared so I multiply the top by sine squared and the bottom by sine squared and just remember to distribute this sine squared over both terms.

Now on top I'm going to get cancellation, so I'll end up with cosine squared on top. Cosine squared theta minus I have a minus sign here so minus sine squared theta, sine squared and on the bottom I just got 1, well that's nice. One more thing I need to end up with 1 minus 2 sine squared theta Pythagorean identity, I want to use the Pythagorean identity on this and it's a form of the original Pythagorean identity we had up here; cosine squared theta equals 1 minus sine squared theta so let me make that substitution.

Cosine squared becomes 1 minus sine squared and I have another minus sine squared and then we have it, final answer 1 minus 2 sine squared theta. That's it so we've proved that this ugly mess is actually 1 minus 2 sine squared theta and we put a little box at the end of our proof.