# Using the Sine and Cosine Addition Formulas to Prove Identities - Problem 3

###### Transcript

Here is a particularly challenging, but kind of cool example that involves the sine and cosine of a difference formulas. The problem says derive the tangent of a difference identity, that means derive an identity for tangent of alpha minus beta.

Now the first thing I know about tangent is that tangent is sine over cosine, so this is sine of alpha minus beta over cosine of alpha minus beta and then I can use the sine and cosine of our difference formulas to fill this out. I have sine, cosine, cosine, sine; sine alpha, cosine beta minus cosine alpha, sine beta over and for cosine I have cosine, cosine, sine, sine. Cosine alpha, cosine beta oops and the minus becomes a plus, sine alpha, sine beta.

Now this is a mess and it requires a trick to simplify and the trick is to multiply the top and bottom by 1 over cosine alpha, cosine beta, 1 over cosine alpha, cosine beta and this will give us a really nice result, watch what happens here. So this 1 over cosine alpha and cosine beta is going to get distributed over both of these terms in the numerator, sine alpha, cosine beta over cosine alpha, cosine beta the cosine betas will cancel and I'll have sine alpha over cosine alpha minus. So that's what happens to the first term.

The second term, I get cosine alpha, sine beta over cosine alpha, cosine beta the cosine alphas cancel and I'm left with sine beta over cosine beta, this is good. The first term in the denominator cosine alpha, cosine beta over cosine alpha cosine beta 1 plus and finally sine alpha sine, sine beta over cosine alpha, cosine beta. Sine alpha, sine beta nothing cancels here cosine alpha, cosine beta, but here is what we get, this is tangent alpha, tangent beta over 1 plus tangent alpha times tangent beta, and that's our tangent of a difference formula. Tangent of alpha minus beta is tangent alpha minus tangent alpha over 1 plus tangent alpha tangent beta.

Let's use this in a really easy example. The problem says find the tangent of 15 degrees? So you first want to express the 15 degrees as some difference that involves nice angles, angles that are special. Tangent of 45 minus 30 for example. Actually let me go with 60 minus 45, let me tell you why. If you remember the tangent of 30 degrees it's 1 over root 3 and the tangent of 60 degrees is root 3, I don't want to have any 1 over root 3s in my answers so I'm going to stays away from that one, but will give me a slightly nicer answer, so it's tangent of 60 degrees minus 45 degrees.

Now from the formula it's tangent alpha minus tangent beta so tangent 60 minus tangent 45 over 1 plus the product of the two, tangent 60, tangent 45 and that's going to give me tangent of 60 it's nice, it's root 3, tangent of 45 is 1, 1 plus and then root 3 times 1 not bas that's a pretty good answer, but usually we don't like to have radicals in the denominator so let's rationalize this.

So in order to rationalize the denominator, do you remember the trick of multiplying by the conjugate? That's what we're going to do here except I'm going to multiply in the form, let's see 1 minus root 3, 1 minus root 3. In the denominator, I'm going to get 1² minus root 3², that's 1 minus 3. In the numerator and this is the same as well let me multiply it through, I get root 3 times 1, root 3 root 3 times negative root 3 minus 3, -1 times 1, -1 and -1 times negative root 3 plus root 3. I have a -2 on the bottom and on top I've got two root 3's minus 4. If I divide that all by -2, I get negative root 3 plus 2 and that's my answer. 2 minus root 3, that's the tangent of 15 degrees.