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# The Half-Angle Identities - Problem 3

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Here is a slightly more challenging problem. We want to prove that tangent alpha equals sine 2 alpha over 1 plus cosine 2 alpha. The reason for this is we're going to try to develop a tangent half-angle formula in a second.

First let's prove this and when you're proving an identity, you can start from either side and work towards the other so I'm going to start on the right hand side. Sine 2 alpha over 1 plus cosine 2 alpha. Here I need both of the double angle identities.

Now for sine, it's 2 sine alpha, cosine alpha. Now here I have 1 plus and then I have cosine 2 alpha. Cosine 2 alpha is cosine squared alpha minus sine squared alpha. Now if you look at the denominator, recall your Pythagorean identity, Pythagorean identity says cosine squared alpha plus sine squared alpha equals 1, so 1 minus sine squared alpha will equal cosine squared alpha, so let me take this guy and replace it with cosine squared. 2 sine alpha, cosine alpha over cosine squared, this cosine squared, plus another one and that's 2 sine alpha cosine alpha over 2 cosine squared alpha. And so you get a little it of a cancellation here, the 2s cancel, the cosine cancels leaving a single cosine in the bottom and a sine in the top and that's tangent alpha just like we wanted.

So this result's proved, now how do we get the half-angle identity? All we do is we replace alpha with theta over 2, so here we get tangent of theta over 2 equals and then we had sine 2 alpha which becomes sine of theta over 1 plus cosine 2 alpha which becomes 1 plus cosine theta, that's our half-angle formula for tangent. Let's use it on the tangent of 15 degrees.

Tangent of 15 degrees, so that's half of 30, we have the sine of 30 degrees over 1 plus cosine 30 degrees. Now the sine of 30 degrees is a half, cosine of 30 degrees root 3 over 2. 1/2, 1 plus root 3 over 2, let's simplify this. We multiply by 2 over 2, we get 1 on the top and 2 plus root 3 on the bottom, it's still not simplified enough. We need to rationalize the denominator and so 1 over 2 plus root 3, we have to multiply by the conjugate. The conjugate of 2 plus root 3 is 2 minus root 3.

So 2 minus root 3 over 2 minus root 3 and this is going to end up being a difference of squares we get 4 minus 3, or 1, 4 minus 3 and on the top we get 2 minus root 3 and so this becomes 2 minus root 3 over 1, 2 minus root 3 and that's our tangent of 15 degrees.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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