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# The Euler Formula - 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.

There are a lot of cool things about the Euler formula. One of them is that it provides a connection between exponential functions, and trigonometric functions. Another, is that the base of this exponential function is that strange number e and it shows you that e is really an important number in mathematics.

One of the things that we can do with this formula, is we can actually use it to derive some of our identities, some of our trigonometric identities. And I want to use it to derive the double angle formulas. And just to remind you what the double formulas are. They are the formulas for cosine 2 theta and sine 2 theta.

So we want to be able to say cosine 2 theta equals some function of sine theta and cosine theta. So let’s do that now.

Let’s express cosine 2 theta plus i sine 2 theta as a complex number. This would be the same as 1 times cosine 2 theta plus i sine 2 theta. And I can put this in exponential form as 1 times e to the i, 2 theta, by the Euler formula. This is the same as e to the i theta squared. Convince yourself that that’s true. When you have an exponential expression squared, you can pull the square inside as a product write e to the i2 theta.

Let me switch back really quickly from e to the i theta to cosine theta, plus i sine theta squared. Really easy to square, it’s just a binomial. I get cosine squared theta, plus i² sine squared theta. And then I get 2, i sine theta cosine theta. So plus i times 2 sine theta cosine theta.

Now i² is -1, so this is cosine squared minus sine squared. Plus i times 2 sine theta cosine theta. So let’s take a look at what we started with. Let me erase this.

For two complex numbers to be equal, the real parts have to be equals and the imaginary parts have to be equal. Let’s look at our real and imaginary parts here. The real part here equals the real part here, that’s your double angle identity for cosine. And the imaginary part has to equal the imaginary part. And that is your double angle identity for sine.

So sine 2 theta, equals 2 sine theta cosine theta. And cosine 2 theta, equals cosine squared theta minus sine squared theta.

The really cool thing about this proof, is you get both of the double angle identities at once.

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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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