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# Using Trigonometric Identities - Problem 2

###### 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.

I'm proving trig identities and I'm going to use a form of the Pythagorean identity in one of my examples, but not this one. The cosine squared theta plus sine squared theta equals 1 is the one you'll probably remember, but sometimes I have trouble remembering the others. There's a Pythagorean Identity that involves tangent and secant and how do we get that?

Well I can get tangent and secant if I divide both sides of this equation by cosine squared, so I'm going to do that now, cosine squared theta plus sine squared theta equals 1, divide everything by cosine squared theta and what you get is 1 plus tangent squared theta equal secant squared theta. So this is the Pythagorean identity that involves tangent and secant and we'll meet a form of this in the next problem.

So here is the problem; prove the identity, oh simplify, sorry, we're going to simplify this, even better, but this is still a pretty hard problem. To simplify this, I'm going to need a trick that I used before let me start with tangent theta, secant theta plus 1, I'm going to leave a little space plus tangent theta, secant theta minus 1. The trick is multiplying by the conjugate.

The idea is I want to multiply the thing that will make this a difference squares, secant theta minus 1 will give me a difference of squares and it will give me secant squared theta minus 1.

I want to do the same thing here except I'm going to multiply by secant theta plus 1, I want to get the same denominator after all. Whenever you're dealing with two separate fractions you want to add or subtract them, you need to get a common denominator, so this will give me a common denominator and simultaneously will give me a difference of squares.

So in the bottom I'm going to get secant squared theta minus 1 and the top I get tan-theta secant theta minus 10 theta plus tan-theta secant theta plus tan-theta which is nice because the tan-theta's cancel, so we cross that out and I end up with 23 I'll write it over here. 2 tan-theta secant theta on top and what about the bottom? Secant theta minus 1?

Well this looks a lot like the identity I just proved except it's not quite the same form, I just need to subtract 1 from both sides tangent theta, tangent squared theta input secant squared theta minus 1, so I can convert secant squared theta minus 1 to tangent squared theta and I'm going to do that now. And here we just cancel the tangents, I'm left with one tangent on the bottom.

Now when you're simplifying trigonometric expressions, one of the things you want to think about is having as few operations and as few functions in your answer as possible. Now right now I've got division and I've got multiplication. I also have two trig functions and I want to see if I can get this down even smaller.

So when I use the trick of switching to sines and cosines, which you often do when you're dealing with reciprocal functions. So 2 times 1 over cosine theta over sine theta over cosine theta and now to get rid of the cosines I'm going to multiply by cosine theta over cosine theta.

So you see on top here the cosines will completely cancel and I'll have a 2 and on the bottom the cosines will completely cancel and I have the sine theta, 2 over sine theta that's much simpler than this. And now I recognize that one over sine theta is co-secant, so this is going to end up being 2 co-secant theta, that's nice and simple, much simpler than what we started with, so this is our final simplified answer.

So remember we use two tricks, one we multiplied by the conjugate, the conjugate is you take a sum and you make it a difference or a difference, make it a sum because multiplying those will give you a difference of squares. The second trick we're switching from reciprocal trig functions to their sine and cosine form. Switching to sine and cosine sometimes allows you to do some algebraic simplifications.

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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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