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# The Reciprocal Trigonometric Functions - Problem 1

###### 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.

Let's find the value of all six trig functions at theta equals 7 pi over 4. I've drawn 7 pi over 4 in the unit circle. The first thing I have to do is identify the reference angle.

Now the reference angle is the angle between the terminal side of 7 pi over 4, and the x axis and that is pi over 4. Then what I need to do is, find the values of sine, cosine and tangent at pi over 4. Sine of pi over 4 is root 2 over 2, cosine of 4 over 2 is also root 2 over 2, and tangent of pi over 4, remember that tangent is sine over cosine. so we get 1.

Now I'm in the fourth quadrant with the angle 7 pi over 4, and in this quadrant, only cosine is going to be positive, so cosine of 7 pi over 4 will equal root 2 over 2, the others are going to be negative, sine of 7 pi over 4 is the opposite or root 2 over 2, and tangent of 7 pi over 4 is going to be -1.

Now to find the other three trig function values, the reciprocal functions, I just need to take the reciprocals of these; cosecant. Cosecant is the reciprocal of sine, so I need to take the reciprocal of negative root 2 over 2, which happens to be negative root 2.

For secant I take the reciprocal of sine, and so I have the reciprocal of root 2 over 2, 1 over root 2 over 2 which is root, and then for cotangent I take the reciprocal of tangent, so 1 over -1. It truly is just as easy as that.

First identify the reference angle, second find sine, cosine and tangent of the reference angle. Third using the value of the quadrant you're in, decide whether sine, cosine and tangent of 7 pi over 4, or whatever angle you're working with, is positive or negative. And then, use the reciprocal identities to find the values of cosecant, secant and cotangent.

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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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