 ###### 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.

##### Thank you for watching the video.

To unlock all 5,300 videos, start your free trial.

# The Reciprocal Trigonometric Functions - 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.

Share

Recall that cosine squared theta plus sine squared theta equals 1. And that tangent that over cosine theta, secant theta equals 1 over cosine theta. I can use these two identities, plus this one to develop another Pythagorean identity. Let me start with cosine squared theta, plus sine squared theta equals 1.

If I divide both sides of these equation by cosine squared theta, what do I get? I get 1 plus sine squared theta over cosine squared theta, that's tangent squared theta. and 1 over cosine squared theta is secant squared theta. This is another very useful identity. And in fact I'm going to use it in the next example, but I'm going to use this form of it; tangent squared theta equals secant theta minus 1.

So let's do the problem. The problem says secant theta equals -5 over 3 and theta is between pi and 3 pi over 2. Find the other five trig functions values at theta.

Well first, I want to use the identity I just proved to find tangent, tangent squared equals secant squared minus 1. I substitute -5/3 for secant, Now -5/3 squared is 25 over 9, and I can write 1 as 9 over 9, 25 over 9 minus 9 over 9 is 16 over 9, so tangent squared theta is 16 over 9, that means tangent theta is plus or minus 4/3, but which is it?

In order to find out, I have to determine what quadrant theta is in, but I'm actually told that theta is between pi and 3 pi over 2, and that put theta in the third quadrant where tangent is positive, so tangent is 4/3. Now it would be really easy to find cotangent, because cotangent theta is the reciprocal of tangent theta, and that equals 3/4.

Next, I can use the fact that I know secant theta to find cosine theta, cosine theta is 1 over secant theta, they're reciprocals of each other, so cosine theta is 1 over -5 over 3, and that's -3/5.

What about sine? There's a couple of ways I define sine theta, I could use the Pythagorean identity, but I could also use the fact that, tangent theta is sine theta over cosine theta. And so sine theta is tangent theta times cosine theta.

Now I've already found tangent and cosine. Tangent was 4/3, cosine -3/5, cancel the threes and you're left with -4/5 and so sine theta is -4/5. There's one more, cosecant. Remember that cosecant is the reciprocal of sine, and the reciprocal of sine is -5/4. Have I got all six? Let me count tangent; 4/3, cotangent; 3/4, cosine; -3/5, sine; -4/5, cosecant; -5/4, and we started out with secant -5/3.