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.

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Asymptotes of Secant, Cosecant, and Cotangent - 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.

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I want to find the x intercepts and asymptotes of y equals negative one quarter co-secant of pi over 3x minus pi over 3. Now first of all let’s make a little substitution just to make this easier, I’ll call this theta and so my function is -1/4 co-secant theta and that’s the same us -1/4 over sine theta.

Now this function is never going to equal 0, the only way it can equal 0 is if the numerator equals 0 and it's going to ever do that so there are no x intercepts here. However the denominator can equal 0 and that’s going to give us vertical asymptotes. So the vertical asymptotes happen when sine theta equals 0 which is at the integer multiples of pi.

Now we made a substitution theta is pi over 3 x minus pi over 3. So let me re-substitute, pi over 3x minus pi over 3 equals n theta. To figure out what the equation of the asymptote is I’m going to have to solve this for x and so I add pi over 3 and then I multiply everything by 3 over pi.

So x equals nPi times 3 over pi is going to be, pi’s cancel, 3n and pi over 3 times 3 over pi is 1. So x equals 3n plus 1.

This is the pattern of the vertical asymptotes, so the vertical asymptotes are going to be x equals; when n is 0 we are going to get x equals 1. When n is 1 we’ll get x equals 4.when n is 2,7 we will get 10 and so on. But we will also get x equals -2, -5, -8 and so on. So all the vertical asymptotes are 3 units apart and they start at x equals 1.

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