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Using the Inverse Trigonometric Functions - Problem 2
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Every once in a while you will see a problem like this on your homework; simplify secant squared of inverse tangent of x. You are going to get some kind of algebraic expressions as a result of this problem, and they're sort of interesting because it shows you that compositions of trig functions, and inverse trig functions have an algebraic interpretation.
Well, let's make a little substitution. Let's call this theta, and remember the inverse trig functions give you an angle. If theta equals inverse tangent of x, then by the definition of inverse tangent, x equals tangent theta. So what I want to do in order to understand this relationship better, I'm going to draw a right triangle and I'm going to label one of the acute angles theta.
I want to label the sides so that tangent of theta equals x. You can label them however you like, but it's best to label them in the best way possible. Remember that tangent is the side opposite theta over this side adjacent theta. So I could label the sides x and 1, and if I do then tangent theta equals x over 1, that works. And for now I'll just call my hypotenuse h.
Now I have the secant squared of theta that I have to calculate. Well let's look at this picture, what's the secant of theta? Remember that secant is one over cosine, so let's first find the cosine of theta. Now cosine is side adjacent over hypotenuse, it's 1 over h. That means secant is h, and secant squared is h². Now what is h?
We could use the Pythagorean theorem in order to get a value for h, 1² plus x² equals h², and the we substitute, secant squared theta equals 1 plus x². 1² plus x² is 1 plus x², and then we realize that we're pretty much done. Theta is inverse tangent of x, secant squared of inverse tangent x equals 1 plus x², and that's what we wanted to show.
We wanted to find an algebraic interpretation for secant squared of inverse tangent x and it is 1 plus x² and that's it.
The key to a problem like this is to draw a right triangle that has the relationship that you need. Make a substitution for the inverse trig function inside. You can always call it theta, or some other Greek letter, just remember that inverse trig functions always give you angles.
So you can set up a right triangle with that angle in it and then label the triangle appropriately, in this case we needed to make the triangle have tangent theta equal x. So it's kind of interesting that two trig functions and an inverse trig function can give you 1 plus x².
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