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Finding a Trig Function of an Angle

Date: 07/09/2004 at 05:23:09
From: Sara
Subject: trigonometry

I need a short way to find the answer to this kind of question:  If
sin(y) = 8/17, find cot(y).

I know cot(y) is cos(y)/sin(y), so cot(y) = (?/17)/(8/17) and
cot(y) = ?/8

Date: 07/10/2004 at 11:34:10
From: Doctor Willae
Subject: Re: trigonometry

Dear Sara,

There is a whole family of problems like these, and yours is the most
basic.  This is really a question about inverse trig functions.  I'll
use the prefix "arc" to mean an inverse trig function.  Like this:

  arcsin(sin(x)) = x

       arcsin(1) = pi/2   because sin(pi/2) = 1

Going back to your original question, if the sin(y) = 8/17, we can
determine y by just applying the inverse sine function:

  y = arcsin(sin(y)) = arcsin(8/17)

Once we know angle y, we can determine the cotangent of y by just
taking the tangent of y and inverting it:

  cot(y) = 1 / tan(y) = 1 / tan(arcsin(8/17))

Now...before you break out your calculator, you should really learn
how to do this by hand.  Fortunately, there's a trick that works for
all of these problems.  Draw yourself a right triangle.  (I'll put one
here, but you'll have to forgive my artwork.)  Our goal will be to
label all 3 sides of the triangle.  Start by labeling one of the
non-right-angles y.  I'll put it in the bottom right.

           * **
           *   ***
           *      ***
           *         ***
           *            ***
           *               ***
           *                y **

What do we know about the angle y?  We know its sine: sin(y) = 8/17. 
And remember that sine is opposite over hypotenuse.  So we can label
two of the sides of our triangle:

           * **
           *   ***      17
           *      ***
        8  *         ***
           *            ***
           *               ***
           *                y **

What else do we know?  Pythagoras.  If A and B are legs of a right
triangle, and C is its hypotenuse, then:

  C^2 = A^2 + B^2.

We've got the length of one leg and the hypotenuse.  So we can solve
for the other leg:

  B^2 = C^2 - A^2
    B = sqrt(C^2 - A^2)
      = sqrt(17^2 - 8^2)
      = sqrt(289 - 64)
      = 15

Now we can finish labeling the sides of the triangle:

           * **
           *   ***      17
           *      ***
        8  *         ***
           *            ***
           *               ***
           *                y **

We're all but finished.  You can read off the value of any trig
function involving y from this triangle.  You're interested in the
cotangent, so let's look at that:

  cot(y) = adjacent/opposite = 15/8

But you could just as easily use this same technique to get the cosine
of y:

  cos(y) = adjacent/hypotenuse = 15/17

Remember that at the beginning I said there was a whole family of this
type of question.  Yours is the most basic since we aren't told
anything else about y, and therefore we just assume that y is an acute
angle and draw the triangle the way we did.

Sometimes in this type of problem, more information is given.  Suppose
we were also told that tan(y) < 0, or in other words the tangent of y
is negative.  That would tell us that y can't be acute, since all six
trig functions of an acute angle are positive.  But by understanding 
the use of reference angles and the signs of the trig functions by
quadrant, we could determine that if sin(y) = 8/17 and tan(y) < 0,
then y must be an angle in the second quadrant.  We could then still
draw our triangle, but we'd have to construct it in the second
quadrant and the 8 would become -8.  

I don't want to go into more detail on that here since your question
didn't involve it.  I just wanted to mention that there are variations
of this sort of problem where you must give consideration as to where
to draw the triangle.  But you can still use the triangle!

This is a really useful trick to know, so let me know if anything
isn't clear.  I actually just used this same procedure to speed up
some software I'm working on.  Let me know how it goes.

- Doctor Willae, The Math Forum 
Associated Topics:
High School Trigonometry

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