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Sine Function and Unit Circle

Date: 08/15/2003 at 10:05:38
From: Sean
Subject: Trigonometry

I have a problem I am just completely stuck on, and was wondering if 
you could help me solve it.

Solve for x:

sin(pi/x) = square root2/2
   where x is an integer

I have tried the question using formula sin^2 x=cos^2 x-cos2x but am 
having little luck. In my mind I can see it turning out to be a 
quadratic, but I'm not sure.


Date: 08/15/2003 at 18:59:52
From: Doctor Warren
Subject: Re: Trigonometry

Hi Sean,

Before setting out to algebraically manipulate your equation, you
might want to spend some time getting re-acquainted with the sine
function. In particular, you should try to recall the following four
points:

   sin(pi/6) = 1/2
   sin(pi/4) = sqrt(2)/2
   sin(pi/3) = sqrt(3)/2
   sin(pi/2) = 1

Remember that if you draw a one-unit-long line at an angle theta from
the positive x-axis, measured counterclockwise, the y-coordinate of
the endpoint is the sine of the angle. Take a look at this diagram of
the unit circle:

                                  |
                                  |
                              y=1 | sin(pi/2) = 1
      sin(2pi/3) = sqrt(3/2) _..--+--.._   sin(pi/3) = sqrt(3/2)
                          .-'     |     `-.  
 sin(3pi/4) = sqrt(2)/2 ,'        |        `.  sin(pi/4) = sqrt(2)/2
                      ,'          |          `.  
    sin(5pi/6) = 1/2 /            |            \  sin(pi/6): y = 1/2
                    /             |             \ 
                   .              |              .
       sin(pi) = 0 |              |              | sin(0) = 0
                ---+--------------+--------------+---
                   |              |              |
                   .              |              .
                    \             |             /
  sin(7pi/6) = -1/2  \            |            /  sin(11pi/6) = -1/2
                      `.          |          .'
sin(5pi/4) = -sqrt(2)/2 `.        |        .'  sin(7pi/4) = -sqrt(2)/2  
                          `-._    |    _.-'      
      sin(4pi/3) = -sqrt(3)/2 ''--+--''   sin(7pi/6) = -sqrt(3)/2 
                             y=-1 | sin(3pi/2) = -1
                                  |
                                  |

In every case, the sine of each angle shown is the value of the
y-coordinate at that point on the unit circle. The four values I
showed you above get repeated again in each quadrant, and the pattern
is really pretty easy to remember. For example, if I want to know the
sine of (7 pi/4), I do the following:

   1. I reduce the fraction as much as possible. If I'm given 6pi/4,
      for example, I immediately reduce it to 3pi/2.  In this case, I
      cannot reduce 7pi/4.

   2. I think about where 7pi/4 is on the unit circle. It's all the
      way around on the bottom right of the unit circle, beyond 3pi/2.
      In fact, 7pi/4 is an angle of pi/4 beyond 3pi/2, which is
      of course the same as 6pi/4. Just think of this angle as a
      "quarter-pi" angle, some multiple of pi/4.

   3. I know that sines of quarter-pi angles are either sqrt(2)/2 or
      -sqrt(2)/2. (Except of course where the fraction can be reduced
      to a multiple of pi or pi/2, which you should have already done
      at step 1.)

   4. If the angle puts me on the bottom half of the circle, I pick
      the negative choice. If the angle puts me on the top of the
      circle, I pick the positive one. In this case, the angle
      7pi/4 is on the bottom, so I'll pick the negative one. Thus
      sin(7pi/4) = -sqrt(2)/2.

Your question is:

   Solve sin(pi/x) = sqrt(2)/2 for x.

You should recognize the right-hand side, sqrt(2)/2, as one of those
convenient quantities that the sine function spits out for quarter-pi
angles. Which quarter-pi angles? There are two whose sines are 
positive sqrt(2)/2:

   sin(pi/4)
   sin(3pi/4)

Your equation therefore has two solutions. Can you see what they are?
The second one is a little tricky, so I'll give you a hint:

    pi       b pi
  ------- = ------
   (a/b)       a

Let me know if you need any more help!

- Doctor Warren, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Trigonometry

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