Solving Trigonometric Equations in Radians

Date: 12/14/97 at 00:26:48
From: alex r
Subject: Solving trigonometric equations in radians

Solve for theta in radians to 2 decimals if 0 < theta < 2 pi tan theta 
= -0.318. 

I tried inverse tan -0.318 (in radians mode) and it gave me -0.307; 
however, it said between 0 and 2pi so I added 2pi and got one of the 
answers, 5.9. However, I can only get the other 2.8 if I add 1 pi.

Why am I adding one pi? Whenever I do something in degrees, if I am 
given an angle and want to find another angle with the same tan or 
whatever, I add 360. Here pi = 180 - why am I adding pi (180) and not 
2pi (360)?

Please help. Thanks.

Date: 12/14/97 at 04:28:51
From: Doctor Luis
Subject: Re: Solving trigonometric equations in radians

This is a good question. To understand why you only add one pi, you 
need to understand something about a special property of trigonometric 
functions called periodicity.

As you may know, trigonometric functions are periodic. A periodic 
function f(x) has the following special property,

   f(x+P) = f(x)

This means that the function repeats itself (it gives you the same 
values over again) when you add a special number. Since there's an 
infinite number of such numbers, we usually take the smallest positive 
number such that the equation above is still true, and we call that 
number the period of the function f(x).

  let p = period of f(x).

You can see that if you add p to x a lot of times you'll still get the 
same value

  f(x+np) = f(x)  , where n is an integer (negative, positive or zero)

This gives you a problem if you want to solve some equation like the 
following for x (there's an infinite number of values of x that will 
give you the same value of y!)

            f(x) = y

  ---->   x + np = inversef(y)

   or          x = inversef(y) + mp  ,   where m = -n

Since you know that you have an infinite values of x, and you also 
know the period of f(x), we usually take the solution x with m = 0, so 
that
               x = inversef(y)

Notice that, in this case, x will have to be between 0 and p, because 
if it were greater than p you could always subtract p until you got x 
in between 0 and p, or if x were less than 0 you could always add p 
until you got x in between 0 and p. Then,

                 0 < x < p

The solution with this restriction on x is called the principal 
solution of the periodic function f(x) = y

You are already familiar with the trigonometric functions sine and 
cosine: their period is the same and is equal to 2pi radians or 
360 degrees. In fact, the cosine is the same curve as the sine, except 
that it is shifted pi/2 radians to the right (that's why they have the 
same period)

   sin(x+2pi) = sin(x)   and   cos(x+2pi) = cos(x)

Now, the tangent doesn't have a period of 2pi like the sine and 
cosine. It has a period of 1 pi. This is because of the way the 
tangent is defined,

             sin(x)
   tan(x) = --------
             cos(x)


  So,

                  sin(x+pi)
     tan(x+pi) = -----------
                  cos(x+pi)

Since the cosine is just a shifted sine curve,

   cos((pi/2)-x) = sin(x)

This is the same as saying that

   sin((pi/2)-x) = cos(x)

Now, you can rewrite sin(x+pi) and cos(x+pi) like this,

   sin(x+pi) = sin(pi+x)
             = sin((pi/2)-(-x-(pi/2)))
             = cos(-x-(pi/2))
             = cos(-((pi/2)+x))       
             = cos((pi/2)+x)         (b/c cos(-x) = cos(x) )
             = cos((pi/2)-(-x))
             = sin(-x)
             = -sin(x)               (b/c sin(-x) = sin(x) )
and,

   cos(x+pi) = cos(pi+x)
             = cos((pi/2)-(x-(pi/2)))
             = sin(x-(pi/2))
             = -sin((pi/2)-x)         (b/c sin(-x) = sin(x) )
             = -cos(x)

Then,

                  sin(x+pi)
     tan(x+pi) = -----------
                  cos(x+pi)

                  - sin(x)
               = -----------
                  - cos(x)

                  sin(x)
               = --------
                  cos(x)
  
               = tan(x)             (that's just the definition!)

And so, you see

     tan(x+pi) = tan(x)

that's just a periodic function with period pi !!

Now, your problem said: solve for x in

     tan(x) = -0.318   with   0 < x < 2pi

so,

       x  = -0.307     (radians)  

but remember, our solution for a periodic function has to be between 
0 and the period p. In this case, we just proved that the period p of 
the tan(x) function is pi  (radians). So we add pi to x, and get

     x + pi = 2.83         (radians)

This brings our solution to be between 0 and pi (3.14...) and gives us 
the principal solution of the equation tan(x) = -0.318

Remember that a periodic function has an infinite number of solutions? 
Since the problem asks for solutions between 0 and 2 pi there are 
going to be two solutions. One between 0 and pi, and the other one 
between pi and pi+pi=2pi. So, to get the other solution we just add pi 
(pi is the period, remember?) to our principal solution, and get

    (x+pi)+pi = 2.83 + pi
              = 5.97       (radians)

If the problem had asked for a solution with 0 < x < 3pi, there would 
have been three solutions, and we would have had to add pi (the period 
of tan(x)) again,

     (x+2pi)+pi = 5.97 + pi
                = 9.11      (radians)

If the problem had asked for a solution between -pi and 0, we would 
have kept the solution your calculator gave you in the first place.

      x = invtan(-0.318)
        = -0.307

So, you see, knowing the principal solution and the period of a 
periodic function gives you all of the solutions. All you have to do 
is find the solutions that you want (for example, all the solutions 
between -10 and 345).

What if the problem had said, instead:

   solve for x,

        tan(2x) = -0.318        with   0 < x < 2pi

what would the solutions be? (hint: there are four solutions)

Something to think about: If a periodic function f(x) has period p, 
what is the period of the periodic function f(rx+s), where r is any 
positive real number and s any real number?

I hope this helped :)

-Doctor Luis,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/