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Trigonometry Equations

Date: 04/23/2003 at 07:04:48
From: Richard
Subject: Trigonometry Equation involving cos^2 and sin

Solve, for 0 <= theta < 360 
(where <= means less than or equal to)
3cos^2(theta) - 2sin(theta) = 2
giving solutions to 1 decimal place where appropriate.

I've managed to rearrange the equation to

[3sin(theta) - 1] [sin(theta) + 1]

and then find two solutions from this,

19.47 = sin(theta)
  -90 = sin(theta)

Whether they are correct I don't know, but I can't see why they 
wouldn't be. The bit I find difficult is finding all the solutions 
between the set parameters of 0 and 360. I dont know where to start, 
and when to add 180 or 360 and when to subtract, etc.

3cos^2theta - 2sin(theta) = 2

sin^2theta + cos^2theta = 1
cos^2theta = 1 - sin^2theta

3(1-sin^2(theta))-2sin(theta)=2
3-3sin^2(theta) - 2sin(theta) = 2
-3sin^2(theta) - 2sin(theta) + 1 = 0
3sin^2(theta) + 2sin(theta) - 1 = 0
(3sin(theta) - 1) (sin(theta) + 1) = 0

3sin(theta) - 1 = 0
3sin(theta) = 1
sin(theta) = 1/3
theta = sin^-1 (1/3)
theta = 19.47

sin(theta) + 1 = 0
sin(theta) = -1
theta = sin^-1 (-1)
theta = -90

I know I need to understand trig ratios here, but I don't.


Date: 04/23/2003 at 09:04:37
From: Doctor Rick
Subject: Re: Trigonometry Equation involving cos^2 and sin

Hi, Richard.

You've done a good job as far as you've gone. Let's go over the part 
you are unsure of. You have

  sin(theta) = 1/3 OR sin(theta) = -1

You know one solution for each equation:

  sin(theta) = 1/3 ==> theta = sin^-1(1/3) = 19.47 degrees
  sin(theta) = -1  ==> theta = sin^-1(-1) = -90 degrees

To find the other solutions in 0 <= theta < 360 degrees, I visualize 
the sine function:

              *****
          *           *
       *                 *
     *                     *
   *                         *
  *-------------+-------------*-------------+-------------*
  0             90          180*           270           *360
                                 *                     *
                                   *                 *
                                      *           *
                                          *****

With this image in mind, I can see that there are two angles in 
0 <= theta < 360 that have a sine equal to any given value between -1 
and 1. If you know an angle theta that lies between 0 and 180 degrees, 
then the other angle with the same sine is (180-theta); you can see 
the symmetry about 90 degrees. That's easy to work out, and to 
remember: an angle and its supplement have the same sine.

What about if theta lies between 180 and 360 degrees? The same rule 
will give us another angle with the same sine, but it won't be between 
0 and 360 degrees. For instance, if theta = 210 degrees, then 
180-theta = -30 degrees. Then all we have to do is add 360 degrees, 
because sin(theta) repeats with period 360 degrees. The new angle is 
330 degrees. We can put the two steps together to find a formula for 
the case 180 < theta < 360:

  theta2 = (180 - theta) + 360
         = 540 - theta

What about your second solution, theta = -90 degrees? I noted that 
there are two solutions in 0 <= theta < 360 degrees only if sin(theta) 
is between -1 and 1. If it is exactly -1 or 1, then there is only one 
solution. To get -90 inside the desired range, just add 360 degrees.

Have I cleared up the confusion? I haven't handled the case when you 
know a cosine, but by sketching the function you can figure that one 
out in the same way.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 04/26/2003 at 07:10:28
From: Richard
Subject: Thank you (Trigonometry Equation involving cos^2 and sin)

Dear Doctor Math,

Thanks a lot for the reply you gave me on the question. I really 
appreciated it, and I finally get the idea now.

Thanks again,
Richard
Associated Topics:
High School Trigonometry

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