 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Solving Trigonometric Equations - Concept

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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When solving trigonometric equations, we find all the angles that make the equation true. If there is no interval given, use periodicity to show the infinite number of solutions. Two ways to visualize the solutions are (1) the graph in the coordinate plane and (2) the unit circle. The unit circle is the more useful of the two in obtaining an answer.

I want to talk about trigonometric equations. Let's start with a really simple example, sine of theta equals a half. Remember when you're solving equations, you're trying to find the values of the variable that make the equation true so we want to find all the angles for which the sine is one half. Now I've drawn the picture a graph of y equals sine theta and I want to show you that I've also drawn a graph of y equals one half and so you can see that there are infinitely many points where sine of theta does actually equal half. It will have infinitely many points and there are going to be two points per period so expect infinitely many answers and expect to have two per period going into the problem.
Now usually I actually find the solutions on the unit circle, so I've drawn a unit circle and I've also drawn the line y equals one half because remember if I draw an angle the point on the unit circle where the angle crosses that point p its y coordinate is going to be the sine of this angle so in this case the y coordinate it's going to have to be one half so the question is what is this angle theta?
The first solution we get comes from the inverse sine. Inverse sine of a half is going to give us the angle in between -5 over 2 and pi over 2 that's satisfies the equation in this case it pi over 6 this solution. But you could see that within the interval from 0 to 2pi within the first period of 2 of of sine sine theta there are two solutions here's the second one. How is this angle related to the first one? We [IB] probably see by symmetry that this angle here is theta so this angle will have to be pi minus theta, the supplement so that's the second solution.
Always remember this identity the sine of pi minus x equals the sine of x with the sine function supplementary angles have the same sine value so an important property of the sine function so if pi over 6 works, 5pi over 6 is also a solution right? Keeping in mind that 5pi over 6 is pi minus pi over 6 so this is the supplement. That gives us a second solution, now I call these two solutions pi over 6 and 5pi over 6 my principle solutions and I want to get the rest of them by using the periodicity of the sine function. Sine is periodic with period 2pi so I can add any integer multiple of 2pi and get another solution, so my solutions will be of the form of theta equals pi over 6 plus 2n pi and that represents an even multiple of pi I could add any even multiple of pi or subtract any even multiple of pi and get a new solution or 5pi over 6 plus 2n pi. So both of these principles solutions yield infinitely many solutions.
Going back, just remember there are infinitely many solutions, two per period and the two solutions in a given period are related in that they are the supplements of one another. We get one solution by using inverse sine, we get another solution by using the fact that the sine of a supplement is the same as the sine of the angle and finally use periodicity to get the remaining solutions.