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Mathematical Expression Editor
Introduction
Frequently we are in the situation of having to determine precisely which angles
satisfy a particular equation. Something like . We know that , meaning that
is a solution of this equation, but is that the only solution or are there
more?
Let’s look at the graph of the sine function.
Notice that the graph of and the graph of the constant function intersect many
times, not just once. In fact, since sine is a periodic function, these graphs intersect
infinitely many times. Each of these intersections represents a single solution of
the equation . We need a process to identify and write down each of these
solutions. Let’s start by looking at the unit circle. Remember that sine values
correspond to the -coordinate of points on the unit circle. This equation
is asking us to find all the points on the unit circle with a -coordinate of
.
You see that there are two locations on the unit circle with -coordinate equal to , one
in the first quadrant and another in the second. As we mentioned earlier, the first
quadrant angle is . The angle in the second quadrant has reference angle , which
means the angle is . Those are the only two points on the circle with that -coordinate,
but remember that there are many other angles which are coterminal with those. For
instance:
The only solutions are the angles , , and all the angles coterminal with them. Since
the sine function has period , that means any other solution has to be an integer
multiple of away from one of these first two solutions. Putting that together, our
solutions are:
The steps we’ve followed are summarized in the following.
To solve a trigonometric
equation:
(a)
Find the reference angle of the solutions. Typically the standard values
will help identify this.
(b)
Find all solutions on a single period of the function. Use the graph, the
unit circle, and the reference angle to identify these.
(c)
Find all solutions. Use the period of the function to find all requested
solutions.
Solve the equation:
We’ll start by finding the reference angle, , the acute angle
between the terminal side of and the -axis. The reference angle satisfies
and the negative sign will be used to indicate the quadrant of the angle.
From the picture we see . Let’s look at the unit circle.
In one period , there are two angles that have reference angle and have negative sine
value. One is in quadrant 3, and one in quadrant 4. That means the solutions in the
interval are and .
To find all solutions, we have to add all multiples of to these. The solutions are
then
Solve the equation:
We’ll start by finding the reference angle, , the acute angle between the terminal side
of and the -axis. The reference angle satisfies and the negative sign will be used to
indicate the quadrant of the angle. Since tangent is opposite over adjacent, we have
the following triangle.
From the picture we see .
The tangent function goes through one period on the interval . In the interval (which
is Quadrant IV), tangent is negative while in (which is Quadrant I), tangent is
positive. For to be negative in this interval, we need to be in . The only angle in
that interval with reference angle is . This is the only solution on this period.
Remember that the tangent function has period , unlike sine and cosine which have
period . On the period , tangent is one-to-one, so there is exactly one angle which
gives the desired output value. Sine and cosine are not one-to-one across a full
period.
To find all solutions, we have to add all multiples of to this. We use instead of
because the period of tangent is only . The solutions are then
Let’s try one a bit more complicated.
Solve the equation:
We’ll start by simplifying
a bit. Note that we use a rearranged form of the Pythagorean identity, .
Notice that this equation is quadratic in . We can factor it like we try to do to solve
any other quadratic equation: Now, we can set each factor equal to zero and solve
the two resulting equations:
yield the equations and . On the interval , has only one solution, . For , we see that
the reference angle . Since cosine is positive in Quadrants I and IV, we find solutions
and .
