The Cosine Function
The cosine function f(x) = cos(x) has the following properties
 The domain of the cosine function is all the real numbers.
 The period of the function is 2π.
 cos(x) = cos(x + 2π) for all x.
 The function is even: symmetric with respect to the yaxis. This is also stated by saying cos(x) = cos(-x)
 -1 ≤ cos (x) ≤ 1. This is the range of the cosine function.
 The cosine function is not one-to-one. For each y-value
in the interval [-1, 1] there are infinitely many x, such that cos (x) = y.
The inverse of cosine is arccos, f(x) = acos (x). It has the
following properties:
 The function is one to one.
 The domain is -1 ≤ x ≤ 1.
 The range is 0 ≤ acos (x) ≤ π
 If x is in the interval 0 ≤ x ≤ π , then: if cos (x) = y then
acos (y) = x.
When solving cos (x) = k, where k is any real number in the interval [0, 2π)
 if |k| > 1, there are no solutions.
 if |k| = 1, there is one solution.
 if |k| < 1, there are two solutions.
The question is how do you find the solutions without using a graphing calculator to
approximate the solutions using the Calc button? We look at the problem with four cases
k = -1, 1, -1 < k < 0, and 0 ≤ k < 1.
Case 1: cos (x) = -1.
1. Taking the inverse of both sides: x = acos (-1) = π.
Case 2: cos (x) = 1.
1. Taking the inverse of both sides: x = acos (1) = 0.
Case 3: cos (x) = k where -1 < k < 0.
1. For this problem there are two answers. An example of
the problem is shown to the right. The points where the
two graphs intersect are the solutions.
2. To find one of the solutions we calculate acos(k). Since k
is negative this will give us a value that is in the interval
π/2 < x < π. This is the left solution.
3. The other solution is at 2π - acos(k). (cos(2π - a) =
cos(a).) [It is not too difficult to see from the picture on the right that each of the
intersection points are equidistant from 0 and 2 π, respectively.
Case 4: cos(x) = k where 0 ≤ k < 1
1. The first solution is at x = acos(k).
2. The second solution is at x = 2π – acos (k). (again: cos (a) = cos (π –a).)
Notice that case 3 and case 4 are identical.
Finding All Solutions to cos (x) = k
So once you have found the solutions in the interval [0, 2π) how do you get all solutions
that solve cos (x) = k. Let us suppose that you found that cos(x) = k in the interval [0, 2π)
had two solutions called x1 and x2. Every other solution is a multiple of 2π from one of
these solutions. That is, all the solutions are:
x = x1 + 2nπ
x = x2 + 2nπ, where n is any integer.
A more general form of cosine
A more general form of the cosine function is f(x) = A cos(ax – b)
 A is the amplitude. If you were to measure the vertical
displacement from the highest point of the graph to the
lowest point of the graph you would find that this distance
is |2A|. In the graph to the right A = 5. We usually
express the amplitude as a positive number even though A
may be negative.
 a determines the period. To find the period p: p = 2π / |a|.
For the graph to the right the period is p = π / 2.
 a and b are used to find what is called the phase shift. To find the phase shift for
the problem to the right:
o Inside the cosine function is the linear expression 4x – 3
o Factor out the 4.
o 4x – 3 = 4(x – ¾).
o The phase shift is ¾. The phase shift determines how much the cosine
function is shifted left or right. Suppose b is positive:
 The expression cos [a(x – b)] is shifted b units to the right.
 The expression cos [a(x + b)] is shifted b units to the left.
Solving Acos (ax – b) = k
1. Divide by A and you get: cos (ax – b) = k / A.
2. Use the method on page 1 above to find the solutions to cos (x) = k / A. Call
these solutions x1 and x2.
3. Solve ax – b = x1 and ax – b = x2. You obtain x = (x1+b)/a, and x = (x2+b)/a.
These are the two solutions in one complete period of A cos(ax – b) = k.
4. Notice that the period is 2π / a.
5. All solutions are found by:
x = (x1+b)/a + 2nπ / a
x = (x2+b)/a + 2nπ / a where n is any integer