The Tangent Function
The tangent function f(x) = tan(x) has the following properties
 The tangent function is defined as tan(x) = sin(x) /cos (x).
 Since cos (x) is in the denominator, the tangent function
is not defined when cos (x) = 0. This means that the
domain of the tangent function is {x | x ≠ π /2 + nπ where
n is an integer}
 The period of tangent is π.
 tan(x) = tan(x + π) for all x in the domain of the tangent
function.
 -∞ < tan (x) < ∞. This is the range of the tangent function.
 The tangent function is not one-to-one. For each y-value there are infinitely many
x, such that tan (x) = y.
The inverse of tangent is arctan, f(x) = atan (x). It has the
following properties:
 The function is one to one.
 The domain is -∞ < x < ∞.
 The range is π / 2 < atan (x) < π/2
 If x is in the interval -π /2 < x < π /2 , then: if tan (x) = y
then atan (y) = x.
The question is how do you find the solutions without using a graphing calculator to
approximate the solutions using the Calc button?
When solving tan (x) = k, where k is any real number in the interval [0, 2π), there will
always be two solutions. We divide these up into two cases k < 0, and k ≥ 0.
Case 1: k < 0
1. Taking the arctan of both sides we obtain x = atan (x).
However, this value is negative. Since the period of tan (x) is π,
the solutions we want are atan (x) + π, and atan (x) + 2π.
Case 2: k ≥ 0
1. Taking the arctan of both sides we obtain x = atan (x). This is
one solution. The second solution is x = atan (x) + π.
Finding All Solutions to tan (x) = k
So once you have found the solutions in the interval [0, 2π) how do you get all solutions
that solve tan (x) = k. Let us suppose that you found that tan(x) = k in the interval [0, 2π)
had two solutions called x1 and x2. Note that x2 is π units from x1. In fact, all solutions
are an integer multiple of π units from x1. That is, all the solutions are:
x = x1 + nπ
A more general form of tangent
A more general form of the tangent function is f(x) = A tan(ax – b)
 A is the vertical stretch.
 a determines the period. To find the period p: p = π / |a|.
For the graph to the right the period is p = π / 3.
 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 tangent function is the linear expression
3x – 1
o Factor out the 3.
o 3x – 1 = 3(x – 1/3).
o The phase shift is 1/3. The phase shift determines how much the tangent
function is shifted left or right. Suppose b is positive:
 The expression tan [a(x – b)] is shifted b units to the right.
 The expression tan [a(x + b)] is shifted b units to the left.
Solving Atan (ax – b) = k
1. Divide by A and you get: tan (ax – b) = k / A.
2. Use the method on page 1 above to find the solution to tan (x) = k / A. Call this
solution x1.
3. Solve ax – b = x1 You obtain x = (x1+b)/a. This is the solution for one complete
period of A tan(ax – b) = k.
4. Notice that the period is π / a.
5. All solutions are found by:
x = (x1+b)/a + nπ / a.