Phase shifts for tangents

This material is very similar to that in another chapter in which phase shift is explained for the sine and cosine functions.

Phase shift for trigonometric functions means moving the graph of the function to the right or to the left. This transformation occurs when a constant is added to (or subtracted from) the angle to which the function is applied.

For example, the following functions include a phase shift transformation.

• $f\left(\theta\right)=\sin\left(\theta+45^\circ\right)$f(θ)=sin(θ+45°)
• $g\left(x\right)=\cos(x-37.5^\circ)$g(x)=cos(x−37.5°)
• $h\left(\alpha\right)=\tan\left(\alpha+180^\circ\right)$h(α)=tan(α+180°)

A more general idea

What is called phase shift in trigonometric functions is the same as the transformation of any real number function that moves the graph to the left or to the right. Thus, there is no need to think of this transformation as something peculiar to the trigonometric functions. It applies similarly to all kinds of functions.

Suppose $f$f is a function with its domain in the real numbers. Let $g$g be another function such that $g(x)=f(x+h)$g(x)=f(x+h) for all values of the domain variable $x$x, and where $h$h is a fixed positive number. Equivalently, $g(v-h)=f(v)$g(v−h)=f(v).

Consider a function value such that $f(v)=k$f(v)=k and $g(u)=k$g(u)=k. Then, $f(v)=g(u)$f(v)=g(u). But $f(v)=g(v-h)$f(v)=g(v−h). Hence, the function value attaind by $f$f at the point $v$v is attained by $g$g a distance $h$h to the left of $v$v.

Thus, all the function values attained by $f$f at points $x$x are attained by $g$g at points $x-h$x−h, which are a distance $h$h to the left of $x$x.

The conclusion to be drawn from this discussion is that the graph of $f(x+h)$f(x+h) should look the same as the graph of $f(x)$f(x) shifted $h$h units to the left.

Similarly, $f(x-h)$f(x−h) represents a shift to the right by $h$h units.

Since this applies to all real functions, it certainly applies to the trigonometric functions and therefore to the tangent function.

Example 1

Sketch the graph of $f(\theta)=\tan(\theta-45^\circ)$f(θ)=tan(θ−45°).

At what points $\theta$θ is the function $f$f undefined?

With graphing software, we obtain the following representation.

We see that the function $\tan\theta$tanθ has been moved to the right a distance $45^\circ$45°. Since $\tan\theta$tanθ is undefined at $\theta=90^\circ+180^\circ n$θ=90°+180°n for all integers $n$n, the undefined points for $\tan(\theta-45^\circ)$tan(θ−45°) must be $\theta=135^\circ+180^\circ n$θ=135°+180°n.

Example 2

Given a graph that looks identical to the graph of $\tan\alpha$tanα except that it has been shifted to the left by $30^\circ$30°, find the formula for the new function.

The shift is to the left. So, the function must be given by $\tan\left(\alpha+30^\circ\right).$tan(α+30°).

 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3