Graphs of Elementary Trigonometric Functions

Graphs of Elementary Trigonometric Functions

To see the functions in a clear manner, we’ll graph the trigonometric functions on the same graph as their reciprocals

This means that the sine curve and the cosecant curve are graphed on the same graph.

It also means that the cosine curve and the secant curve are graphed on the same graph.

And finally, it means that the tangent curve and the cotangent curve are graphed on the same graph.

Graph of:

f(x)=\sin x

g(x)=\csc x

We notice that the two functions meet at the maximum and minimum values.

The function g(x)=\csc x is not defined when \sin x=0. These are called the vertical asymptotes for the function g(x)=\csc x

Graph of:

f(x)=\cos x

g(x)=\sec x

We also notice that the two functions meet at the maximum and minimum values.

The function g(x)=\sec x is not defined when \cos x=0. These are again the vertical asymptotes for the function g(x)=\sec x

Graph of:

f(x)=\tan x

g(x)=\cot x

We also notice that the two functions intersect only when \sin x=\cos x.

These points are multiple of a single angle when x=\frac{\pi}{4}+n\pi. Each function has an asymptote when the other function is 0.

We can see in these graphs that each function resumes its pattern after some values of x. They are said to be periodic.

The repeated pattern is called a cycle.

Let’s get a summary of the general behavior of these functions

The “sine” function:

f(x)=A\sin (Bx+C)+D

With A,B,C and D being constants, we can see that we can easily graph this complex function using the base function of f(x)=\sin x and the technics we learned from the quadratic functions.

The Amplitude:

The amplitude A is the measure of the maximum or minimum values from the midline D.

The amplitude itself is always positive. |A|. The original \sin x is multiplied by A.

Period:

This is the horizontal width of a single cycle or wave. After this wave, the pattern repeats itself.

We can see that functions y=\sin x , y=\csc x, y=\cos x and y=\sec x repeat after  2\pi.

For the function:

y=A\sin (Bx+C)+D, we know that the period of the original/parent function y= \sin x is 2\pi

The period of this function will be:

period=\frac{2\pi}{|B|}

Frequency:

The frequency of the function is simply the reciprocal of the period.

frequency=\frac{1}{period}

The phase shift:

This is the horizontal shift of the original function and is calculated using:

phase\; shift=-\frac{C}{B}

The vertical shift:

The vertical shift is simply D

All these values will be calculated the same way when we graph the function:

f(x)=A\cos (Bx+C)+D

The period is useful when solving trigonometric equations.

For tangent and cotangent functions, we know that the period of the original/ parent function is \pi.

The phase shift is important for the tangent:

phase\; shift=-\frac{C}{B}. The first cycle begins at the ‘zero’ that is “Phase shift units to the right of the origin”. The asymptotes are at \frac{1}{2}P to the left of the beginning cycle, and another at \frac{1}{2}P to the right of the beginning cycle.

To graph y=A\tan (Bx+C):

For nonzero real numbers A and B

-The period is \frac{\pi}{|B|} and the phase shift is -\frac{C}{B}

Two consecutive asymptotes are found by solving:

-\frac{\pi}{2}<Bx+C<\frac{\pi}{2}

The graphing is similar for the cotangent.

Example1:

Find the amplitude and the period of the function y=3\sin {2x}

In this function:

A=3 and B=2

The period P=\frac{2\pi}{2}=\pi

Example2:

y=2\sin \frac{1}{2}x

Amplitude: A=2

Period: \frac{2\pi}{\frac{1}{2}}=4\pi

Example 3:

y=3\sin \left(2x+\frac{\pi}{2}\right)

A=3, B=2, c=-\frac{\pi}{2} and D=0.

The amplitude: A=3

The period: P=\frac{2\pi}{2}=\pi

Phase shift: \frac{\frac{-\pi}{2}}{2}=-\frac{\pi}{4}

The first interval is [-\frac{\pi}{4}, 3\frac{\pi}{4}] for a period of \pi.

Another way is to use the following rule:

The interval containing one cycle can be found by solving the following inequality:

0\leq Bx+C \leq 2\pi

We get:

0 \leq 2x+\frac{\pi}{2} \leq 2\pi

-\frac{\pi}{4} \leq x \leq \frac{3\pi}{4}

Example 4:

Find the amplitude, period and phase shift of the following equation:

y=2\cos (3x-\pi)

Amplitude:

A=2

Period: P=\frac{2\pi}{3}

Phase shift:

-\frac{-\pi}{3}=\frac{\pi}{3}

A sketch can be:

[\frac{\pi}{3},\pi]

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