# Graph tangent and cotangent

The sketching and graphing of the tangent and
cotangent
functions of the form

y = a tan [ k ( x - d) ] and y = a cot [ k ( x - d) ] are discussed with detailed examples.

## Graphing Parameters of y = tan(x) and y = cot(x)

range: (-? , +?)

Period = ?

Horizontal Shift (translation) = d , to the left if (- d) is positive and to the right if (- d) is negative.

Vertical asymptotes of y = tan(x) at x = ?/2 + k? , k = 0 , ±1, ±2, ...

Vertical asymptotes of y = cot(x) at x = k? , k = 0 , ±1, ±2, ...

We need to know how to sketch basic tangent and cotangent functions using the identities y = tan(x) = sin(x) / cos(x) and y = cot(x) = sin(x) / cos(x) to understand certain properties.

## 1) y = tan(x) = sin(x) / cos(x)

All zeros of the sin(x) are also zeros of tan(x) and all zeros of cos(x) (which is in the denominator) are vertical asymptotes of the tan(x) as shown below.

## 2) y = cot(x) = cos(x) / sin(x)

All zeros of the cos(x) are also zeros of cot(x) and all zeros of sin(x) (which is in the denominator) are vertical asymptotes of the cot(x) as shown below.

## Sketching tangent and cotangent Functions: Examples with Detailed Solutions

### Example 1

Sketch the graph of y = tan(2x + ?/2) over one period.

__Solution__

__Graphing Parameters__

range: (-? , +?)

Period = ?/|k| = ?/2

Vertical asymptotes are found by solving for x the equation: 2x + ?/2 = ?/2 + k? which gives x = k?/2 , k = 0 , ±1, ±2, ...

Horizontal Shift: Because of the term ?/2, the graph is shifted horizontally. We first rewrite the given function as: y = tan [ 2( x + ?/4)] and we can now write the shift as being equal to ?/4 to the left.

We start by skeching tan(2 x) over one period from 0 to ?/2 (blue graph below).

We then sketch y = tan [ 2( x + ?/4)] translating the previous graph ?/4 to the left (red graph below) so that the sketched period starts at - ?/4 and ends at - ?/4 + ?/2 = ?/4 which is an interval over one period equal to ?/2.

### Example 2

Sketch the graph of y = cot(4x - ?/4) over one period.

__Solution__

__Graphing Parameters__

range: (-? , +?)

Period = ?/|k| = ?/4

Vertical asymptotes are found by solving for x the equation: 4x - ?/4 = k? which gives x = (k? + ?/4) / 4 , k = 0 , ±1, ±2, ...

Horizontal Shift: Because of the term - ?/4, the graph is shifted horizontally. We first rewrite the given function as: y = cot [ 4( x - ?/16)] and we can now write the shift as being equal to ?/16 to the right.

We start by skeching cot(4 x) over one period from 0 to ?/4 (blue graph).

We then sketch y = cot [ 4( x - ?/16)] translating the previous graph ?/16 to the right (red graph below) so that the sketched period starts at ?/16 and ends at ?/16 + ?/4 = 5?/16 which is one period.

__More References and links__

Tangent Function

Cotangent Function

Properties of The Six Trigonometric Functions

High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers

Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers

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