Sketch Trigonometric Functions  tangent and cotangent
The sketching 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 = π
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 = π/2 + kπ , k = 0 , ~+mn~1, ~+mn~2, ...
Vertical asymptotes of y = cot(x) at x = kπ , k = 0 , ~+mn~1, ~+mn~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

Sketch the graph of y = tan(2x + π/2) over one period.
Solution
Graphing Parameters
range: (∞ , +∞)
Period = π/k = π/2
Vertical asymptotes are found by solving for x the equation: 2x + π/2 = π/2 + kπ which gives x = kπ/2 , k = 0 , ~+mn~1, ~+mn~2, ...
Horizontal Shift: Because of the term π/2, the graph is shifted horizontally. We first rewrite the given function as: y = tan [ 2( x + π/4)] and we can now write the shift as being equal to π/4 to the left.
We start by skeching tan(2 x) over one period from 0 to π/2 (blue graph below).
We then sketch y = tan [ 2( x + π/4)] translating the previous graph π/4 to the left (red graph below) so that the sketched period starts at  π/4 and ends at  π/4 + π/2 = π/4 which is an interval over one period equal to π/2.

Sketch the graph of y = cot(4x  π/4) over one period.
Solution
Graphing Parameters
range: (∞ , +∞)
Period = π/k = π/4
Vertical asymptotes are found by solving for x the equation: 4x  π/4 = kπ which gives x = (kπ + π/4) / 4 , k = 0 , ~+mn~1, ~+mn~2, ...
Horizontal Shift: Because of the term  π/4, the graph is shifted horizontally. We first rewrite the given function as: y = cot [ 4( x  π/16)] and we can now write the shift as being equal to π/16 to the right.
We start by skeching cot(4 x) over one period from 0 to π/4 (blue graph).
We then sketch y = cot [ 4( x  π/16)] translating the previous graph π/16 to the right (red graph below) so that the sketched period starts at π/16 and ends at π/16 + π/4 = 5π/16 which is one period.

More High School Math (Grades 10, 11 and 12)  Free Questions and Problems With Answers
More Middle School Math (Grades 6, 7, 8, 9)  Free Questions and Problems With Answers
More Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers
Author 
email
Home Page