## Thursday, January 4, 2007

### Precalculus 4.6 notes - Graphs of Other Trigonometric Functions

4.6 Graphs of Other Trigonometric Functions

I. Graph of the Tangent Function: y = a tan (bx - c) + d

A. The tangent function is odd. tan (-x) = - tan x
B. Therefore the graph of the y = tan x is symmetric with respect to the origin.
C. Period = π
D. Domain: all the reals except x ≠ π /2 + n π
E. Range: all the reals
F. Vertical Asymptotes: x = π /2 + n π

II. To sketch the graph y = a tan (bx - c) + d
A. Locate key points that identify the intercepts and asymptotes.
1. to find the asymptotes:

bx - c = -π /2 and bx - c = π /2

2. The midpoint between two consecutive asymptotes is an x-intercept of the graph.

3. The period of the tangent function is the distance between two consecutive asymptotes. This can also be found by: period = π /b

4. The amplitude of a tangent function is not defined. But if “a” is a negative number, the graph is a reflection in the x-axis

Example 1: y = tan 2x

bx - c = -π /2 OR bx - c = π / 2
2x = -π /2 OR 2x = π /2
x = -π /4 OR x = π /4

(-π /2 + π /2)/2 = 0 so the midpoint is (0,0)

period = π /b = π /2

So therefore some of the key points are: (-π /2 , 0 ), (0,0) , (π /2 , 0)

and some of the asymptotes are x = -3π /4 , x = -π /4, x = π /4, x = 3π /4

Example 2: y = -3 tan (x/4)

bx - c = -π /2 OR bx - c = π /2
x/4 = -π /2 OR x/4 = π /2
x = -2π OR x = 4π

(-2π + 2π )/2 = 0 so the midpoint is (0,0)

period = π /b = π /(1/4) = 4π Or (2π --2π ) = 4π

So therefore some of the key points are: (-π , 3 ), (0,0) , (π , -3) and some of the asymptotes are
x = -2π , x = 2π

III. Graph of the Cotangent function y = cot x = (cos x)/(sin x)

has vertical asymptotes wherever sin x = 0 so... x = n π

A. The cotangent function is odd. cot (-x) = -cot x
B. Therefore the graph of the y = cot x is symmetric with respect to the origin.
C. Period = π
D. Domain: all the reals except x ≠ n π
E. Range: all the reals
F. Vertical Asymptotes: x = n π

Some key points are: (-7π/4 , 1), (-3π/2, 0), (-5π/4, -1), (-3π/4, 1), (-π/2, 0), (-π/4, -1)

IV. Cosecant function is y = csc x = 1/(sin x)
so to sketch the graph of the cosecant function, first make a sketch of the sine curve, then take the reciprocals of the y-coordinates to obtain the points needed.

The vertical asymptotes are where sin x = 0 so therefore at x = nπ

A. The cosecant function is odd. csc (-x) = -csc x
B. Therefore the graph of the y = csc x is symmetric with respect to the origin.
C. Period = 2π
D. Domain: all the reals except xn π
E. Range: all the reals except y ≠ [-1,1]
F. Vertical Asymptotes: x = n π

V. Secant function is y = sec x = 1/(cos x)
so to sketch the graph of the secant function, first make a sketch of the cosine curve, then take the reciprocals of the y-coordinates to obtain the points needed.

The vertical asymptotes are where cos x = 0 so therefore at x = π/2 + nπ

A. The secant function is even. sec (-x) = sec x
B. Therefore the graph of the y = sec x is symmetric with respect to the y - axis.
C. Period = 2π
D. Domain: all the reals except x ≠ π/2 + n π
E. Range: all the reals except y ≠ [-1,1]
F. Vertical Asymptotes: x = π/2 + n π

VI. Damped Trigonometric Graphs:

A product of 2 functions can be graphed using properties of the individual functions.

Example 1: f(x) = e-x cos (x)

key points are (-3 π /2 , 0), (-3.93, -35.89), (-π/2 , 0), (0,1), (π/2 , 0)

The damping factor is e-x because:

-e-x ≤ e-x cos x ≤ e-x

As the y-values approach zero, the x-values approach infinity.

Example: h(x) = 2(-x/4) sin x

graph this and see what happens as x approaches zero and as y approaches zero.

Homework: #33; pg. 341; #1 - 17 odd, 41, 43, 47, 49, 51-55, 57, 63 - 67 odd, 75,
pg. 330; #12, 15, 21, 29, 45, 99, 101