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| 13 Theory slides |
| 9 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Trigonometric Function | Reciprocal Trigonometric Function |
---|---|
y=sinx | y=sinx1 |
y=cosx | y=cosx1 |
y=tanx | y=tanx1 |
This lesson will explore the graphs of these reciprocal trigonometric functions.
Here are a few recommended readings before getting started with this lesson.
A massive communication tower is anchored to the ground with wires.
Let P be the point of intersection of the terminal side of an angle in standard position and the unit circle. The cotangent function, denoted by cot, is defined as the ratio of the x-coordinate to the y-coordinate of P.
cotθ=sinθcosθ
Since division by 0 is not defined, the graph of the parent cotangent function y=cotx has vertical asymptotes where sinx=0. This means that the graph has vertical asymptotes at every multiple of π. The graph of y=cotx can be drawn by making a table of values.
Consider now the general form of a cotangent function.
y=acotbx
Here, a and b are non-zero real numbers and x is measured in radians. The properties of the cotangent function are stated below.
Properties of y=acotbx | |
---|---|
Amplitude | No amplitude |
Number of Cycles in [0,2π] | 2∣b∣ |
Period | ∣b∣π |
Domain | All real numbers except multiples of ∣b∣π |
Range | All real numbers |
Asymptotes occur at the end of each cycle. Therefore, the given function has asymptotes at θ=0 and θ=2π.
Divide the period into fourths and locate the three equidistant points between the asymptotes. The period for this function goes from 0 to 2π, so a table of values will be made for θ=2π, θ=π, and θ=23π.
θ | 3cot(21θ) | y |
---|---|---|
2π | 3cot(21⋅2π) | 3 |
π | 3cot(21⋅π) | 0 |
23π | 3cot(21⋅23π) | -3 |
The points found in the table are (2π,3), (π,0), and (23π,-3).
Finally, the points can be connected with a smooth curve to draw the graph for one cycle.
Once the graph for one cycle is drawn, it can be replicated as many times as desired to draw more cycles. Here, another cycle is graphed.
Find the period and cycle, then graph the asymptotes and plot some points. Finally, sketch the curve.
Next, divide the period into fourths and locate three equidistant points between the asymptotes. Since a period goes from 0 to 2π, a table of values will be made for x=8π, x=4π, and x=83π.
x | 21cot2x | y |
---|---|---|
8π | 21cot2(8π) | 21 |
4π | 21cot2(4π) | 0 |
83π | 21cot2(83π) | -21 |
The points found in the table are (8π,21), (4π,0), and (83π,-21). These three points can be plotted on the plane.
The points are then connected with a smooth curve to graph one period of the function.
Finally, the cycle can be replicated as many times as desired. In this case, the graph will be drawn for values of x between 0 and 2π.
Let P be the point of intersection of the terminal side of an angle in standard position and the unit circle. The secant function, denoted as sec, is defined as the reciprocal of the x-coordinate of P.
secθ=cosθ1
Since division by 0 is not defined, the graph of the parent secant function y=secx has vertical asymptotes where cosx=0. This means that the graph has vertical asymptotes at odd multiples of x=2π. The graph of y=secx can be drawn by making a table of values.
Consider the general form of a secant function.
y=asecbx
Here, a and b are non-zero real numbers and x is measured in radians. The properties of the secant function are be stated in the table below.
Properties of y=asecbx | |
---|---|
Amplitude | No amplitude |
Number of Cycles in [0,2π] | ∣b∣ |
Period | ∣b∣2π |
Domain | All real numbers except odd multiples of 2∣b∣π |
Range | (-∞,-∣a∣]∪[∣a∣,∞) |
Here, the period is 4π and the asymptotes occur every 2π radians. Therefore, the asymptotes are located not only at the beginning and at the end of each period, but also in the middle of each period. Divide the interval between two asymptotes in fourths by the following pattern.
Notice that the middle point in the pattern, the maximum or minimum point, was already plotted in the previous step. The remaining four points can be found for the interval that goes from -π to 3π by making a table of values.
θ | 2sec21θ | y |
---|---|---|
-2π | 2sec(21(-2π)) | 22≈2.83 |
2π | 2sec(21⋅2π) | 22≈2.83 |
23π | 2sec(21⋅23π) | -22≈-2.83 |
25π | 2sec(21⋅25π) | -22≈-2.83 |
Finally, the points can be connected with a smooth curve to draw the graph for one cycle.
Once the graph for one cycle is drawn, it can be replicated as many times as desired to draw more cycles. Here, one more cycle will be graphed.
Graph the related cosine function and recall that the asymptotes of the secant function occur at the zeros of the cosine function.
x | 2secx | y |
---|---|---|
-45π | 2sec(-45π) | -22≈-2.83 |
-43π | 2sec(-43π) | -22≈-2.83 |
-4π | 2sec(-4π) | 22≈2.83 |
4π | 2sec4π | 22≈2.83 |
43π | 2sec43π | -22≈-2.83 |
45π | 2sec45π | -22≈-2.83 |
47π | 2sec47π | 22≈2.83 |
49π | 2sec49π | 22≈2.83 |
The points found in the table can now be plotted. Finally, each set of points can be connected with smooth curves.
The graph of the curve of the bowl has been drawn.
Let P be the point of intersection of the terminal side of an angle in standard position and the unit circle. The cosecant function, denoted as csc, is defined as the reciprocal of the y-coordinate of P.
cscθ=sinθ1
Since division by 0 is not defined, the graph of the parent cosecant function y=cscx has vertical asymptotes where sinx=0. This means that the graph has vertical asymptotes at multiples of π. The graph of y=cscx can be drawn by making a table of values.
Consider the general form of a cosecant function.
y=acscbx
Here, a and b are non-zero real numbers and x is measured in radians. The properties of the cosecant function are stated below.
Properties of y=acscbx | |
---|---|
Amplitude | No amplitude |
Number of Cycles in [0,2π] | ∣b∣ |
Period | ∣b∣2π |
Domain | All real numbers except multiples of ∣b∣π |
Range | (-∞,-∣a∣]∪[∣a∣,∞) |
Here, the period is 4π and the asymptotes occur every 2π radians. Therefore, the asymptotes are located not only at the beginning and at the end of each period, but also in the middle of each period. Divide the interval between two asymptotes in fourths by the following pattern.
Notice that the middle point, which is the maximum or minimum point, was already plotted in the previous step. Four more points can be found for the interval that goes from -2π to 2π by making a table of values.
θ | 2csc21θ | y |
---|---|---|
-23π | 2csc(21(-23π)) | -22≈-2.83 |
-2π | 2csc(21(-2π)) | -22≈-2.83 |
2π | 2csc(21⋅2π) | 22≈2.83 |
23π | 2csc(21⋅23π) | 22≈2.83 |
Finally, each set of points can be connected with a smooth curve to draw the graph for one cycle.
Once the graph for one cycle is drawn, it can be replicated as many times as desired to draw more cycles. Another cycle is graphed below.
Graph the related sine function and recall that the asymptotes of the cosecant function occur at the zeros of the sine function.
x | 21cscx | y |
---|---|---|
-35π | 21csc(-35π) | 33≈0.58 |
-34π | 21csc(-34π) | 33≈0.58 |
-32π | 21csc(-32π) | -33≈-0.58 |
-3π | 21csc(-3π) | -33≈-0.58 |
3π | 21csc3π | 33≈0.58 |
32π | 21csc32π | 33≈0.58 |
34π | 21csc34π | -33≈-0.58 |
35π | 21csc35π | -33≈-0.58 |
The points found in the table can now be plotted. Finally, connect the sets of points with smooth curves.
Find the period of the following functions. Round the answers to two decimal places.
The challenge presented at the beginning can be solved with the topics covered in this lesson. It was given that a massive communication tower is anchored to the ground with wires.
Graph the cosecant function and pay close attention to the y-value for θ=4π.
θ | 5cscθ | y |
---|---|---|
-43π | 5csc(-43π) | -52≈-7.07 |
-4π | 5csc(-4π) | -52≈-7.07 |
4π | 5csc4π | 52≈7.07 |
43π | 5csc43π | 52≈7.07 |
Finally, the points found in the table are be plotted and each set of points connected with smooth curves.
Now that the function has been graphed, the point at θ=4π can be located and its y-coordinate identified.
The exact y-coordinate cannot be found in the graph, but its nearest integer can be identified as 7. Therefore, the length of the wire that makes an angle of 4π radians with the ground is about 7 meters.
Let's start by recalling that the period of a cotangent function is the quotient of π and the absolute value of b. Period: π/|b| We know that the period of the given function is 12. Therefore, the above quotient is equal to 12.
We found that the absolute value of b is 2π. Therefore, b can be either - 2π or 2π. |b|=2π ↙ ↘ b=- 2π or b=2π
Order the functions from the least average rate of change to the greatest average rate of change over the interval -4π≤x≤4π.
We are asked to order the given functions with respect to their average rate of change over the interval - π4 ≤ x ≤ π4. First, let's recall the general formula for the average rate of change. Average Rate of Change = f(x_2)-f(x_1)/x_2-x_1 Here, x_1 and x_2 are the endpoints of the interval, and f(x_1) and f(x_2) are the values of the function at these points. We are going to calculate the average rate of change for each function one at a time.
We want to evaluate the average rate of change over the interval - π4 ≤ x ≤ π4. To do that, let's first add the endpoints of the interval to the graph.
To calculate the average rate of change in the desired interval, we will use the points (- π4, - 2) and ( π4,2). We can substitute the coordinates of these point into the formula for the average rate of change and then evaluate the resulting numeric expression.
The average rate of change of the function in choice A is 8π.
By following the same steps, we can evaluate the average rate of change of the remaining functions.
Endpoints | Substitute | Evaluate | |
---|---|---|---|
Function A | ( - π/4, - 2) and ( π/4, 2) | 2-( - 2)/π4-( - π4) | 8/π≈ 2.55 |
Function B | ( - π/4, 1) and ( π/4, - 1) | - 1- 1/π4-( - π4) | - 4/π≈ - 1.27 |
Function C | ( - π/4, - 1/2) and ( π/4, 1/2) | 12-( - 12)/π4-( - π4) | 2/π≈ 0.64 |
Function D | ( - π/4, 2) and ( π/4, - 2) | - 2- 2/π4-( - π4) | - 8/π≈ -2.55 |
We can now order the functions from the least average rate of change to the greatest average rate of change.
Average Rate of Change | ||||
---|---|---|---|---|
Value | - 8/π | - 4/π | 3/π | 8/π |
Function | D | B | C | A |
Ramsha made a mistake when finding the period of y=cot5x.
The period of a cotangent function with the form y=a cot bx is the quotient of π and |b|. Cotangent Function & Period y = a cot bx & π/| b| In our function, we know that b=5. Let's substitute this number in the above formula to find the period of y=cot 5x.
The period of the function is π5. The mistake Ramsha made was using the incorrect formula. She thought that the period of a cotangent function was 2π|b| instead of π|b|.