Post on 07-Apr-2022
Trigonometric Functions Sine Function A periodic function repeats the same value at equal intervals of the independent variable (the pattern repeats). Ex: A sine function is a periodic function. Knowing this will help graph and/or solve sine functions. The basic rule of a sine function is:
π¦ = sin π₯
The period is the length of one cycle (2π)
The frequency is how often the cycle happens per unit time β the reciprocal of the period (1
2π)
The amplitude is the distance from the center to the highest (or lowest) point (1) The maximum is 1. The minimum is β1. So the range is [β1, 1] The domain is ] β β, β[ The zeros are integer multiples of π
The transformed sine function is π¦ = π sin(π(π₯ β β)) + π
The amplitude is βaβ and it is a vertical stretch and/or reflection over the x-axis.
π =ππππ₯βππππ
2
The frequency is βbβ and it is a horizontal stretch and/or reflection over the y-axis.
The period is calculated using ππππππ = 2π
|π|
βhβ is a horizontal translation (phase shift) βkβ is a vertical translation (h,k) is a βstartingβ point on the curve The maximum is πππ₯ = π + |π| and the minimum is πππ = π β |π| If ππ > 0, then start by going up from (h,k) If ππ < 0, then start by going down from (h,k)
Sketching a Sine Function To sketch a sine function:
1) Determine a, b, h, k 2) Find period, max, min, and starting point 3) Determine if going up or down from starting point 4) Use period and min/max to label graph 5) Plot starting point and additional points for one cycle 6) Sketch and graph additional cycles as necessary
Ex: Sketch the function π¦ = 2 sin π₯ π = 2, π = 1, β = 0, π = 0 πππ = π β |π|, πππ₯ = π + |π| πππ = 0 β |2|, πππ₯ = 0 + |2| πππ = β2, πππ₯ = 2
ππππππ =2π
1= 2π
π π‘πππ‘πππ πππππ‘: (0,0) ππ > 0 π π πππππππ πππ
Ex: Sketch the function π¦ = sin 2π₯ π = 1, π = 2, β = 0, π = 0 πππ = π β |π|, πππ₯ = π + |π| πππ = 0 β |1|, πππ₯ = 0 + |1| πππ = β1, πππ₯ = 1
ππππππ =2π
2= π
π π‘πππ‘πππ πππππ‘: (0,0) ππ > 0 π π πππππππ πππ
Ex: Sketch the function π¦ = 3 sin(1
2( π₯ β π)) + 1
π = 3, π =1
2, β = π, π = 1
πππ = π β |π|, πππ₯ = π + |π| πππ = 1 β |3|, πππ₯ = 1 + |3| πππ = β2, πππ₯ = 4
ππππππ =2π
1/2= 4π
π π‘πππ‘πππ πππππ‘: (π, 1π) ππ > 0 π π πππππππ πππ
Ex: Sketch the function π¦ = β2 sin (Ο
4( π₯ β 3)) β 2
π = β2, π =π
4, β = 3, π = β2
πππ = π β |π|, πππ₯ = π + |π| πππ = β2 β |β2|, πππ₯
= β2 + |β2| πππ = β2 β 4, πππ₯ = 0
ππππππ =2π
π/4= 8
π π‘πππ‘πππ πππππ‘: (3, β2) ππ < 0 π π πππππππ πππ
Solving Sine Functions To solve for y given x, plug in x and solve (in radians)
Ex: Given π(π₯) = 4 sin (π
6(π₯ + 1)) β 3, determine the initial value of π(π₯).
π(π₯) = 4 sin (π
6(π₯ + 1)) β 3
π(0) = 4 sin (π
6(0 + 1)) β 3
π(0) = 4 sin (π
6(1)) β 3
π(0) = 4 sin (π
6) β 3
π(0) = 4 (1
2) β 3
π(0) = 2 β 3
π(0) = β1 To solve for x given y:
1. Isolate sin(π(π₯ β β)) and let (π(π₯ β β)) = π if π β 1 ππ β β 0
2. Use CAST rule or unit circle to solve for π
3. Let (π(π₯ β β)) = π and solve for x
4. Solve for x over the given domain
Ex: Solve 2 sin π₯ β β3 = 0 over 0 β€ π β€ 2π
2 sin π₯ β β3 = 0
2 sin π₯ = β3
sin π₯ =β3
2
π₯ =π
3 πππ
2π
3
Ex: Solve 4 sin π₯ + 3 = 0 over 0 β€ π β€ 2π
4 sin π₯ + 3 = 0 4 sin π₯ = β3
sin π₯ = β3
4 in quadrants 3 and 4
ππ = 0.8481 π1 = π + 0.8481 = 3.9897
π2 = 2π β 0.8481 = 5.4351 So π₯ = 3.9897 πππ 5.4351
Ex: Solve sin π₯ + 6 = 2 0 over 0 β€ π β€ 2π
sin π₯ + 6 = 2 sin π₯ = β4
NO SOLUTION
Ex: Solve 3 sin(π₯ β π) + 3 = 0 over 0 β€ π β€ 8π
3 sin(π₯ β π) + 3 = 0 3 sin(π₯ β π) = β3 sin(π₯ β π) = β1
sin π = β1
π =3π
2
π₯ β π =3π
2
π₯ =5π
2
ππππππ =2π
|π|=
2π
1= 2π
So
π₯ =5π
2,5π
2+ 2π,
5π
2+ 4π
Also
π₯ =5π
2β 2π
Therefore:
π₯ =π
2,5π
2,9π
2,13π
2
Ex: Solve 6 sin(π(π₯ β 1)) = β3
6 sin(π(π₯ β 1)) = β3
sin(π(π₯ β 1)) = β1
2
sin π = β1
2
π =7π
6 πππ
11π
6
π(π₯ β 1) =7π
6 πππ π(π₯ β 1) =
11π
6
π₯ β 1 =7
6 πππ π₯ β 1 =
11
6
π₯ =13
6 πππ π₯ =
17
6
ππππππ =2π
|π|=
2π
π= 2
Therefore:
π₯ =13
6+ 2π πππ π₯ =
17
6+ 2π
Ex: Given π(π₯) = 4 sin (π
6(π₯ + 8)) + 4 over a domain of [0, 30]
For what values of x is π¦ = 6?
4 sin (π
6(π₯ + 8)) + 4 = 6
4 sin (π
6(π₯ + 8)) = 2
sin (π
6(π₯ + 8)) =
1
2
sin π =1
2
π =π
6 πππ
5π
6
π
6(π₯ + 8) =
π
6 πππ
π
6(π₯ + 8) =
5π
6
π₯ + 8 = 1 πππ π₯ + 8 = 5
π₯ = β7 πππ π₯ = β3
ππππππ =2π
|π|=
2π
π/6= 12
π₯ = β7 + 12 = 5 π₯ = 5 + 12 = 17
π₯ = 17 + 12 = 29
π₯ = β3 + 12 = 9 π₯ = 9 + 12 = 21
Therefore:
π₯ = 5, 9, 17, 21, 29
Sine Function Inequalities Solving sine function inequalities is very similar to solving sine functions. The only difference is we will need to look at a sketch (or use a test point) to determine the intervals for the solution. Steps:
1) Change inequality to equality 2) Solve as before 3) Use sketch or test point to determine solution interval 4) Solve over domain (or generalize if no domain is given)
Ex: Where is the function π(π₯) = β sin 3π₯ +1
2 greater than 0?
a) Over 0 β€ π₯ β€ 2π b) Over ββ β€ π₯ β€ β
SOLVE
β sin 3π₯ +1
2> 0
β sin 3π₯ +1
2= 0
β sin 3π₯ = β1
2
sin 3π₯ =1
2
sin π =1
2
π =π
6 πππ
5π
6
3π₯ =π
6 πππ 3π₯ =
5π
6
π₯ =π
18 πππ π₯ =
5π
18
SKETCH Use period to solve for all critical points.
ππππππ =2π
3=
12π
18
Consult sketch to determine possible intervals. SOLUTION
a) [0,π
18[ βͺ ]
5π
18,
13π
18[ βͺ ]
17π
18,
25π
18[ βͺ ]
29π
18, 2π]
b) ]5π
18+
12π
18π,
13π
18+
12π
18π[
(π
18, 0) (
5π
18, 0)
(13π
18, 0)
(17π
18, 0)
(25π
18, 0) (
29π
18, 0)
Finding the Rule of Sine Functions From a Graph To find the rule of a sine function from a graph:
1) Find the center line (k): max +min
2
2) Find h (use any point where π¦ = π)
3) Determine period (one full cycle) and b: 2π
ππππππ
4) Determine a: πππ₯ β π 5) Determine signs on a and b (does the function increase or decrease from h,k) 6) Write the function
Note: There are an infinite number of possibilities when writing the rule based on the h,k you choose and the signs on a and b. Ex: Determine the rule for the sine function graphed below.
π =β1 β β5
2= β3
Choose h at 2π
(β, π) = (2π, β3)
ππππππ = 4π
π =2π
4π=
1
2
|π| = πππ₯ β π = β1 β β3 = 2
Function decreases from (h,k) so a and b must have opposite signs
β΄ π(π₯) = 2 sin (β1
2(π₯ β 2π)) β 3
Finding the Rule of Sine Functions From Words Recall π¦ = π sin[π(π₯ β β)] + π Where:
a is the amplitude (a vertical stretch and/or reflection over the x-axis) π =ππππ₯βππππ
2
b is the frequency (a horizontal stretch and/or reflection over the y-axis) ππππππ = 2π
|π|
h is the horizontal translation/phase shift (the x-value when π¦ = π)
βkβ is the vertical translation π =ππππ₯+ππππ
2
(h,k) is a βstartingβ point on the curve The maximum is πππ₯ = π + |π| and the minimum is πππ = π β |π| If ππ > 0, then start by going up from (h,k) If ππ < 0, then start by going down from (h,k) Ex: Lobster fishermen dock at a small fishing harbor on the Bay of Fundy. The depth of the water varies according to the tides. A sinusoidal function can be used to predict the waterβs depth. The Bay of Fundy has the highest tidal range in the world. At low tide, the depth of the water is 2m. Six hours later, at high tide, the depth of the water is 14m. The fishermen leave from, and return to, the harbor when the depth of the water is at least 5 m. For how many hours can the fishermen be at sea between two consecutive low tides?
π =ππππ₯ β ππππ
2=
14 β 2
2= 6
ππππππ = 12 π π π =2π
12=
π
6
h: 12 hour period. If low tide is at π₯ = 0
β =12
4= 3
π =ππππ₯ + ππππ
2=
14 + 2
2= 8
Increasing to start, so a and b same signs
π(π₯) = 6 sin [π
6(π₯ β 3)] + 8
Now we can use the function to determine when the water depth is at least 5 m.
π(π₯) = 6 sin [π
6(π₯ β 3)] + 8
6 sin [π
6(π₯ β 3)] + 8 β₯ 5
6 sin [π
6(π₯ β 3)] + 8 = 5
6 sin [π
6(π₯ β 3)] = β3
sin [π
6(π₯ β 3)] = β
1
2
sin π = β1
2
π =7π
6 πππ
11π
6
π
6(π₯ β 3) =
7π
6 πππ
π
6(π₯ β 3) =
11π
6
(π₯ β 3) = 7 πππ (π₯ β 3) = 11 π₯ = 11 πππ π₯ = 14
Looking at our solutions, we know ? 2 = 11 πππ ? 3 = 14. We also know ? 1 =? 3 β πππ ππππππ = 2 The question states that the fishermen leave and return between two consecutive low tides, so they are never at sea during a low tide. means they must leave at ?1 (2) and return at ?2 (10). The fishermen are away for 8 hours.
Cosine Functions
A cosine function is a sine function shifted by π
2
cos π₯ = sin (π₯ +π
2) ππ cos π₯ = sin (π₯ β
3π
2)
sin π₯ = cos (π₯ βπ
2) ππ sin π₯ = cos (π₯ +
3π
2)
Basic function
π(π₯) = cos π₯
Domain: ]ββ, β[ Range: [β1, 1] Period: 2π Amplitude: 1 Initial Value: 1
Parameters (a, b, h, k) are the same as in the sine function: a: amplitude
b: 2π
ππππππ
h: horizontal translation (phase shift) k: vertical translation Whatβs new?
1) The sign of b has no effect on the graph since the graph is symmetrical over the y-axis. a. If π > 0, decreasing to βstartβ (βstartβ at max) b. If π < 0 increasing to βstartβ (βstartβ at min)
2) (h, k) is on the graph a. If βstartβ is max, then βstarting pointβ is (β, max) = h, k + |π|) b. If βstartβ is min, then βstarting pointβ is (β, min) = h, k β |π|)
Tangent Function
Recall that tan π₯ =πππ
πππ, but we also know that tan π₯ =
sin π₯
cos π₯
Basic Function
If we look at the unit circle, we find that tan π₯ = 1 ππ‘π
4 πππ
5π
4 and tan π₯ = β1 ππ‘
3π
4 πππ
7π
4
We also know that tan π₯ is 0 when sin π₯ = 0 (0 πππ π).
We can also determine that tan π₯ is undefined when cos π₯ = 0 (π
2πππ
3π
2). This means there
must be vertical asymptotes at π
2πππ
3π
2.
Key Information ππππππ = π
π·πππππ: β / {π
2+ ππ}
π ππππ: β
π΄π π¦πππ‘ππ‘ππ : π
2+ ππ
πππππ : π + ππ πΌπππ‘πππ ππππ’π: 0 No max, min, or amplitude
Transformed Tangent Function
π(π₯) = π tan(π(π₯ β β)) + π ππ π(π₯) = π (sin(π(π₯ β β))
cos(π(π₯ β β))) + π
If ππ > 0 the function is increasing If ππ < 0 the function is decreasing
ππππππ =π
|π|
π΄π π¦πππ‘ππ‘ππ : π πππ£π cos(π(π₯ β β)) = 0
Sketching a Tangent Function To sketch a tangent function:
β’ Find inflection point (β, π)
β’ Find period: π
|π|
β’ Find asymptotes: cos(π(π₯ β β)) = 0
β’ Find zeros
β’ Determine increasing/decreasing
β’ Sketch
Ex: Sketch π(π₯) = 2 tan(2(π₯ β π))
Inflection point (π, 0)
Period: π
2
Asymptotes: cos(2(π₯ β π)) = 0
2(π₯ β π) =π
2 and 2(π₯ β π) =
3π
2
(π₯ β π) =π
4 and (π₯ β π) =
3π
4
π₯ =5π
4 and x=
7π
4
So asymptotes at: π₯ =5π
4+
π
2π
Zeros: 2 tan(2(π₯ β π)) = 0
tan(2(π₯ β π)) = 0
sin(2(π₯ β π))
cos(2(π₯ β π))= 0
sin(2(π₯ β π)) = 0
2(π₯ β π) = 0 πππ 2(π₯ β π) = π
(π₯ β π) = 0 πππ (π₯ β π) =π
2
π₯ = π πππ π₯ =3π
2
So zeros at: π₯ = π +π
2π
ππ > 0 so increasing
Solving Tangent Functions We can add tangent values to the unit circle.
Just like the sine function and cosine function, we can use the unit circle to solve tangent functions, or we can solve for x using π‘ππβ1, using CAST.
πππ = 0 πππ = 0
πππ = π’ππππππππ
πππ = π’ππππππππ
πππ = 1
πππ = 1
πππ = β1
πππ = β1
πππ =β3
3
πππ =β3
3
πππ = ββ3
3
πππ =β3
3
πππ = β3
πππ = β3
πππ = ββ3
πππ = ββ3
Ex: Solve β3 tan (π
3(π₯ + 2)) = β3
β3 tan (π
3(π₯ + 2)) = β3
tan (π
3(π₯ + 2)) = β
β3
3
(π
3(π₯ + 2)) =
5π
6 πππ (
π
3(π₯ + 2))
=11π
6
π₯ + 2 = (5π
6) (
3
π) πππ π₯ + 2
= (11π
6) (
3
π)
π₯ + 2 = (5
2) πππ π₯ + 2 = (
11
2)
π₯ =1
2 πππ π₯ =
7
2
Period:
ππππππ =ππ3
= π (3
π) = 3
β΄ π₯ =1
2+ 3π
Ex: Solve 3 tan(β4(π₯ β 6)) = 1 when 6 β€ π₯ β€ 7
3 tan(β4(π₯ β 6)) = 1
tan(β4(π₯ β 6)) =1
3
tan(π) =1
3
ππ = 0.3218 π1 = 0.3218 π2 = π + 0.3218
π1 = 0.3218 π2 = 3.4633 β4(π₯ β 6) = 0.3218 πππ β 4(π₯ β 6) = 3.4633
(π₯ β 6) = β0.0805 πππ (π₯ β 6) = β0.8658 π₯ = 5.9195 πππ π₯ = 5.1342
ππππππ =π
4
β΄ π₯ = 6.7049
Solving Tangent Inequalities To solve inequalities:
1. Replace inequality with = 2. Solve 3. Consult sketch for solution set
Ex: Determine the intervals over which the function π(π₯) = 3 tan (2 (π₯ +π
2)) + 3 is positive.
3 tan (2 (π₯ +π
2)) + 3 β₯ 0
3 tan (2 (π₯ +π
2)) + 3 = 0
3 tan (2 (π₯ +π
2)) = β3
tan (2 (π₯ +π
2)) = β1
2 (π₯ +π
2) =
3π
4 πππ 2 (π₯ +
π
2) =
7π
4
(π₯ +π
2) =
3π
8 πππ (π₯ +
π
2) =
7π
8
π₯ = βπ
8 πππ π₯ =
3π
8
β΄ π(π₯)ππ πππ ππ‘ππ£π ππ£ππ [3π
8+
π
2π,
3π
4+
π
2π]
ππππππ =π
2
Asymptotes:
cos (2 (π₯ +π
2)) = 0
2 (π₯ +π
2) =
π
2 πππ 2 (π₯ +
π
2) =
3π
2
(π₯ +π
2) =
π
4 πππ (π₯ +
π
2) =
3π
4
π₯ = βπ
4 πππ π₯ =
π
4
ππ > 0 π π πππππππ πππ
Finding the Rule of a Tangent Function To find the rule of a tangent function:
1. Find the coordinates of the inflection point (h, k) and another point (x, y) 2. Use graph to determine period and solve for b 3. Substitute h, k, x, y, and b, then solve for a 4. Determine signs on a and b 5. Write rule
Ex: Find the rule of the tangent function graphed below.
Inflection point: (3, 2) Additional point: (0.5, 0) Period: 2.5 β 0.5 = 2
π =π
2
π(π₯) = π tan(π(π₯ β β)) + π
0 = π tan (π
2(0.5 β 3)) + 2
0 = π tan (π
2(β2.5)) + 2
β2 = π tan (β5π
4)
β2 = π(β1) π = 2
Increasing, so ππ > 0
β΄ π(π₯) = 2 tan (π
2(π₯ β 3)) + 2