Please show your work
1. If sin 0= √2/2 Which could not be the value of 0
225∘
135∘
12∘
2. A gazebo is located in the center of a large, circular lawn with a diameter of 200 feet. Straight paths extend from the gazebo to a sidewalk around the lawn. If two of the paths form a 75° angle, how far would you have to travel around the sidewalk to get from one path to the other? Round your answer to the nearest foot if necessary.
a. 183 ft
b. 262 ft
c. 131 ft
d. 3,125 ft
3. A sine function can be used to model light waves. Green light has a wavelength, or period, of about 510 nanometers (nm). Which equations best model green light?
A y = sin π/225θ
By = sin 510/π θ
1. We know that sin 45° = √2/2, so if sin 0 = √2/2, then 0 must be a quadrantal angle (0°, 90°, 180°, or 270°). Therefore, 12° could not be the value of 0.
2. We can use the Law of Cosines to find the distance around the sidewalk:
c^2 = a^2 + b^2 - 2ab cos C
where c is the distance around the sidewalk, a and b are the lengths of the two paths, and C is the angle between them (75°). Since the two paths are equal, we can let a = b = r (the radius of the lawn).
c^2 = 2r^2 - 2r^2 cos 75°
c^2 = 2r^2 - r^2 √3 + r^2 / 2
c^2 = 3r^2 / 2 - r^2 √3
c^2 = r^2 (3/2 - √3)
c = r √(3/2 - √3)
Substituting the diameter (200 feet) for 2r, we get:
c = 100 √(3/2 - √3) ≈ 131 feet
So the answer is c.
3. The equation for a sine function is y = A sin(Bx - C) + D, where A is the amplitude, B is the period (in radians), C is the phase shift, and D is the vertical shift.
Since the period of green light is 510 nm, or 5.10 x 10^-7 meters, we can use the formula for the period of a sine function:
period = 2π / B
5.10 x 10^-7 = 2π / B
B = 2π / 5.10 x 10^-7 ≈ 1.23 x 10^7
So the equation for green light would be:
y = A sin (1.23 x 10^7θ - C) + D
Since we're not given any other information about the function (such as its amplitude, phase shift, or vertical shift), neither of the given equations would be the best model for green light.
Please show your work
12. Find the period, range, and amplitude of the cosine function. Y= -4 cos 8x
13. For a particular angle θ the cosine function f(x) = c cos b (θ) has the following values within one cycle of the function.
f (0) = 2 f(π/4) = 0, f(π./2) = -2, f (3π/4) = 0 f(π) = 2
12. The general form of a cosine function is y = A cos(Bx - C) + D, where A is the amplitude, B is the period (in radians), C is the phase shift, and D is the vertical shift.
In the given function, y = -4 cos 8x, we can see that B = 8. To find the period, we use the formula:
period = 2π / B
period = 2π / 8 = π/4
So the period is π/4.
To find the amplitude, we see that A = -4. Since the cosine function oscillates between 1 and -1, carrying a minus sign implies that the cosine function is being reflected in the x-axis, so the amplitude is 4.
To find the range, we need to find the minimum and maximum values of the function. Since the amplitude is 4 and the function is being reflected across the x-axis, the maximum value is 4 and the minimum value is -4. So the range is [-4, 4].
Therefore, the period is π/4, the amplitude is 4, and the range is [-4, 4].
13. We're given a cosine function f(x) = c cos b(θ) and some values of the function within one cycle.
At θ = 0, f(0) = c cos b(0) = 2
At θ = π/4, f(π/4) = c cos b(π/4) = 0
At θ = π/2, f(π/2) = c cos b(π/2) = -2
At θ = 3π/4, f(3π/4) = c cos b(3π/4) = 0
At θ = π, f(π) = c cos b(π) = 2
Since we have the values of cosine for each of these angles within one cycle of the function, we can determine the values of b and c.
At θ = 0 and θ = π, the cosine function equals 2, so:
c cos b(0) = 2
c cos b(π) = 2
Since cosine has a period of 2π, these two equations are equivalent to:
c cos 0 = 2
c cos π = 2
c = ±2
If c = -2, then we have the function f(x) = -2 cos b(θ). Plugging in the values of π/4 and π/2, we get:
At θ = π/4, f(π/4) = -2 cos b(π/4) = 0
At θ = π/2, f(π/2) = -2 cos b(π/2) = -2
These values are inconsistent with the given values, so c cannot be -2.
Therefore, c = 2, and we have the function f(x) = 2 cos b(θ). Plugging in the values of π/4 and π/2, we get:
At θ = π/4, f(π/4) = 2 cos b(π/4) = 0
At θ = π/2, f(π/2) = 2 cos b(π/2) = -2
We can solve for b using these two equations:
2 cos b(π/4) = 0
2 cos b(π/2) = -2
cos b(π/4) = 0
cos b(π/2) = -1
The only angle between 0 and 2π whose cosine is 0 is π/2, so b(π/4) = π/2. We can use this to solve for b:
cos (π/2) = cos b(π/4)
b(π/4) = arcsin(0) or b(π/4) = π - arcsin(0)
b = 2nπ + π/2 or b = 2nπ + 3π/2, where n is an integer.
So the function f(x) = 2 cos[(2nπ + π/2)θ] or f(x) = 2 cos[(2nπ + 3π/2)θ] satisfies the given conditions.
13. For a particular angle θ the cosine function f(x) = c cos b (θ) has the following values within one cycle of the function.
f (0) = 2 f(π/4) = 0, f(π./2) = -2, f (3π/4) = 0 f(π) = 2
We can use the given function and values to find the amplitude, period, and standard form of the cosine function.
The general form of a cosine function is y = A cos(Bx - C) + D, where A is the amplitude, B is the period (in radians), C is the phase shift, and D is the vertical shift.
Since we're given values for the cosine function at several angles within one cycle, we can use these to find the amplitude and period of the function.
f(0) = c cos b(0) = 2
f(π/2) = c cos b(π/2) = -2
Since cosine reaches its maximum value of 1 at 0 radians and its minimum value of -1 at π/2 radians, we can see that the amplitude of the function is 2.
f(0) = c cos b(0) = 2
f(π) = c cos b(π) = 2
Since cosine has a period of 2π, these two values tell us that the period of the function is 2π.
To determine the phase shift and vertical shift, we need to write the function in standard form. We can do this by using the general form of the cosine function:
y = A cos(Bx - C) + D
Since the amplitude is 2, we have A = 2.
Since the period is 2π, we have B = 2π / (2π) = 1.
To find C, we need to know the phase shift. We can find this by looking at the given value of f(0). Since the cosine function is at its maximum value of 2 at θ = 0, we know that the phase shift is 0. Therefore, C = 0.
To find D, we can use the midpoint formula:
D = (f(0) + f(π)) / 2 = (2 + 2) / 2 = 2
So the standard form of the function is:
f(x) = 2 cos(x) + 2
What is the value of tan (-5π/6)
A √3/3
B 1
C 1/2
We can use the unit circle to find the value of tan(-5π/6).
First, we need to determine the point on the unit circle that corresponds to the angle of -5π/6. We know that the angle is in the third quadrant, so the x-coordinate will be negative and the y-coordinate will be negative.
Next, we can use the Pythagorean theorem to find the radius of the unit circle:
r^2 = (-1)^2 + (-1/2)^2
r^2 = 1 + 1/4
r^2 = 5/4
r = √5/2
So the point on the unit circle is (-√5/2, -1/2).
Finally, we can use the definition of the tangent function:
tan(-5π/6) = y/x
tan(-5π/6) = (-1/2) / (-√5/2)
tan(-5π/6) = √5/5
Therefore, the answer is not among the options given. The correct answer is √5/5.
Show your work
18. What is the value of cos π/6
√3/2
√3/3
We can start by using the unit circle to find the exact value of cos(π/6).
Recall that the unit circle is a circle with a radius of 1 centered at the origin of the coordinate plane. To evaluate trigonometric functions using the unit circle, we draw a right triangle whose hypotenuse lies on the unit circle and whose other two sides lie along the x- and y-axes.
We want to find the cosine of the angle π/6, which corresponds to the acute angle between the positive x-axis and the line passing through the origin and the point on the unit circle that intersects the terminal side of the angle.
Since this angle is π/6, we know that the coordinates of the point on the unit circle are (cos(π/6), sin(π/6)) = (x, y), where x is the value of the cosine and y is the value of the sine.
To find x, we can draw the right triangle as shown:
```
|\
| \
| \
| \
hyp | \ opp
| \
| \
|_______\
adj
```
From the triangle, we can see that the adjacent side has length 1/2 and the hypotenuse has length 1 (since the point lies on the unit circle). Using the Pythagorean theorem, we can find the length of the opposite side:
opp² + (1/2)² = 1²
opp² = 1 - (1/4)
opp = √3/2
Therefore, cos(π/6) = adj/hyp = 1/2.
So the answer is not among the options given. The exact value of cos(π/6) is 1/2.