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The difference in length of a spring on a pogo stick from its non-compressed length when a teenager is jumping on it after θ seconds can be described by the function f of theta equals 2 times cosine theta plus radical 3 period Part A: Determine all values where the pogo stick's spring will be equal to its non-compressed length. (5 points)Part B: If the angle was doubled, that is θ became 2θ, what are the solutions in the interval [0, 2π)? How do these compare to the original function? (5 points)

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Answer:

Hope this helps ^^

Explanation:

Part A:

To determine the values where the pogo stick's spring will be equal to its non-compressed length, we need to set the function f(θ) equal to zero since the non-compressed length is represented by zero in this case.

Given:

f(θ) = 2cos(θ) + √3

To find the values where f(θ) = 0, we solve the equation:

2cos(θ) + √3 = 0

Subtracting √3 from both sides:

2cos(θ) = -√3

Dividing both sides by 2:

cos(θ) = -√3/2

Using the unit circle or trigonometric values, we find that θ can take on the values of π/6 and 11π/6.

Therefore, the values where the pogo stick's spring will be equal to its non-compressed length are θ = π/6 and θ = 11π/6.

Part B:

If we double the angle, θ becomes 2θ. So the function becomes:

f(2θ) = 2cos(2θ) + √3

In the interval [0, 2π), we need to find the solutions for f(2θ) = 0.

Simplifying the equation:

2cos(2θ) + √3 = 0

Dividing the equation by 2:

cos(2θ) = -√3/2

Using the double-angle identity for cosine, we have:

cos(2θ) = 2cos^2(θ) - 1

Replacing cos(2θ) in the equation:

2cos^2(θ) - 1 = -√3/2

Multiplying the equation by 2:

4cos^2(θ) - 2 = -√3

Adding √3/2 to both sides:

4cos^2(θ) = -√3/2 + 1

Simplifying further:

cos^2(θ) = (-√3 + 2)/4

Taking the square root of both sides:

cos(θ) = ±√((-√3 + 2)/4)

Using the unit circle or trigonometric values, we can find the corresponding angles in the interval [0, 2π). These angles will be the solutions for f(2θ) = 0.

Comparing to the original function, the solutions for f(2θ) = 0 will have different values compared to the solutions for f(θ) = 0. The angles will be different due to the doubling of the angle, resulting in a different function.

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