The difference in length of a spring on a pogo stick from its non-compressed length when a teenager is jumping on it after $\theta$ seconds can be described by the function $y = 2\sin(\theta) - \sqrt{3}$ . Part A: Determine, algebraically, all values where the pogo stick's spring will be equal to its non-compressed length. Show all necessary calculations. ( $8$ points) Part B: If the angle was doubled, that is $\theta$ became $2\theta$ , what are the solutions in the interval $[0, 2\pi)$ ? Solve algebraically and show all necessary calculations. How do these compare to the original function? ( $9$ points) Part C: A toddler is jumping on another pogo stick whose length of their spring can be represented by the function $y = -\sqrt{3} \cos(2\theta)$ . At what times are the springs from the original pogo stick and the toddler's pogo stick lengths equal? Solve algebraically and show all necessary calculations. ( $8$ points)

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Algebra 2

Solve trigonometric equations I

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Q. The difference in length of a spring on a pogo stick from its non-compressed length when a teenager is jumping on it after

\theta

seconds can be described by the function

y = 2\sin(\theta) - \sqrt{3}

. Part A: Determine, algebraically, all values where the pogo stick's spring will be equal to its non-compressed length. Show all necessary calculations. (

8

points) Part B: If the angle was doubled, that is

\theta

became

2\theta

, what are the solutions in the interval

[0, 2\pi)

? Solve algebraically and show all necessary calculations. How do these compare to the original function? (

9

points) Part C: A toddler is jumping on another pogo stick whose length of their spring can be represented by the function

y = -\sqrt{3} \cos(2\theta)

. At what times are the springs from the original pogo stick and the toddler's pogo stick lengths equal? Solve algebraically and show all necessary calculations. (

8

points)

Identify Non-Compressed Length: Identify when the spring is at its non-compressed length, which means $y = 0$ . Set up the equation from Part A: $y = 2\sin(\theta) - \sqrt{3} = 0$ .
Solve for $\theta$ : Solve for $\theta$ : $2\sin(\theta) = \sqrt{3}$ . Then, $\sin(\theta) = \frac{\sqrt{3}}{2}$ .
Find $\theta$ Interval: Find $\theta$ in the interval $[0, \pi]$ since $\sin(\theta) = \frac{\sqrt{3}}{2}$ at $\theta = 60^\circ$ and $\theta = 120^\circ$ .
Replace $\theta$ with $2\theta$ : For Part B, replace $\theta$ with $2\theta$ in the original function: $y = 2\sin(2\theta) - \sqrt{3}$ . Set $y = 0$ for non-compressed length: $2\sin(2\theta) - \sqrt{3} = 0$ .
Solve for $2\theta$ : Solve for $2\theta$ : $2\sin(2\theta) = \sqrt{3}$ , so $\sin(2\theta) = \sqrt{3} / 2$ .
Find $2\theta$ Interval: Find $2\theta$ in the interval $[0, 2\pi): 2\theta = 60°, 120°, 420°, 480°$ . Divide by $2$ to find $\theta: \theta = 30°, 60°, 210°, 240°$ .