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The difference in length of a spring on a pogo stick from its non-compressed length wen a teenager is jumping on it after θ seconds can be described by the function f of theta equals 2 times sine theta plus radical 3 period Part A: Determine hall 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) Part C: A toddler is jumping on another pogo stick whose length of their spring can be represented by the function g of theta equals 1 minus cosine squared theta plus radical 3 period At what times are the springs from the original pogo stick and the toddler's pogo stick lengths equal? (5 points)

Sagot :

Answer:

Refer to your notes from module 6 on this.

Step-by-step explanation:

Complete this test question just like you did the practice question from the module. Remember your test is an open note test, but it is not a copy from the internet test. You can do it!!!

AshishdwivedilVTidn

The values of [tex]\theta[/tex] for different conditions were calculated with the help of trigonometric functions.

What are the solutions to trigonometric functions?

The solution of trigonometric functions is the values of angles where the function value becomes zero.

Given function is:

A)[tex]f(\theta ) = 2Sin\theta +\sqrt{3}[/tex]

[tex]f(\theta ) = 2Sin\theta +\sqrt{3} = 0\\2Sin\theta = -\sqrt{3} \\Sin\theta =-\frac{\sqrt{3} }{2}[/tex]

[tex]\theta =n\pi +\frac{\pi }{3}[/tex] where n=1,3,5,7......

So, at [tex]\theta =n\pi +\frac{\pi }{3}[/tex]  pogo stick's spring will be equal to its non-compressed length.

B)If the angle was doubled, the function will look like this,

[tex]f(\theta ) = 2Sin2\theta +\sqrt{3}[/tex]

[tex]f(\theta ) = 2Sin2\theta +\sqrt{3} =0[/tex]

[tex]2Sin2\theta +\sqrt{3} = 0\\2Sin2\theta = -\sqrt{3} \\Sin2\theta =-\frac{\sqrt{3} }{2}[/tex]

[tex]2\theta =n\pi +\frac{\pi }{3}\\\theta = \frac{n\pi }{2} +\frac{\pi }{6}[/tex]where  n=1,3,5,7....

if n=1, [tex]\theta =\frac{2\pi }{3}[/tex]

n=2, [tex]\theta =\frac{7\pi }{6}[/tex]

n=3, [tex]\theta =\frac{5\pi }{3}[/tex]

So,  [tex]\theta =\frac{2\pi }{3}, \frac{7\pi }{6} ,\frac{5\pi }{3}[/tex] will be solution in [0,2π}.

C)[tex]g(\theta) =1-cos^{2} \theta +\sqrt{3}[/tex]

[tex]g(\theta) =Sin^{2} \theta + \sqrt{3}[/tex]

[tex]g(\theta) =f(\theta)[/tex]

[tex]sin^{2} \theta +\sqrt{3} = 2Sin\theta+\sqrt{3}[/tex]

[tex]Sin^2\theta = 2Sin\theta[/tex]

[tex]Sin^2\theta-2Sin\theta = 0\\Sin\theta(Sin\theta-2) =0\\Sin\theta =2(NotPOSSIBLE)\\Sin\theta =0[/tex]

[tex]\theta=n\pi[/tex] where n=1,2,3.....

So, [tex]\theta=n\pi[/tex], the lengths of springs from the original pogo stick and the toddler's pogo stick are equal.

Hence, the values of [tex]\theta[/tex] for different conditions were calculated with the help of trigonometric functions.

To get more about trigonometric functions visit:

https://brainly.com/question/2291993

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