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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 sine theta plus radical 2 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)

Part C: A toddler is jumping on another pogo stick whose length of its spring can be represented by the function g of theta equals 1 minus cosine squared theta plus radical 2 period At what times are the springs from the original pogo stick and the toddler's pogo stick lengths equal? (5 points)

Answer :

BWPENNEY27GO

Answer:

0

Step-by-step explanation:

So,0= the lengths of springs from the original pogo stick and the toddler's pogo stick are equal.

Hence, the values for different conditions were calculated with the help of trigonometric functions.

Part A

θ=kπ

Where k is an integer k=0,1,2..

Part B

θ=[tex]k\frac{\pi }{2}[/tex]

Where k is an integer k=0,1,2..

Part C

θ=kπ

Where k is an integer k=0,1,2..

What is trigonometric function?

The trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths

Part A

Given that,

f(x)=2sinθ+[tex]\sqrt{2}[/tex]

Non compressed length = [tex]\sqrt{2}[/tex]

Then

2sinθ+[tex]\sqrt{2}[/tex]=[tex]\sqrt{2}[/tex]

2sinθ=0

sinθ=0

Therefore θ is the integer multiple of π

θ=kπ

Where k is an integer k=0,1,2..

Part B

Here angle is doubled

Then,

2sin2θ+[tex]\sqrt{2}[/tex]=[tex]\sqrt{2}[/tex]

2sin2θ=0

sin2θ=0

Here θ will be the integer multiple of [tex]\frac{\pi }{2}[/tex]

θ=[tex]k\frac{\pi }{2}[/tex]

Where k is an integer k=0,1,2..

Part C

g(x)=1-[tex]cos^{2}[/tex]θ+[tex]\sqrt{2}[/tex]

Here

f(x)=g(x)

2sinθ+[tex]\sqrt{2}[/tex]=1-[tex]cos^{2}[/tex]θ+[tex]\sqrt{2}[/tex]

2sinθ=1-[tex]cos^{2}[/tex]θ

we know that [tex]1-cos^{2}[/tex]θ=[tex]sin^{2}[/tex]θ

2sinθ=[tex]sin^{2}[/tex]θ

sinθ=0

Therefore θ is the integer multiple of π

θ=kπ

Where k is an integer k=0,1,2..

Hence, we have solved that

Part A

θ=kπ

Where k is an integer k=0,1,2..

Part B

θ=[tex]k\frac{\pi }{2}[/tex]

Where k is an integer k=0,1,2..

Part C

θ=kπ

Where k is an integer k=0,1,2..

Learn more about Trigonometric functions here

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