Mathematics High School

30 POINTS!!! ANSWER ASAP!!
A person is on the outer edge of a carousel that is rotating counterclockwise. Using the unit circle to model the carousel, what is the exact position of the rider after the carousel rotates π/12 radians?

30 POINTS!!! ANSWER ASAP!!A Person Is On The Outer Edge Of A Carousel That Is Rotating Counterclockwise.

Answers

Answer 1

Answer:

the last answer or d....

Step-by-step explanation:

30 POINTS!!! ANSWER ASAP!!A Person Is On The Outer Edge Of A Carousel That Is Rotating Counterclockwise.
Answer 2

The exact position of the rider after the carousel rotates π/12 radians is option D ([tex]\frac{\sqrt{2} +\sqrt{6} }{4}[/tex], [tex]\frac{\sqrt{6} -\sqrt{2} }{4}[/tex]).

What is Unit Circle?

Unit circle is a circle whose radius is 1 unit.

The coordinates of any point on the unit circle which rotates to θ to anticlockwise direction is x = r cos θ and y = r sin θ, where r is the radius.

Since r = 1 for unit circle, x = cos θ and y = sin θ.

Here θ = π/12 radians

So the exact position of the rider after the carousel rotates π/12 radians is,

(cos π/12, sin π/12).

We know, π/12 = π/3 - π/4

Using the trigonometric identities,

sin (x - y) = sin x cos y - cos x sin y

cos (x - y) = cos x cos y + sin x sin y

we have,

cos π/12 = cos (π/3 - π/4)

              = cos (π/3) cos (π/4) + sin (π/3) sin (π/4)

              = ([tex]\frac{1}{2}[/tex] × [tex]\frac{1}{\sqrt{2} }[/tex] ) + ([tex]\frac{\sqrt{3} }{2}[/tex] × [tex]\frac{1}{\sqrt{2} }[/tex] )

              = [tex]\frac{1+\sqrt{3} }{2\sqrt{2} }[/tex]

Multiplying throughout by √2,

cos π/12 = [tex]\frac{\sqrt{2} +\sqrt{6} }{4}[/tex]

Similarly,

sin π/12 = sin (π/3 - π/4)

             = sin(π/3) cos(π/4) - cos(π/3) sin(π/4)

             = ([tex]\frac{\sqrt{3} }{2}[/tex] × [tex]\frac{1}{\sqrt{2} }[/tex] ) - ([tex]\frac{1}{2}[/tex] × [tex]\frac{1}{\sqrt{2} }[/tex] )

             = [tex]\frac{\sqrt{6} -\sqrt{2} }{4}[/tex]

Hence the position of the rider is ([tex]\frac{\sqrt{2} +\sqrt{6} }{4}[/tex], [tex]\frac{\sqrt{6} -\sqrt{2} }{4}[/tex]).

                   

Learn more about Unit Circles here :

https://brainly.com/question/30591251

#SPJ2

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Answers

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