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stepan [7]
3 years ago
12

The point (Negative StartFraction StartRoot 2 EndRoot Over 2 EndFraction, StartFraction StartRoot 2 EndRoot Over 2 EndFraction)

is the point at which the terminal ray of angle Theta intersects the unit circle. What are the values for the cosine and cotangent functions for angle Theta?
Mathematics
2 answers:
kotegsom [21]3 years ago
8 0

Answer:

A

Step-by-step explanation:

sveticcg [70]3 years ago
4 0

The values of cosine Ф and cotangent Ф are \frac{-\sqrt{2} }{2} and -1

Step-by-step explanation:

When a terminal side of an angle intersect the unit circle at

point (x , y), then:

  • The x-coordinate is equal to cosine the angle between the positive part of x-axis and the terminal side
  • The y-coordinate is equal to sine the angle between the positive part of x-axis and the terminal side
  • If x and y coordinates are positive, then the angle lies in the 1st quadrant
  • If x-coordinate is negative and y-coordinate is positive, then the angle lies in the 2nd quadrant
  • If x and y coordinates are negative, then the angle lies in the 3rd quadrant
  • If x-coordinate is positive and y-coordinate is negative, then the angle lies in the 4th quadrant

∵ The terminal ray of angle Ф intersects the unit circle at point (\frac{-\sqrt{2} }{2},\frac{\sqrt{2} }{2})

- According to the 1st and 2nd notes above

∴ cosФ = x-coordinate of the point

∴ sinФ = y-coordinate of the point

∵ The x-coordinate of the point is negative

∵ They-coordinate of the point is positive

- According the the 4th note above

∴ Angle Ф lies in the 2nd quadrant

∵ x-coordinate = \frac{-\sqrt{2} }{2}

∴ cosФ = \frac{-\sqrt{2} }{2}

∵ y-coordinate = \frac{\sqrt{2} }{2}

∴ sinФ = \frac{\sqrt{2} }{2}

- cotФ is the reciprocal of tanФ

∵ tanФ = sinФ ÷ cosФ

∴ cotФ = cosФ ÷ sinФ

∴ cotФ = \frac{-\sqrt{2} }{2} ÷ \frac{\sqrt{2} }{2}

∴ cotФ = -1

The values of cosine Ф and cotangent Ф are \frac{-\sqrt{2} }{2} and -1

Learn more:

You can learn more about the trigonometry function in brainly.com/question/4924817

#LearnwithBrainly

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