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Let (2 -7) be a point on the terminal side of theta. Find the exact values of cos0, csc0, tan0.

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Answers:


\cos(\theta) = (2)/(√(53)) = (2√(53))/(53)\\\\\tan(\theta) = -(7)/(2)\\\\\csc(\theta) = -(√(53))/(7)\\\\

=======================================================

Step-by-step explanation

The terminal point is located at (x,y) = (2,-7)

Let's calculate the distance from the origin to the terminal point.


r = √(x^2+y^2)\\\\r = √(2^2+(-7)^2)\\\\r = √(4+49)\\\\r = √(53)\\\\

From there we would write the following:


\cos(\theta) = \frac{\text{x}}{\text{r}} = (2)/(√(53)) = (2√(53))/(53)\\\\\tan(\theta) = \frac{\text{y}}{\text{x}} = -(7)/(2)\\\\\csc(\theta) = \frac{\text{r}}{\text{y}} = -(√(53))/(7)\\\\

Side notes:

  • The terminal point (2,-7) is in quadrant 4. This is the southwest quadrant.
  • Cosine is positive in quadrant 4, while tangent and cosecant are negative in this quadrant.
  • Rationalizing the denominator may be optional.
  • x = adjacent
  • y = opposite
  • r = hypotenuse
  • cosecant is the reciprocal of sine
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