Question

Precalculus. Please help and show all your work. The answer must be written as a fraction.

Let y=-6/7 be the y-coordinate of the point P(x,y), where the terminal side of angle θ (in standard position) meets the unit circle. If P is in Quadrant III, what is cos(θ)?

Answer 1

cos(θ) = x/1

Since P is in Quadrant III, x must be negative.

We can use the Pythagorean Theorem to find x:

x^2 + (-6/7)^2 = 1

x^2 + 36/49 = 1

x^2 = 1 - 36/49

x^2 = 13/49

x = ±√13/7

Since P is in Quadrant III, x must be negative, so x = -√13/7

cos(θ) = x/1 = -√13/7