Since P is in Quadrant III, the angle θ must be in the range of 270° < θ < 360°.
cos(θ) = x
We can use the Pythagorean Theorem to find x:
x2 + (-6/7)2 = 12
x2 + 36/49 = 1
x2 = 1 - 36/49
x2 = 13/49
x = ±√(13/49)
Since P is in Quadrant III, x must be negative.
x = -√(13/49)
cos(θ) = -√(13/49)
Please help and show all your work.
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)?
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