Geometry test..

Question

Given sin θ = − 1/6 and angle θ is in Quadrant IV, what is the exact value of cos θ in simplest form? Simplify all radicals if needed.

Answer 1

Solution: Given sin θ = -1/6 , Take sin^-1 of both sides, sin^-1 sinθ = sin^-1( -1/6) , and θ = -0.1674 rad

Take cos θ = cos ( -0.1674 rad) = 0.986

Solving this in degrees:

θ = -0.1674 rad ( 180°)/ π rad = -9.59° , In a trigonometric circle rotating from 0 clockwise 9.59° give the position of θ in the fourth Quadrant Quad IV, so if we take the cosine of the angle θ, its projection is on the positive x axis is cos( -9.59°) = 0.986

another method to solve this is by using trigonometric formula sin²θ + cos²θ = 1, cos²θ = 1 - sin²θ,

sin θ = sin(-0.1674 rad) = -0.1666 , cos²θ = 1 - (-0.1666)² = 1 - 0.02775 = 0.972

cos θ = √(0.972) = 0.986 same answer as above

Answer 2

sin(theta)^2 + cos(theta)^2 = 1 so

1/36 + cos(theta)^2 = 1

cos(theta)^2 = 35/36

cos(theta) = sqrt(35)/6 (positive since in Q IV)