Given sine, theta, equals, one sixthsinθ=
6
1
and angle thetaθ is in Quadrant II, what is the exact value of cosine, thetacosθ in simplest form? Simplify all radicals if needed.
Since sinθ = 1/6 and θ is in Quadrant II, we know that cosθ will be negative.
To find the value of cosθ, we can use the Pythagorean identity:
sin²θ + cos²θ = 1
We know sinθ = 1/6, so let's substitute that into the equation:
(1/6)² + cos²θ = 1
1/36 + cos²θ = 1
Subtract 1/36 from both sides:
cos²θ = 1 - 1/36
cos²θ = 36/36 - 1/36
cos²θ = 35/36
Since cosθ is negative in Quadrant II, we take the negative square root:
cosθ = -√(35/36)
Simplifying the radical, we get:
cosθ = -√35/√36
cosθ = -√35/6
Therefore, the exact value of cosθ in simplest form is -√35/6.