In Quadrant III, cosine is negative. Since we are given that cosθ = -6/7, we can determine the value of sinθ.
Using the Pythagorean identity, we know that sinθ = √(1 - cos^2θ).
Substituting in our value for cosθ:
sinθ = √(1 - (-6/7)^2)
sinθ = √(1 - 36/49)
sinθ = √(49/49 - 36/49)
sinθ = √(13/49)
sinθ = √13/7
Therefore, the exact value of sinθ is √13/7.
Given cosine, theta, equals, minus, start fraction, 6, divided by, 7, end fractioncosθ=−
7
6
and angle thetaθ is in Quadrant III, what is the exact value of sine, thetasinθ in simplest form? Simplify all radicals if needed.
1 answer