David is told that an angle has a cosine of −3√/2 and is in Quadrant III. Use the Pythagorean identity sin2(θ)+cos2(θ)=1 and the quadrant to find the sine.(1 point)
In Quadrant III, both the sine and the cosine are negative.
Given that the cosine is `-3√/2`, we can find the sine using the Pythagorean identity `sin^2(θ) + cos^2(θ) = 1`.
Let's substitute the value of the cosine:
`sin^2(θ) + (-3√/2)^2 = 1`
`sin^2(θ) + 9/4 = 1`
`sin^2(θ) = 1 - 9/4`
`sin^2(θ) = 4/4 - 9/4`
`sin^2(θ) = -5/4`
Since both the sine and the cosine are negative in Quadrant III, the sine of the angle is `-√(5/4)` or `-√5/2`.