Find the EXACT value of cosθ when sinθ= (-3/5) and tanθ>0?
Im having trouble solving this problem. I tried it with the Pythagorean theorem but then realized they didnt specify that it is a right triangle.
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This is in the third quadrant.
sin²θ + cos²θ = 1
(-3/5)² + cos²θ = 1
cos²θ = 25/25 - 9/25
cos²θ = 16/25
cosθ = ± 4/5
Since tangent is > 0 and sine is < 0, then this is quadrant III and cosine is negative in quadrant III.
cosθ = -4/5
since sin(θ) < 0 and tan(θ) > 0, cos(θ) < 0.
cos(θ) = -â(1 - sin²(θ)) = -â(1 - 9/25) = -4/5