cos θ = ± √ ( 1 - sin² θ )
In quadrant IV sine is negative , cosine is positive.
So in quadrant IV cos θ = √ ( 1 - sin² θ )
In this case:
cos θ = √ ( 1 - sin² θ ) = √ [ 1 - ( - 2 / 3 )² ] = √ ( 1 - 4 / 9 ) =
√ ( 9 / 9 - 4 / 9 ) = √ ( 5 / 9 ) = √ 5 / √ 9 = √ 5 / 3
Find the exact value of costheta if ,sintheta=-2/3 and theta is in standard position with its terminal side in Quadrant IV.
2 answers
in QIV,
y = -2
r = 3
x = √5
cosθ = x/r = √5/3
y = -2
r = 3
x = √5
cosθ = x/r = √5/3