Let Θ be the radian angle measure Θ satisfies sin^2Θ - sinΘ = -1/4 for 0<Θ<π/2. What is cosΘ ?
F) 1/16
G) 1/4
H) 1/2
J) sqrt3/2
K) sqrt15/4
please help and add an explanation if you oculd, thank you!
—just had a brainwave and ive figured it out haha –
I didn’t immediately see a nice way involving trig identities (maybe someone else can point one out) but you can easily solve for sin θ since the given equation is quadratic in terms of sin θ:
sin^2 θ - sin θ + 1/4 = 0
4 sin^2 θ - 4 sin θ + 1 = 0
This factors to (2 sin θ - 1)^2 = 0, or sin θ = 1/2. cos θ is also positive, and cos θ = sqrt(1 - sin^2 θ) = sqrt(1 - 1/4) = sqrt(3)/2 or J.
Whoops I must’ve posted right when you figured out…lol whoops.
thanks for the help though!!! you’re the best 