If sin(θ) = 1/4 and θ is in the 2nd quadrant, find cos (θ)

Question

If sin(θ) = 1/4 and θ is in the 2nd quadrant, find cos (θ)

What would cos(θ) be?

Answer 1

sin² θ + cos² θ = 1

cos² θ = 1 - sin² θ

cos² θ = 1 - 1/16

cos² θ = 5/16

cos θ ∈ {-√5 / 16, √5 / 16}

In QII cos θ < 0

Answer 2

Recall that sin²x + cos²x = 1. So, cosx = ±√(1 - sin²x).

Since θ lies in quadrant 2, cosθ < 0. So, cosθ = -√ (1 - sin²θ) = -√15 / 4

Answer 3

sin(θ) = 1/4

" = 14.477.......°

θ third quad is the best I can do=194.477......°; .25 is now negative.