If cos(θ)= 3/5 and 270°<θ<360°, what is sin(θ)?

Question

How to solve?

Answer 1

cos²θ + sin²θ = 1

So, sinθ = ±√(1 - cos²θ)

Since θ lies in Quadrant 4, sinθ < 0

So, sinθ = -√(1 - (3/5)²) = -√(16/25) = - 4/5

Answer 2

if cosT = 3/5 and T is in quadrant IV (let T = theta)

then sinT = -4/5

cosT = adjacent side over hypotenuse= 3/5

sinT = opposite side over hypotenuse= -4/5

opposite side = square root of (hypotenuse squared minus opposite side squared) = sqr(5^2 -3^2) = sqr(25-9)= sqr16 = -4 (ignore the positive square root as sines are all negative in quadrant 4)

use the Pythagorean Theorem

draw a unit circle and a right triangle in quadrant 4, with sides 3,4 and 5