Given sin x = 0.9, what is cos x ?
(As a decimal rounded to the nearest hundreth)
sin x = 9/10 = y/r
So....cos x / r
To find x
x = ±√[r^2- y^2 ] = ±√[ 10^2 - 9^2] = ±√[100 - 81 ] = ±√19
So....depending upon where x lies
cos x = √19 /100 ≈ .04 or cos x ≈ -.04
+121 
