jika sin teta = 1/4√6, maka cos teta =

sin²a + cos²a = 1
maka
Cos a = √(1-sin²a)
=√(1-(¼√6)²)
=√(1-(6/16))
=√(16-6 / 16)
=√(10/16)
=√10 / √16
=√10 / 4
=¼√10

Sin teta = 1/4 V6 , cos teta?

Cos^2 theta - sin^2 theta = 1
Cos^2 theta -(V6/4)^2 = 1
Cos^2 theta - 6/16 = 1
Cos^2 theta = 1 - 6/16
Cos^2 theta = 16/16 - 6/16
Cos^2 theta = 10/16
Cos theta = V10 / V16
Cos theta = V10 / 4
Cos theta = 1/4 V10

Semoga membantu;)