if the arcsin is 3/5 how do i get cos theta?

Question

i know its 4/5 but how?
method please =D

Answer 1

ok draw a right angled triangle label one of the acute angles theta.

if sin is 3/5
then label the hypotenuse 5
and the opposite side to the theta 3
as you have a right angled triangle you can consider pythagorus to find the missign side.
in this case you have a very famous triangle the 3-4-5 triangle 3^2 +4^2 = 5^2
so the adjacent side to the angle is 4

definition of the cos ratio is = adjacent / hypotenuse = 4/5

Answer 2

I think you mean the SINE (not the arcsin) is 3/5. (If the arcsin were 3/5, the cos would be 0.8253.)

If you know the sine, you can use the following formula to figure out the cosine:

sin²θ + cos²θ = 1

Solve for cosθ

cosθ = sqrt(1 − sin²θ)

Since you know that sinθ = 3/5, substitute that:

cosθ = sqrt(1 − (3/5)²)

Answer 3

arcsin is the same as inverse sine or sin^-1
so you know that the sin of theta equals 3/5
sin is opposite over hypotenuse (SOH CAH TOA)
Now call the adjacent side x
5^2=x^2+3^2
25=x^2+9
x^2=16
x=4
So, cos is adjacent over hypotenuse so it is 4/5

Answer 4

There are several ways but I would do it by definition:

sine ^2 + cos^2 =1
1-sin^2=cos^2
1-9/25=cos^2
25/25 -9/25=cos^2
16/25=cos^2
4/5=cos