if sin x = 1/3, what is cos x?

Answer 1

Suppose you were to draw a right triangle where one of the angles was x, and you wanted its sine to be 1/3. Remember the right angle relations: sine = opposite leg/hypotenuse. So your triangle would have one leg - that opposite x - of length 1 and a hypotenuse of length 3. How long would the second leg be?

Pythagorean Theorem:

1^2 + b^2 = 3^2
1 + b^2 = 9
b^2 = 8

b = 2sqrt(2)

Now, the cosine is the adjacent over the hypotenuse. You now know both, so

cos x = 2sqrt(2)/3

Answer 2

sin is y/r cos is x/r
x^2+y^2 =r^2
so if sin is 1/3 then 3^2 -1^2 = r^2 so 9-1=8 so r = sqrt 8 = 2 root 2 so when sin x = 1/3 the cos x= (2 root 2)/3

Answer 3

The key is that (sinx)^2 + (cosx)^2 = 1.
So you have (1/3)^2 + (cosx)^2 = 1 or 1/9 + (cosx)^2 = 1
Subtract 1/9, and you have (cosx)^2 = 8/9
You need to take the square root.
That gives the square root of 8 over 9
or (what your teacher probably wants) 2 times the square root of two all over 9.

Answer 4

sinx = 1/3

(sinx)^2=1/9

(sinx)^2 + (cosx)^2 = 1

(cosx)^2 = 8/9

cos x = (2sqrt(2))/3

Answer 5

It is cosx = sqrt(1 - senx^2) = sqrt(8)/3