Question

Question #18d08

Answer 1

`cosA=+-(2sqrt2)/3`

Explanation:

`•color(white)(x)sin^2x+cos^2x=1`

`rArrcosx=+-sqrt(1-sin^2x)`

`rArrcosA=+-sqrt(1-(1/3)^2)`

`color(white)(rArrcosA)=+-sqrt(8/9)`

`color(white)(rArrcosA)=+-(2sqrt2)/3`

`"the sign of "cosA"` will be dependent on which quadrant A is in.

Answer 2

`cosA = +-(2sqrt2)/3`

Explanation:

The sine of a right triangle is the ratio of the lengths opposite side to the hypotenuse, and the cosine is the ratio of the adjacent side to the hypotenuse.

From the given `sin`, we know that the ratio gives us

• `"opposite" = 1`

• `"hypotenuse" = 3`

We can use the Pythagorean theorem to find the length of the adjacent side:

`"adjacent" = sqrt(3^2 - 1^2) = color(red)(ul(+-2sqrt2`

And since

`cos = "adjacent"/"hypotenuse"`

We have

`color(blue)(ulbar(|stackrel(" ")(" "cos = +-(2sqrt2)/3" ")|)`

depending on the quadrant it lies in.