Question

How do you find the value for `sin(arccos(-1/3))`?

Answer 1

Use calculator to evaluate `theta = arccos(-1/3)`, allow for `theta+pi` (since `cos(theta) = cos(theta+pi)`); use calculator to evaluate the two values of theta (within the `0` to `2pi` range.

Explanation:

Using caluclator
`color(white)("XXXX")` `arccos(-1/3) = 1.910633`
`color(white)("XXXX")` `color(white)("XXXX")`i.e `cos(1.910633) = -1/3`
`color(white)("XXXX")`note that
`color(white)("XXXX")` `color(white)("XXXX")` `cos(1.910633+pi)` also `= -1/3`

Using calculator evaluate:
`color(white)("XXXX")` `sin(1.910633) = 0.942809`
and
`color(white)("XXXX")` `sin(1.910633+pi) = sin(4.372552) = -0.942809`

Answer 2

Find`sin (arccos (-1/3))`

Explanation:

`cos x = -1/3`--> arc x?

Calculator gives `x = +- 109.47`.
Now, find `sin (+- 109.47)`

sin (109.47) = 0.94
sin (-109.47) = -0.94