Question

How do you find two exact values of `sin((cos^-1)(sqrt5/6))`?

Answer 1

The two exact values of `sin(cos^(-1)(sqrt(5)/6))` are
`sqrt(31)/6` and `-sqrt(31)/6`

Explanation:

Let's consider the right triangle with angle `x` where `cos(x) = sqrt(5)/6`

Then `cos^(-1)(sqrt(5)/6) = x` and so
`sin(cos^(-1)(sqrt(5)/6)) = sin(x) = y/6`

By the Pythagorean theorem,

`(sqrt(5))^2 + y^2 = 6^2 => y = sqrt(36 - 5) = sqrt(31)`

So we have one value of `sin(cos^(-1)(sqrt(5)/6))` as `sqrt(31)/6`

But assuming we are looking at `cos^(-1)` as multivalued, looking at it geometrically loses one solution, as we only look at one possibility for `cos^(-1)(sqrt(5)/6)`. To get the other, note that as as the cosine function is even,

`cos(-x) = cos(x) = sqrt(5)/6`

So we need to consider `-x` as our other value. Then, as the sine function is odd,

`sin(-x) = -sin(x) = -sqrt(31)/6`

Thus the two exact values of `sin(cos^(-1)(sqrt(5)/6))` are `sqrt(31)/6` and `-sqrt(31)/6`