Question

If `cosx = p`, how do you derive expressions for `sinx`, `tanx` and `sin(90˚ - x)` in terms of `p`?

Answer 1

`sinx = +-sqrt(1 - p^2)`
`tanx = +-sqrt(1 - p^2)/p`
`sin(90˚ - x) = p`

Explanation:

We know that `cos^2x+ sin^2x = 1`, therefore `sinx = +- sqrt(1 - cos^2x)`.

Thus:

`sinx = +- sqrt(1 - p^2)`

Knowing that `tanx = sinx/cosx`, we know that

`tanx = +-sqrt(1 - p^2)/p`

As for `sin(90 - x)`, we can write as `sin(90˚)cosx - sinxcos90˚` because `sin(A - B) = sinAcosB - sinBcosA`.

The above expression can be simplified to `cosx`, or `p`, because `sin(90˚) = 1` and `cos(90˚) = 0`.

Hopefully this helps!