cos(sin^-1(5/x)) ?
$$\cos{
(
\sin^{-1}(
\frac{5}{x}
)
)
} = \pm \sqrt{ 1-
\left(
\dfrac{5}{x}
\right)
^2 }$$
I always do these with a triangle to aid me.
if x is positive then the angle is in the 1st or 2nd quads and cos of the angle could be pos or neg.
If xis neg then the angle is in the 3rd or 4th quads and cos of the angle could be pos or neg.
now I'll look at the first quad.
$$cos\left(sin^{-1}\left(\frac{5}{x}\right)\right)$$
$$I let $\theta =sin^{-1}\left(\frac{5}{x}\right)$\\
I then found the third side using Pythagoras' Theorum.\\
Hence\\
$cos\left(sin^{-1}\left(\frac{5}{x}\right)\right)=\pm \frac{\sqrt{x^2-25}}{x}$$$
$$ok, let me see if I can make this look like Heureka's answer\\\\
$\pm \frac{\sqrt{x^2-25}}{x}$\\\\
=$\pm \sqrt{\frac{{x^2-25}}{x^2}$\\\\
=$\pm \sqrt{1-\frac{{25}}{x^2}$\\\\
=$\pm \sqrt{1-\frac{{25}}{x^2}$\\\\
=$\pm \sqrt{1-\left(\frac{{5}}{x}\right)^2$\\\\$$
There you go
cos(sin^-1(5/x))
$$\cos{(\sin^{-1}{( \frac{5}{x} )})} = \pm \sqrt{ 1 - \left(
\frac{5}{x} \right)^2 }$$
or
$$\cos{(\sin^{-1}{( \dfrac{z}{1} )})} = \pm \sqrt{ 1 - \left(z
\right)^2 }$$
It may not be clear as to why we need to take both roots in this answer......
Remember that....sin2(Θ) + cos2(Θ) = 1 ......so we have.....using Melody's triangle as a reference.......
(5/x)2 + cos2(Θ) = 1
cos2(Θ) = 1 - (5/x)2
cos2(Θ) = (x2 - 25) / x2 ......take the square root of both sides......remember we need to use the square root property......!!!!
cos(Θ) = ±√ (x2 - 25) / x