$\cos(\arcsin(x)) = \cdots $ I've been asked to prove $$y=\frac{\sqrt{3}} 2 x+\frac 1 2 \sqrt{1-x^2}$$
given $x=\sin(t)$ & $y=\sin(t+\frac \pi 6)$
I did $t=\arcsin(x)$ and plugged that into the $y$ equation.
Used the $\sin(a+b)$ identity to get:
$$y=x\cos\left(\frac \pi 6\right)+\frac{\cos(\arcsin(x))}2 = \frac{\sqrt3} 2 x+\frac{\cos(\arcsin(x))} 2$$
Now I'm sure there must be an identity for $cos(arcsin(x))$ however I'm unaware of it.
I'm also unaware of how to prove it.
I did a quick google and I found this page which says:

which is seemingly exactly what I need to complete the question; however, it wouldn't be proving that $y = \text{answer}$ if I didn't show how to get to this result.
Is there a "more correct" way to complete this question without having to fiddle with this formula / arcsins etc.

$$\cos^2(\arcsin(x))+\sin^2(\arcsin(x))=1\\\cos^2(\arcsin (x))+x^2=1\\\cos^2(\arcsin(x))=1-x^2\\\cos(\arcsin(x))=\sqrt{1-x^2}$$
The last step is okay because $-\frac{\pi}{2}\leq\arcsin x\leq\frac{\pi}{2}$ and $\cos$ is positive on this interval.
