$\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.
 A: $$\cos^2(\arcsin(x))+\sin^2(\arcsin(x))=1\\\cos^2(\arcsin (x))+x^2=1\\\cos^2(\arcsin(x))=1-x^2\\\cos(\arcsin(x))=\sqrt{1-x^2}$$
The last step is okay because $-\frac{\pi}{2}\leq\arcsin x\leq\frac{\pi}{2}$ and $\cos$ is positive on this interval.
A: The usual proof involves drawing a triangle.  The opposite side will be called $\sin(y)=x$, the hypotenuse will be $1$, and the angle will be $\arcsin x$.  Can you use the pythagorean theorem to find $\cos(\arcsin x)$?

A: $$y=\sin\left(t+{\pi\over 6}\right)=\cos{\pi\over 6}\sin t+\sin{\pi\over 6}\cos t\\={\sqrt3\over 2}x+{1\over 2}\sqrt{1-x^2}$$
A: Draw a right triangle in which the "opposite" side has length $x$ and the hypotenuse has length $1$. Then the sine of the angle having that "opposite" side is $\sin=\dfrac{\text{opposite}}{\text{hypotenuse}} =\dfrac x 1 = x.$ So that angle is $\arcsin x$.
Now use the Pythagorean theorem to show that the "adjacent" side has length $\sqrt{1-x^2}$.  Then we have
$$
\cos\arcsin x = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{1-x^2}} 1 = \sqrt{1-x^2}.
$$
A: Prove $$y=\frac{\sqrt{3}} 2 x+\frac 1 2 \sqrt{1-x^2}$$
given $$x=\sin(t) 
        \qquad \text{and} \qquad 
        y=\sin\left(t+\frac \pi 6\right)$$

It's wrong to use  $t = \arcsin x$.
Using $t = \arcsin x$ means that whatever you find out is only true for
$-\dfrac{\pi}{2} \le t \le \dfrac{\pi}{2} $; not for ALL values of $t$.
If you start with $y=\sin\left(t+\frac \pi 6\right)$, you can say
\begin{align}
   y &=\sin\left(t+\frac \pi 6\right) \\
     &=  \sin(t) \cos\left(  \frac \pi 6 \right)
       + \cos(t) \sin\left(  \frac \pi 6 \right) \\
     &= \dfrac{\sqrt 3}{2} \sin t + \dfrac 12 \cos t \\
     &= \dfrac{\sqrt 3}{2} x + \dfrac 12 \cos t \\
\end{align}
Where $x = \sin t$. But what are you going to do about $\cos t$? Because $\sin^2 t + \cos^2 t = 1$, it follows that
$$\cos^2 t = 1 - \sin^2 t = 1 - x^2$$
And so $ \cos t = \pm \sqrt{1 - x^2}$.
So, when $\cos t \ge 0$, then
$y = \dfrac{\sqrt 3}{2} x + \dfrac 12 \sqrt{1-x^2}$. Which is what you had to prove.
But, when $\cos t < 0$, then
$y = \dfrac{\sqrt 3}{2} x - \dfrac 12 \sqrt{1-x^2}$. Which is something different.
