$\cos(\arcsin(x)) = \cdots $

Question

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.

Answer 1

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 $\cos(\arcsin x)$?

Answer 2

$$y=\sin\left(t+{\pi\over 6}\right)=\cos{\pi\over 6}\sin t+\sin{\pi\over 6}\cos t\\={\sqrt3\over 2}x+{1\over 2}\sqrt{1-x^2}$$

Answer 3

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

Answer 4

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 $$ \cos\arcsin x = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{1-x^2}} 1 = \sqrt{1-x^2}. $$

Answer 5

Prove $$y=\frac{\sqrt{3}} 2 x+\frac 1 2 \sqrt{1-x^2}$$

given $$x=\sin(t) \qquad \text{and} \qquad y=\sin\left(t+\frac \pi 6\right)$$

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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.