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

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.

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.

• Nice, algebraic, solution!
– Tobi
Dec 11, 2016 at 18:48

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 $\cos(\arcsin x)$?

• This makes a lot of sense, I guess the - relative - complication of the question intimidated me...
– Tobi
Dec 11, 2016 at 18:46
• @Tobi As is the nature of trigonometry sometimes, but often it ends up beautiful :) Dec 11, 2016 at 22:17
• I find there to be so many directions in which I can take a question, it's often a matter of chance. Maybe with experience I'll be able to tell, faster, which identities to use.
– Tobi
Dec 11, 2016 at 23:54
• @Tobi Yes, definitely practice your trig identities. Personally, the most important ones are the Pythagorean identities (what we have here) and the sum of angles formula. Combined, you can get most of the other formulas. Learning how to use the formulas will become especially useful in some parts of calculus btw. Dec 12, 2016 at 0:35
• The solutions can be so illusive sometimes, simple rearrangement / substitution can often take me forever to spot. It's like there's no instant way to do them, just brute force.
– Tobi
Dec 12, 2016 at 0:42

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

• What is this showing?
– Tobi
Dec 11, 2016 at 18:53
• @Tobi Can't this be a more correct way to do this question rather than to fiddle with the arcsins? Dec 11, 2016 at 18:55
• How do you go from $\cos{t}$ to $\sqrt{1-x^2}$
– Tobi
Dec 11, 2016 at 18:58
• @Tobi $$\cos t=\sqrt {1-\sin^2 t}=\sqrt{1-x^2}$$ Dec 11, 2016 at 19:00
• Oh I see, very illusive.
– Tobi
Dec 11, 2016 at 19:00

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

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.