$ \sin(\arccos(\frac1{\sqrt3})) = \frac{\sqrt2}{\sqrt3} $ proof

Question

$$ \sin\left(\arccos\left(\frac1{\sqrt3}\right)\right) = \frac{\sqrt2}{\sqrt3} $$

Can someone explain why this is the case?

Answer 1

Let $\theta=\arccos(1/\sqrt{3})$, so $\cos \theta=1/\sqrt{3}$. You want to find $\sin \theta$. Use $\sin^2\theta+\cos^2\theta=1$.

Answer 2

$ \sin^2(\arccos(\frac1{\sqrt3}))+ \cos^2(\arccos(\frac1{\sqrt3}))= \sin^2(\arccos(\frac1{\sqrt3}))+\frac{1}{3}=1$.

Answer 3

Solution sketch: Draw a right triangle, with $1$ as the hypotenuse and $\frac1{\sqrt3}$ as one of the legs (or $\sqrt3$ as the hypotenuse and $1$ as one of the legs, same thing). Then see what $\arccos(1/\sqrt 3)$ corresponds to in that triangle, then see what $\sin$ of that represents. Once you know what you are actually after, you do the calculations.

Answer 4

So... we first look at $\arccos(\frac{1}{\sqrt{3}})$ and ask ourselves "what sort of right triangle might the angle we get back describe?"

Well, remembering the mnemonic "SOH-CAH-TOA" we say that the angle we get back can be used to describe a right triangle whose adjacent side is of length $1$ and whose hypotenuse is of length $\sqrt{3}$.

Now, armed with the knowledge that our triangle has a side of length $1$ and a hypotenuse of length $\sqrt{3}$, we can learn what the length of the remaining side is via the pythagorean theorem. The length should satisfy $a^2+b^2=c^2$

$$(1)^2 + b^2 = (\sqrt{3})^2 \implies b^2 = 3-1 \implies b = \sqrt{2}$$

Now, we know that $\sin$ of that angle will be referring to the ratio of the length of the opposite side (which we learned to be $\sqrt{2}$) over the length of the hypotenuse (which we learned to be $\sqrt{3}$), so we arrive at the value of $\sin(\arccos(\frac{1}{\sqrt{3}}))=\dfrac{\sqrt{2}}{\sqrt{3}}$.