Solving $\sin(\arccos(x/2))$

Question

I am having trouble solving $\sin(\arccos(x/2))$. I can see how to do this equation when the question is just $\arccos(x)$ but having the $x/2$ is throwing me off a bit. I still know one side of the triangle is $\sqrt{1-x^2}$ but can't figure out how to go from there.

Answer 1

Hint:

Put $p=x/2$ and do what you normally do to reduce $\sin(\arccos(p))$. Then put $p=x/2$ back in.

Answer 2

let $\arccos(x/2) = t.$ then $0 \le t \le \pi$ and $\cos t = x/2.$ therefore $\sin t = \sqrt{1 - \cos^2 t} = \sqrt{1 - x^2/4}$

Answer 3

Let $\arccos\left(\dfrac{x}{2}\right) = \theta$. Then $\cos\theta = \dfrac{x}{2}$, where $0 \leq \theta \leq \pi$. Since $\sin^2\theta + \cos^2\theta = 1$, we obtain \begin{align*} \sin^2\theta & = 1 - \cos^2\theta\\ & = 1 - \left(\frac{x}{2}\right)^2\\ & = 1 - \frac{x^2}{4}\\ & = \frac{4 - x^2}{4} \end{align*} Taking square roots yields $$|\sin\theta| = \frac{\sqrt{4 - x^2}}{2}$$ Since $\sin\theta \geq 0$ if $0 \leq \theta \leq \pi$, $$\sin\theta = \frac{\sqrt{4 - x^2}}{2}$$

Note: You expressed an interest in determining the answer by drawing a triangle. If we let $\arccos\left(\dfrac{x}{2}\right) = \theta$, we can draw a right triangle with acute angle $\theta$ with adjacent leg of length $|x|$ and hypotenuse of length $2$. By the Pythagorean Theorem, the opposite leg then has length $\sqrt{4 - x^2}$, as shown in the diagram below. We can then use the right triangle to determine that $$\sin\theta = \frac{\sqrt{4 - x^2}}{2}$$

Caveats: I have drawn a right triangle in the first quadrant. However, the triangle collapses to a line segment if $\theta = 0, \dfrac{\pi}{2}, \text{or}~\pi$. Also, if $\dfrac{\pi}{2} < \theta < \pi$, the right triangle would be drawn in the second quadrant, which is why I denoted the length of the adjacent leg by $|x|$ rather than $x$. The algebraic argument given above does not require such caveats.

Answer 4

Let $\theta = \arccos(\frac x2)$. Then $0 \le \theta \le \pi$ and $\theta$ corresponds to the point $(x,y) = (x,\sqrt{4-x^2})$ with amplitude $r = 2$. Note that the y-coordinate is non negative since $\theta$ is in quadrants I and II.

It follows that $\sin(\arccos(\frac x2)) = \sin \theta = \dfrac yr = \dfrac{\sqrt{4-x^2}}{2}.$