Is it possible to cancel out the nested $\arccos$ inside the $\cos$ and $\sin$ in this expression?

Question

I have the expression $\sqrt{\frac{23 + 16 × \sqrt{2}}{68}} × \sin\left(a × \arccos\left(\frac{\sqrt{8} - 1}{4}\right)\right) - \frac{1}{2} × \cos\left(a × \arccos\left(\frac{\sqrt{8} - 1}{4}\right)\right)$. I'd like to get rid of the $\arccos$ inside the regular trig functions if possible, but the extra $a$ variable inside the trig functions is confounding me.

I know that $\cos(\arccos(x)) = x$ and that $\sin(\arccos(x)) = \sqrt{1 - x^2}$ (over the domain $-1 ≤ x ≤ 1$, at least), but is it possible to similarly cancel the regular and inverse trig functions when there's an extra factor inside the regular trig function but outside the inverse function?

Answer 1

In general, the answer is no; you cannot cancel the inverse and regular functions in the presence of a scale factor in the argument. For example, if $a=2$, \begin{align}\sin\left(2\arccos\left(\frac{\sqrt{8} - 1}{4}\right)\right)&=2\sin\left(\arccos\left(\frac{\sqrt{8} - 1}{4}\right)\right)\cos\left(\arccos\left(\frac{\sqrt{8} - 1}{4}\right)\right)\\&=\frac{2\sqrt{2}-1}{4}\sqrt{1-\frac{(2\sqrt{2}-1)^2}{16}}\end{align} By the double-angle formula.

Answer 2

Some simplification possible another way, arccos is abbreviated but still included:

$$\frac{1}{1} \sin\left(a × \arccos\left(\frac{\sqrt{8} - 1}{4}\right)\right) - \frac{1}{2} × \cos\left(a × \arccos\left(\frac{\sqrt{8} - 1}{4}\right)\right)\tag 1$$

$$ \text{Let}\cos \alpha =\frac{1}{\sqrt {5}}, \sin \alpha =\frac{2}{\sqrt 5}, t=\arccos\left(\frac{\sqrt{8} - 1}{4}\right), hypo= \sqrt {5}/2 $$

First expression is $$\frac{1}{1} \sin at - \frac{1}{2} \cos at \tag 2$$ $$\left(\frac{2}{\sqrt 5} \sin at - \frac{1}{\sqrt 5} \cos at\right) hypo \tag 3$$ $$\left(\sin \alpha \sin at - \cos \alpha \cos at\right) hypo \tag 4$$

$$- hypo \cos(at+\alpha) \tag 5$$