Is there a way to simplify $\cos(\arcsin(x) - \arcsin(a*x))$?

Question

I worked out you can do $\cos(\arcsin(x)) = \sqrt{1-x^2}$ but can it be done if there is a difference between inverse sines?

Answer 1

On the one hand, you have the formula $$ \cos(\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha\sin \beta\tag{1} $$ On the other hand, for $\alpha=\arcsin(x)$ and $\beta=\arcsin(ax)$ you have $$ \sin\alpha = x,\quad \sin\beta = ax.\tag{2} $$ Also, note that $$ \cos\theta = \sqrt{1-\sin^2\theta}\tag{3} $$ for $\theta\in[-\pi/2,\pi/2]$.

Now combine (1) (2) and (3) to get your answer.

Answer 2

$$\begin{align*} \cos(\arcsin(x) - \arcsin(ax)) &= \sin(\arcsin(x))\sin(\arcsin(ax)) + \cos(\arcsin(x))\cos(\arcsin(ax)) \\ &= ax^2 + \sqrt{(1-x^2)(1-a^2x^2)}\end{align*}$$