I am having problems understanding how to write an algebraic expression in $x$ for:
$$\sin\left(\arcsin(x)-\arctan\left(\frac{2}{x}\right)\right)$$
I am having problems understanding how to write an algebraic expression in $x$ for:
$$\sin\left(\arcsin(x)-\arctan\left(\frac{2}{x}\right)\right)$$
Assuming that $x\in(0,1)$, we have: $$\sin\left(\arcsin x-\arctan\frac{2}{x}\right)= x\cdot\cos\arctan\frac{2}{x}-\sqrt{1-x^2}\cdot\sin\arctan\frac{2}{x}$$ by the sine addition formulas and the Pythagorean theorem in the form $\cos\arcsin x=\sqrt{1-x^2}$. Since: $$\cos\arctan\frac{2}{x} = \frac{x}{\sqrt{4+x^2}},\qquad \sin\arctan\frac{2}{x}=\frac{2}{\sqrt{4+x^2}},$$ it follows that:
$$\sin\left(\arcsin x-\arctan\frac{2}{x}\right)=\frac{x^2-2\sqrt{1-x^2}}{\sqrt{4+x^2}}.$$
If $\arctan\dfrac2x=y,\tan y=\dfrac2x,-\dfrac\pi2\le y\le\dfrac\pi2\implies\cos y\ge0$
$\cos y=\dfrac1{\sqrt{1+\tan^2y}}=\cdots$
$\implies\sin y=\tan y\cdot\cos y=\cdots$
Now $\arcsin(-z)=-\arcsin(z)$
Finally use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
For arcs in the first quadrant, we can put: $$ \arcsin x=\alpha \iff x=\sin \alpha \iff \cos \alpha=\sqrt{1-x^2} $$ $$ \arctan (2/x)= \beta \iff \tan \beta=2/x $$ so that: $$ \cos \beta=\dfrac{1}{\sqrt{1+\tan^2 \beta}}=\dfrac{x}{\sqrt{x^2+4}} $$
$$ \sin \beta=\dfrac{\tan \beta}{\sqrt{1+\tan^2 \beta}}=\dfrac{2}{\sqrt{x^2+4}} $$
Now using $$ \sin (\alpha-\beta)= \sin \alpha \cos \beta-\cos \alpha \sin \beta $$ and substituing you fine the result.