<s>[INST] Simplify $\sin{(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}})}.$ First I denote $x=\arctan{2}$ and $y=\arcsin{\frac{1}{\sqrt{10}}}$ and then use the addition formula for sine:
$$\sin{(x-y)}=\sin{x}\cos{y}-\cos{x}\sin{y}=\sin{x}\cos{y}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Now I use the fact that $\cos{y}=1-\sin^2{y}$ which gives
$$\cos{y}=1-\sin^2{\left(\arcsin{\frac{1}{\sqrt{10}}}\right)}=1-\frac{1}{10}=\frac{9}{10}$$
So I have reduced the problem to the following
$$\sin{x}\cdot \frac{9}{10}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Here I'm stuck. I cant just divide this expression by $(\sin{x}\cos{x})$ because I'd still be left with sine and cos terms and I'd also change the value of the expression. Everything now boils down to compute $\sin(\arctan{2})$ and $\cos{\arctan2}.$ I also tried rewriting $\cos(\arctan{2})$ as $1-\sin^2(\arctan{2}),$ but to no avail.
Any suggestions on 


*

*how to proceed from where I left;

*how tocompute this by means more effective;

*both of the above.

 [/INST] $x=\arctan 2$ means $\tan x=2$ thus
$\sin x = \dfrac{\sqrt{\sin^2 x}}{\sqrt{1}}=\dfrac{\sqrt{\sin^2 x}}{\sqrt{\sin^2 x+\cos^2 x}}=\dfrac{\sqrt{\frac{\sin^2 x}{\cos^2 x}}}{\sqrt{\frac{\sin^2 x}{\cos^2 x}+\frac{\cos^2 x}{\cos^2 x}}}=\dfrac{\tan x}{\sqrt{\tan^2 x+1}}$
$\sin x=\dfrac{2}{\sqrt 5}$ and $\cos x =\sqrt{1-\sin^2 x}=\dfrac{1}{\sqrt 5}$
$y=\arcsin \dfrac{1}{\sqrt{10}}$ means $\sin y=\dfrac{1}{\sqrt{10}}$ and $\cos y=\sqrt{1-\sin^2 y}=\dfrac{3}{\sqrt{10}}$
Therefore $\sin{\left(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}}\right)}=\sin(x-y)=\sin x\cos y -\cos x\sin y=$
$=\dfrac{2}{\sqrt 5}\cdot \dfrac{3}{\sqrt{10}}-\dfrac{1}{\sqrt 5}\cdot \dfrac{1}{\sqrt{10}}=\dfrac{5}{\sqrt{50}}=\dfrac{5}{5\sqrt 2}=\dfrac{1}{\sqrt 2}$
</s>