arccos,sin

[Гост]

Да се докаже, че :
arcsin(x)+arccos(y)=arcsin(xy+[tex]\sqrt{1-x^2}\sqrt{1-y^2}[/tex])
sin(2arctg(x))=[tex]\frac{2x}{1+x^2}[/tex]

[Добромир Глухаров]

$x=sin\alpha;\ y=cos\beta$

$sin(\alpha+\beta)=sin\alpha cos\beta+cos\alpha sin\beta=xy+\sqrt{1-x^2}.\sqrt{1-y^2}$

$\alpha+\beta=arcsin(xy+\sqrt{1-x^2}.\sqrt{1-y^2})$

$arcsinx+arccosy=arcsin(xy+\sqrt{1-x^2}.\sqrt{1-y^2})$

$x=tg\alpha\Rightarrow\alpha=arctgx$

$sin2\alpha=\frac{2tg\alpha}{1+tg^2\alpha}\Rightarrow sin(2arctgx)=\frac{2x}{1+x^2}$