Да се пресметне

[Гост]

Някой има ли идея как ще се пресметне това?

[s.karakoleva]

[tex]f\left(\frac{2\sqrt{2}+1}{7}\right)=\arcsin\frac{\frac{2}{7}(2\sqrt2+1)}{\frac{2\sqrt2+1}{7}+1}=\arcsin\frac{\frac{2}{7}(2\sqrt2+1)}{\frac{2\sqrt2+8}{7}}[/tex]

[tex]=\arcsin\frac{\frac{2}{7}(2\sqrt2+1)}{\frac{2}{7}(\sqrt2+4)}=\arcsin\frac{2\sqrt2+1}{\sqrt2+4}=\arcsin\frac{(2\sqrt2+1)(\sqrt2-4)}{(\sqrt2+4)(\sqrt2-4)}[/tex] [tex]=\arcsin\frac{\sqrt2}{2}=\frac{\pi}{4} \ .[/tex]

[Гост]

[tex]f\left(\frac{2\sqrt{2}+1}{7}\right)=\arcsin\frac{\frac{2}{7}(2\sqrt2+1)}{\frac{2\sqrt2+1}{7}+1}=\arcsin\frac{\frac{2}{7}(2\sqrt2+1)}{\frac{2\sqrt2+8}{7}}[/tex]

[tex]=\arcsin\frac{\frac{2}{7}(2\sqrt2+1)}{\frac{2}{7}(\sqrt2+4)}=\arcsin\frac{2\sqrt2+1}{\sqrt2+4}=\arcsin\frac{(2\sqrt2+1)(\sqrt2-4)}{(\sqrt2+4)(\sqrt2-4)}[/tex] [tex]=\arcsin\frac{\sqrt2}{2}=\frac{\pi}{4} \ .[/tex]

Бърз и точен както винаги. Благодаря!