347.

Inverzne trigonometrijske funkcije

TEKST ZADATKA

Izračunati:

arctg⁡(1+2)+arctg⁡(1−2)\arctg{(1+\sqrt{2})}+\arctg{(1-\sqrt{2})}
REŠENJE ZADATKA

Pretpostaviti da je:

α=arctg⁡(1+2),β=arctg⁡(1−2)\alpha=\arctg{(1+\sqrt{2})}, \beta=\arctg{(1-\sqrt{2})}

Odatle sledi da je:

tg⁡α=1+2,tg⁡β=1−2\tg{\alpha}=1+\sqrt{2}, \tg{\beta}=1-\sqrt{2}

Početni izraz zameniti sa α\alpha i β:\beta:

arctg⁡(1+2)+arctg⁡(1−2)=α+β\arctg{(1+\sqrt{2})}+\arctg{(1-\sqrt{2})}=\alpha+\beta

Primeniti formulu za tangens zbira dva ugla: tg⁡(α+β)=tg⁡α+tg⁡β1−tg⁡αtg⁡β,α=/ π2+πk,β=/ π2+πn,k,n∈Z,tg⁡αtg⁡β=/ 1 \tg{(\alpha+\beta)}=\frac {\tg{\alpha}+\tg{\beta}} {1-\tg{\alpha}\tg{\beta}} , \alpha {=}\mathllap{/\,} \frac {\pi} 2 +\pi k, \beta {=}\mathllap{/\,} \frac {\pi} 2+\pi n , k,n \in Z, \tg{\alpha}\tg{\beta}{=}\mathllap{/\,}1 i uvrstiti prethodno dobijene vrednosti:

tg⁡(α+β)=tg⁡α+tg⁡β1−tg⁡αtg⁡β=1+2+1−21−(1+2)(1−2) \tg{(\alpha+\beta)}=\frac {\tg{\alpha}+\tg{\beta}} {1-\tg{\alpha}\tg{\beta}}=\frac {1+\sqrt{2}+1-\sqrt{2}} {1-(1+\sqrt{2})(1-\sqrt{2})}

Srediti izraz:

tg⁡(α+β)=21−(1−2)=22=1\tg{(\alpha+\beta)}=\frac 2 {1-(1-2)}=\frac 2 2=1

Pošto je tg⁡(α+β)=1,\tg{(\alpha+\beta)}=1 , to znači da je:

α+β=arctg⁡1=π4\alpha+\beta=\arctg1=\frac {\pi} 4

Konačno rešenje:

arctg⁡(1+2)+arctg⁡(1−2)=π4\arctg{(1+\sqrt{2})}+\arctg{(1-\sqrt{2})}=\frac {\pi} 4

Balkan Tutor Sva Prava Zadržana © 2025

Politika privatnosti