Trigonometrijske funkcije poluugla
Formule za poluugao izvode se direktno iz formula za dvostruki ugao: ako znamo cos 2 β \cos 2\beta cos 2 β β = α 2 \beta = \frac{\alpha}{2} β = 2 α cos α \cos\alpha cos α α 2 \frac{\alpha}{2} 2 α
∣ sin α 2 ∣ = 1 − cos α 2 ∣ cos α 2 ∣ = 1 + cos α 2 \left|\sin\frac{\alpha}{2}\right| = \sqrt{\frac{1 - \cos\alpha}{2}} \qquad \left|\cos\frac{\alpha}{2}\right| = \sqrt{\frac{1 + \cos\alpha}{2}} sin 2 α = 2 1 − c o s α cos 2 α = 2 1 + c o s α
∣ tg α 2 ∣ = 1 − cos α 1 + cos α ∣ ctg α 2 ∣ = 1 + cos α 1 − cos α \left|\operatorname{tg}\frac{\alpha}{2}\right| = \sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}} \qquad \left|\operatorname{ctg}\frac{\alpha}{2}\right| = \sqrt{\frac{1 + \cos\alpha}{1 - \cos\alpha}} tg 2 α = 1 + c o s α 1 − c o s α ctg 2 α = 1 − c o s α 1 + c o s α
Iz formula za dvostruki ugao slede i korisni oblici koji se često koriste za uprošćavanje:
1 − cos α = 2 sin 2 α 2 1 + cos α = 2 cos 2 α 2 1 - \cos\alpha = 2\sin^2\frac{\alpha}{2} \qquad 1 + \cos\alpha = 2\cos^2\frac{\alpha}{2} 1 − cos α = 2 sin 2 2 α 1 + cos α = 2 cos 2 2 α
Univerzalna trigonometrijska smena (izražavanje sin α \sin\alpha sin α cos α \cos\alpha cos α tg α 2 \operatorname{tg}\frac{\alpha}{2} tg 2 α
sin α = 2 tg α 2 1 + tg 2 α 2 cos α = 1 − tg 2 α 2 1 + tg 2 α 2 \sin\alpha = \frac{2\operatorname{tg}\frac{\alpha}{2}}{1 + \operatorname{tg}^2\frac{\alpha}{2}} \qquad \cos\alpha = \frac{1 - \operatorname{tg}^2\frac{\alpha}{2}}{1 + \operatorname{tg}^2\frac{\alpha}{2}} sin α = 1 + tg 2 2 α 2 tg 2 α cos α = 1 + tg 2 2 α 1 − tg 2 2 α
Treba primetiti da sve formule sadrže apsolutnu vrednost.
Apsolutne zagrade se uklanjaju tako što se ispred korena stavi + + + − - − α 2 \frac{\alpha}{2} 2 α
sin α 2 = ± 1 − cos α 2 \sin\frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{2}} sin 2 α = ± 2 1 − c o s α
Koji znak se bira, određuje se posebno za svaki zadatak. Na primer, ako je α 2 \frac{\alpha}{2} 2 α + + + − - −
Primeri
Izračunavanje funkcija poluugla iz poznatih vrednosti
Zadatak. Ako je tg α = 24 7 \operatorname{tg}\alpha = \dfrac{24}{7} tg α = 7 24 α ∈ ( π , 3 π 2 ) \alpha \in \left(\pi, \dfrac{3\pi}{2}\right) α ∈ ( π , 2 3 π ) sin α 2 \sin\dfrac{\alpha}{2} sin 2 α cos α 2 \cos\dfrac{\alpha}{2} cos 2 α tg α 2 \operatorname{tg}\dfrac{\alpha}{2} tg 2 α
Korak 1. Određujemo cos α \cos\alpha cos α 1 + tg 2 α = 1 cos 2 α 1 + \operatorname{tg}^2\alpha = \dfrac{1}{\cos^2\alpha} 1 + tg 2 α = cos 2 α 1 cos 2 α = 49 625 ⟹ cos α = − 7 25 \cos^2\alpha = \frac{49}{625} \implies \cos\alpha = -\frac{7}{25} cos 2 α = 625 49 ⟹ cos α = − 25 7
Korak 2. Određujemo kvadrant za α 2 \frac{\alpha}{2} 2 α π < α < 3 π 2 \pi < \alpha < \frac{3\pi}{2} π < α < 2 3 π π 2 < α 2 < 3 π 4 \frac{\pi}{2} < \frac{\alpha}{2} < \frac{3\pi}{4} 2 π < 2 α < 4 3 π α 2 \frac{\alpha}{2} 2 α sin α 2 > 0 \sin\frac{\alpha}{2} > 0 sin 2 α > 0 cos α 2 < 0 \cos\frac{\alpha}{2} < 0 cos 2 α < 0
Korak 3. Primenjujemo formule:
sin α 2 = 1 − ( − 7 25 ) 2 = 32 50 = 16 25 = 4 5 \sin\frac{\alpha}{2} = \sqrt{\frac{1 - \left(-\frac{7}{25}\right)}{2}} = \sqrt{\frac{32}{50}} = \sqrt{\frac{16}{25}} = \frac{4}{5} sin 2 α = 2 1 − ( − 25 7 ) = 50 32 = 25 16 = 5 4
cos α 2 = − 1 + ( − 7 25 ) 2 = − 18 50 = − 9 25 = − 3 5 \cos\frac{\alpha}{2} = -\sqrt{\frac{1 + \left(-\frac{7}{25}\right)}{2}} = -\sqrt{\frac{18}{50}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5} cos 2 α = − 2 1 + ( − 25 7 ) = − 50 18 = − 25 9 = − 5 3
tg α 2 = sin α 2 cos α 2 = 4 5 − 3 5 = − 4 3 \operatorname{tg}\frac{\alpha}{2} = \frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}} = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3} tg 2 α = c o s 2 α s i n 2 α = − 5 3 5 4 = − 3 4
Dokazivanje identiteta
Zadatak. Dokazati: tg α 2 = 1 − cos α sin α \operatorname{tg}\dfrac{\alpha}{2} = \dfrac{1 - \cos\alpha}{\sin\alpha} tg 2 α = sin α 1 − cos α
Zamenjujemo 1 − cos α = 2 sin 2 α 2 1 - \cos\alpha = 2\sin^2\frac{\alpha}{2} 1 − cos α = 2 sin 2 2 α sin α = 2 sin α 2 cos α 2 \sin\alpha = 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2} sin α = 2 sin 2 α cos 2 α
2 sin 2 α 2 2 sin α 2 cos α 2 = sin α 2 cos α 2 = tg α 2 ✓ \frac{2\sin^2\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}} = \frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}} = \operatorname{tg}\frac{\alpha}{2} \checkmark 2 s i n 2 α c o s 2 α 2 s i n 2 2 α = c o s 2 α s i n 2 α = tg 2 α ✓
Zadatak. Dokazati: 1 − 2 sin x 2 − cos x 1 + 2 sin x 2 − cos x = sin x 2 − 1 sin x 2 + 1 \dfrac{1 - 2\sin\dfrac{x}{2} - \cos x}{1 + 2\sin\dfrac{x}{2} - \cos x} = \dfrac{\sin\dfrac{x}{2} - 1}{\sin\dfrac{x}{2} + 1} 1 + 2 sin 2 x − cos x 1 − 2 sin 2 x − cos x = sin 2 x + 1 sin 2 x − 1
Zamenjujemo 1 − cos x = 2 sin 2 x 2 1 - \cos x = 2\sin^2\frac{x}{2} 1 − cos x = 2 sin 2 2 x
2 sin 2 x 2 − 2 sin x 2 2 sin 2 x 2 + 2 sin x 2 = 2 sin x 2 ( sin x 2 − 1 ) 2 sin x 2 ( sin x 2 + 1 ) = sin x 2 − 1 sin x 2 + 1 ✓ \frac{2\sin^2\frac{x}{2} - 2\sin\frac{x}{2}}{2\sin^2\frac{x}{2} + 2\sin\frac{x}{2}} = \frac{2\sin\frac{x}{2}\left(\sin\frac{x}{2} - 1\right)}{2\sin\frac{x}{2}\left(\sin\frac{x}{2} + 1\right)} = \frac{\sin\frac{x}{2} - 1}{\sin\frac{x}{2} + 1} \checkmark 2 s i n 2 2 x + 2 s i n 2 x 2 s i n 2 2 x − 2 s i n 2 x = 2 s i n 2 x ( s i n 2 x + 1 ) 2 s i n 2 x ( s i n 2 x − 1 ) = s i n 2 x + 1 s i n 2 x − 1 ✓
Univerzalna trigonometrijska smena
Kada izraz sadrži sin α \sin\alpha sin α cos α \cos\alpha cos α tg α 2 \operatorname{tg}\frac{\alpha}{2} tg 2 α
Zadatak. Izračunati A = sin x + 2 cos x tg x − ctg x A = \dfrac{\sin x + 2\cos x}{\operatorname{tg} x - \operatorname{ctg} x} A = tg x − ctg x sin x + 2 cos x tg x 2 = 2 \operatorname{tg}\dfrac{x}{2} = 2 tg 2 x = 2
Smenom t = 2 t = 2 t = 2
sin x = 2 ⋅ 2 1 + 4 = 4 5 cos x = 1 − 4 1 + 4 = − 3 5 \sin x = \frac{2 \cdot 2}{1 + 4} = \frac{4}{5} \qquad \cos x = \frac{1 - 4}{1 + 4} = -\frac{3}{5} sin x = 1 + 4 2 ⋅ 2 = 5 4 cos x = 1 + 4 1 − 4 = − 5 3
tg x = − 4 3 ctg x = − 3 4 \operatorname{tg} x = -\frac{4}{3} \qquad \operatorname{ctg} x = -\frac{3}{4} tg x = − 3 4 ctg x = − 4 3
A = 4 5 − 6 5 − 4 3 + 3 4 = − 2 5 − 7 12 = 2 5 ⋅ 12 7 = 24 35 A = \frac{\frac{4}{5} - \frac{6}{5}}{-\frac{4}{3} + \frac{3}{4}} = \frac{-\frac{2}{5}}{-\frac{7}{12}} = \frac{2}{5} \cdot \frac{12}{7} = \frac{24}{35} A = − 3 4 + 4 3 5 4 − 5 6 = − 12 7 − 5 2 = 5 2 ⋅ 7 12 = 35 24
Zadatak. Izraziti A = 2 5 − 4 sin α + 3 cos α A = \dfrac{2}{5 - 4\sin\alpha + 3\cos\alpha} A = 5 − 4 sin α + 3 cos α 2 z = tg α 2 z = \operatorname{tg}\dfrac{\alpha}{2} z = tg 2 α
Zamenjujemo sin α = 2 z 1 + z 2 \sin\alpha = \dfrac{2z}{1+z^2} sin α = 1 + z 2 2 z cos α = 1 − z 2 1 + z 2 \cos\alpha = \dfrac{1-z^2}{1+z^2} cos α = 1 + z 2 1 − z 2 1 + z 2 1 + z^2 1 + z 2
A = 2 5 ( 1 + z 2 ) − 8 z + 3 ( 1 − z 2 ) 1 + z 2 = 2 ( 1 + z 2 ) 2 z 2 − 8 z + 8 = 2 ( 1 + z 2 ) 2 ( z − 2 ) 2 = 1 + z 2 ( z − 2 ) 2 A = \frac{2}{\dfrac{5(1+z^2) - 8z + 3(1-z^2)}{1+z^2}} = \frac{2(1+z^2)}{2z^2 - 8z + 8} = \frac{2(1+z^2)}{2(z-2)^2} = \frac{1+z^2}{(z-2)^2} A = 1 + z 2 5 ( 1 + z 2 ) − 8 z + 3 ( 1 − z 2 ) 2 = 2 z 2 − 8 z + 8 2 ( 1 + z 2 ) = 2 ( z − 2 ) 2 2 ( 1 + z 2 ) = ( z − 2 ) 2 1 + z 2