Do sada su trigonometrijske vrednosti bile poznate uglavnom za standardne uglove: 30 ° 30° 30° 45 ° 45° 45° 60 ° 60° 60°
Adicione formule proširuju tu mogućnost: iz poznatih vrednosti za uglove α \alpha α β \beta β α + β \alpha + \beta α + β α − β \alpha - \beta α − β
Sadržaj
Pregled formula Primeri
Sinus zbira i razlike:
sin ( α + β ) = sin α cos β + cos α sin β \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta sin ( α + β ) = sin α cos β + cos α sin β
sin ( α − β ) = sin α cos β − cos α sin β \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta sin ( α − β ) = sin α cos β − cos α sin β
Kosinus zbira i razlike:
cos ( α + β ) = cos α cos β − sin α sin β \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta cos ( α + β ) = cos α cos β − sin α sin β
cos ( α − β ) = cos α cos β + sin α sin β \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta cos ( α − β ) = cos α cos β + sin α sin β
Tangens zbira i razlike:
tg ( α + β ) = tg α + tg β 1 − tg α tg β \operatorname{tg}(\alpha + \beta) = \frac{\operatorname{tg}\alpha + \operatorname{tg}\beta}{1 - \operatorname{tg}\alpha\operatorname{tg}\beta} tg ( α + β ) = 1 − tg α tg β tg α + tg β
tg ( α − β ) = tg α − tg β 1 + tg α tg β \operatorname{tg}(\alpha - \beta) = \frac{\operatorname{tg}\alpha - \operatorname{tg}\beta}{1 + \operatorname{tg}\alpha\operatorname{tg}\beta} tg ( α − β ) = 1 + tg α tg β tg α − tg β
Kotangens zbira i razlike:
ctg ( α + β ) = ctg α ctg β − 1 ctg α + ctg β \operatorname{ctg}(\alpha + \beta) = \frac{\operatorname{ctg}\alpha\operatorname{ctg}\beta - 1}{\operatorname{ctg}\alpha + \operatorname{ctg}\beta} ctg ( α + β ) = ctg α + ctg β ctg α ctg β − 1
ctg ( α − β ) = ctg α ctg β + 1 ctg β − ctg α \operatorname{ctg}(\alpha - \beta) = \frac{\operatorname{ctg}\alpha\operatorname{ctg}\beta + 1}{\operatorname{ctg}\beta - \operatorname{ctg}\alpha} ctg ( α − β ) = ctg β − ctg α ctg α ctg β + 1
2. Primeri
2.1 Izračunavanje vrednosti za nestandardni ugao
Ugao koji nije u standardnoj tabeli vrednosti (30 ° 30° 30° 45 ° 45° 45° 60 ° 60° 60°
Zadatak. Naći vrednosti svih trigonometrijskih funkcija ugla 15 ° 15° 15°
Korak 1. Pišemo 15 ° = 45 ° − 30 ° 15° = 45° - 30° 15° = 45° − 30°
Korak 2. Sinus:
sin 15 ° = sin ( 45 ° − 30 ° ) = sin 45 ° cos 30 ° − cos 45 ° sin 30 ° \sin 15° = \sin(45° - 30°) = \sin 45°\cos 30° - \cos 45°\sin 30° sin 15° = sin ( 45° − 30° ) = sin 45° cos 30° − cos 45° sin 30° = 2 2 ⋅ 3 2 − 2 2 ⋅ 1 2 = 6 − 2 4 = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4} = 2 2 ⋅ 2 3 − 2 2 ⋅ 2 1 = 4 6 − 2
Korak 3. Kosinus:
cos 15 ° = cos ( 45 ° − 30 ° ) = cos 45 ° cos 30 ° + sin 45 ° sin 30 ° \cos 15° = \cos(45° - 30°) = \cos 45°\cos 30° + \sin 45°\sin 30° cos 15° = cos ( 45° − 30° ) = cos 45° cos 30° + sin 45° sin 30° = 2 2 ⋅ 3 2 + 2 2 ⋅ 1 2 = 6 + 2 4 = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4} = 2 2 ⋅ 2 3 + 2 2 ⋅ 2 1 = 4 6 + 2
Korak 4. Tangens:
tg 15 ° = tg 45 ° − tg 30 ° 1 + tg 45 ° tg 30 ° = 1 − 3 3 1 + 3 3 = 3 − 3 3 + 3 \operatorname{tg} 15° = \frac{\operatorname{tg} 45° - \operatorname{tg} 30°}{1 + \operatorname{tg} 45°\operatorname{tg} 30°} = \frac{1 - \frac{\sqrt{3}}{3}}{1 + \frac{\sqrt{3}}{3}} = \frac{3 - \sqrt{3}}{3 + \sqrt{3}} tg 15° = 1 + tg 45° tg 30° tg 45° − tg 30° = 1 + 3 3 1 − 3 3 = 3 + 3 3 − 3
Racionalizujemo imenilac množenjem sa 3 − 3 3 − 3 \dfrac{3 - \sqrt{3}}{3 - \sqrt{3}} 3 − 3 3 − 3
tg 15 ° = ( 3 − 3 ) 2 9 − 3 = 9 − 6 3 + 3 6 = 12 − 6 3 6 = 2 − 3 \operatorname{tg} 15° = \frac{(3-\sqrt{3})^2}{9 - 3} = \frac{9 - 6\sqrt{3} + 3}{6} = \frac{12 - 6\sqrt{3}}{6} = 2 - \sqrt{3} tg 15° = 9 − 3 ( 3 − 3 ) 2 = 6 9 − 6 3 + 3 = 6 12 − 6 3 = 2 − 3
Korak 5. Kotangens je recipročna vrednost tangensa:
ctg 15 ° = 1 2 − 3 ⋅ 2 + 3 2 + 3 = 2 + 3 4 − 3 = 2 + 3 \operatorname{ctg} 15° = \frac{1}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3} ctg 15° = 2 − 3 1 ⋅ 2 + 3 2 + 3 = 4 − 3 2 + 3 = 2 + 3
Za racionalizaciju imenilaca oblika a ± b a \pm \sqrt{b} a ± b a ∓ b a \mp \sqrt{b} a ∓ b
2.2 Uprošćavanje prepoznavanjem obrasca
Neki izrazi izgledaju složeno, ali direktno odgovaraju desnoj strani neke adicione formule. Tada ne razvijamo, već sklapamo unazad.
Zadatak. Uprostiti: cos 7 π 10 cos π 5 + sin 7 π 10 sin π 5 \cos\dfrac{7\pi}{10}\cos\dfrac{\pi}{5} + \sin\dfrac{7\pi}{10}\sin\dfrac{\pi}{5} cos 10 7 π cos 5 π + sin 10 7 π sin 5 π
Prepoznajemo obrazac cos α cos β + sin α sin β = cos ( α − β ) \cos\alpha\cos\beta + \sin\alpha\sin\beta = \cos(\alpha - \beta) cos α cos β + sin α sin β = cos ( α − β )
cos ( 7 π 10 − π 5 ) = cos ( 7 π 10 − 2 π 10 ) = cos 5 π 10 = cos π 2 = 0 \cos\left(\frac{7\pi}{10} - \frac{\pi}{5}\right) = \cos\left(\frac{7\pi}{10} - \frac{2\pi}{10}\right) = \cos\frac{5\pi}{10} = \cos\frac{\pi}{2} = 0 cos ( 10 7 π − 5 π ) = cos ( 10 7 π − 10 2 π ) = cos 10 5 π = cos 2 π = 0
Zadatak. Uprostiti: tg ( π 4 + α ) − tg α 1 + tg ( π 4 + α ) tg α \dfrac{\operatorname{tg}\!\left(\dfrac{\pi}{4} + \alpha\right) - \operatorname{tg}\alpha}{1 + \operatorname{tg}\!\left(\dfrac{\pi}{4} + \alpha\right)\operatorname{tg}\alpha} 1 + tg ( 4 π + α ) tg α tg ( 4 π + α ) − tg α
Prepoznajemo obrazac formule za tg ( x − y ) \operatorname{tg}(x - y) tg ( x − y ) x = π 4 + α x = \dfrac{\pi}{4} + \alpha x = 4 π + α y = α y = \alpha y = α
tg ( π 4 + α − α ) = tg π 4 = 1 \operatorname{tg}\!\left(\frac{\pi}{4} + \alpha - \alpha\right) = \operatorname{tg}\frac{\pi}{4} = 1 tg ( 4 π + α − α ) = tg 4 π = 1
2.3 Uprošćavanje razvijanjem i sređivanjem
Zadatak. Uprostiti: tg ( π 4 + α ) − tg ( π 4 − α ) \operatorname{tg}\!\left(\dfrac{\pi}{4} + \alpha\right) - \operatorname{tg}\!\left(\dfrac{\pi}{4} - \alpha\right) tg ( 4 π + α ) − tg ( 4 π − α )
Razvijamo oba tangensa:
tg ( π 4 + α ) = 1 + tg α 1 − tg α tg ( π 4 − α ) = 1 − tg α 1 + tg α \operatorname{tg}\!\left(\frac{\pi}{4} + \alpha\right) = \frac{1 + \operatorname{tg}\alpha}{1 - \operatorname{tg}\alpha} \qquad \operatorname{tg}\!\left(\frac{\pi}{4} - \alpha\right) = \frac{1 - \operatorname{tg}\alpha}{1 + \operatorname{tg}\alpha} tg ( 4 π + α ) = 1 − tg α 1 + tg α tg ( 4 π − α ) = 1 + tg α 1 − tg α
Svodimo na zajednički imenilac ( 1 − tg 2 α ) (1 - \operatorname{tg}^2\alpha) ( 1 − tg 2 α )
( 1 + tg α ) 2 − ( 1 − tg α ) 2 1 − tg 2 α \frac{(1 + \operatorname{tg}\alpha)^2 - (1 - \operatorname{tg}\alpha)^2}{1 - \operatorname{tg}^2\alpha} 1 − tg 2 α ( 1 + tg α ) 2 − ( 1 − tg α ) 2
Razvijamo brojilac koristeći razliku kvadrata ( a 2 − b 2 ) = ( a − b ) ( a + b ) (a^2 - b^2) = (a-b)(a+b) ( a 2 − b 2 ) = ( a − b ) ( a + b )
= 4 tg α 1 − tg 2 α = 2 ⋅ 2 tg α 1 − tg 2 α = 2 tg ( 2 α ) = \frac{4\operatorname{tg}\alpha}{1 - \operatorname{tg}^2\alpha} = 2 \cdot \frac{2\operatorname{tg}\alpha}{1 - \operatorname{tg}^2\alpha} = 2\operatorname{tg}(2\alpha) = 1 − tg 2 α 4 tg α = 2 ⋅ 1 − tg 2 α 2 tg α = 2 tg ( 2 α )
Zadatak. Uprostiti: cos ( π 3 + α ) cos ( π 3 − α ) − cos 2 α \cos\!\left(\dfrac{\pi}{3} + \alpha\right)\cos\!\left(\dfrac{\pi}{3} - \alpha\right) - \cos^2\alpha cos ( 3 π + α ) cos ( 3 π − α ) − cos 2 α
Razvijamo koristeći cos ( α ± β ) \cos(\alpha \pm \beta) cos ( α ± β )
( cos π 3 cos α − sin π 3 sin α ) ( cos π 3 cos α + sin π 3 sin α ) − cos 2 α \left(\cos\frac{\pi}{3}\cos\alpha - \sin\frac{\pi}{3}\sin\alpha\right)\left(\cos\frac{\pi}{3}\cos\alpha + \sin\frac{\pi}{3}\sin\alpha\right) - \cos^2\alpha ( cos 3 π cos α − sin 3 π sin α ) ( cos 3 π cos α + sin 3 π sin α ) − cos 2 α
Prepoznajemo razliku kvadrata ( A − B ) ( A + B ) = A 2 − B 2 (A - B)(A + B) = A^2 - B^2 ( A − B ) ( A + B ) = A 2 − B 2
= cos 2 π 3 cos 2 α − sin 2 π 3 sin 2 α − cos 2 α = \cos^2\frac{\pi}{3}\cos^2\alpha - \sin^2\frac{\pi}{3}\sin^2\alpha - \cos^2\alpha = cos 2 3 π cos 2 α − sin 2 3 π sin 2 α − cos 2 α
Zamenjujemo cos π 3 = 1 2 \cos\frac{\pi}{3} = \frac{1}{2} cos 3 π = 2 1 sin π 3 = 3 2 \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} sin 3 π = 2 3
= 1 4 cos 2 α − 3 4 sin 2 α − cos 2 α = − 3 4 cos 2 α − 3 4 sin 2 α = − 3 4 ( cos 2 α + sin 2 α ) = − 3 4 = \frac{1}{4}\cos^2\alpha - \frac{3}{4}\sin^2\alpha - \cos^2\alpha = -\frac{3}{4}\cos^2\alpha - \frac{3}{4}\sin^2\alpha = -\frac{3}{4}(\cos^2\alpha + \sin^2\alpha) = -\frac{3}{4} = 4 1 cos 2 α − 4 3 sin 2 α − cos 2 α = − 4 3 cos 2 α − 4 3 sin 2 α = − 4 3 ( cos 2 α + sin 2 α ) = − 4 3
Zadatak. Uprostiti: sin ( α + β ) − sin β cos α sin ( α − β ) + sin β cos α \dfrac{\sin(\alpha + \beta) - \sin\beta\cos\alpha}{\sin(\alpha - \beta) + \sin\beta\cos\alpha} sin ( α − β ) + sin β cos α sin ( α + β ) − sin β cos α
Razvijamo sinus zbira i razlike:
sin ( α + β ) = sin α cos β + cos α sin β \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta sin ( α + β ) = sin α cos β + cos α sin β sin ( α − β ) = sin α cos β − cos α sin β \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta sin ( α − β ) = sin α cos β − cos α sin β
Uvrštavamo u razlomak. U brojiocu cos α sin β \cos\alpha\sin\beta cos α sin β − sin β cos α -\sin\beta\cos\alpha − sin β cos α
sin α cos β sin α cos β = 1 \frac{\sin\alpha\cos\beta}{\sin\alpha\cos\beta} = 1 s i n α c o s β s i n α c o s β = 1