Zbir ili razliku trigonometrijskih funkcija možemo pretvoriti u proizvod.
Njihova primena olakšava transformaciju i sređivanje trigonometrijskih izraza.
sin α + sin β = 2 sin α + β 2 cos α − β 2 \sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} sin α + sin β = 2 sin 2 α + β cos 2 α − β
sin α − sin β = 2 cos α + β 2 sin α − β 2 \sin\alpha - \sin\beta = 2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} sin α − sin β = 2 cos 2 α + β sin 2 α − β
cos α + cos β = 2 cos α + β 2 cos α − β 2 \cos\alpha + \cos\beta = 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} cos α + cos β = 2 cos 2 α + β cos 2 α − β
cos α − cos β = − 2 sin α + β 2 sin α − β 2 \cos\alpha - \cos\beta = -2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} cos α − cos β = − 2 sin 2 α + β sin 2 α − β
Jedino kod razlike kosinusa stoji minus ispred 2 2 2
Sve četiri formule dobijaju se direktno iz formula za pretvaranje proizvoda u zbir, zamenom promenljivih.
Formula za sin α + sin β \sin\alpha + \sin\beta sin α + sin β
Polazimo od formula za pretvaranje proizvoda u zbir:
sin x cos y = 1 2 [ sin ( x + y ) + sin ( x − y ) ] \sin x \cos y = \frac{1}{2}[\sin(x+y) + \sin(x-y)] sin x cos y = 2 1 [ sin ( x + y ) + sin ( x − y )]
Odatle:
2 sin x cos y = sin ( x + y ) + sin ( x − y ) 2\sin x\cos y = \sin(x+y) + \sin(x-y) 2 sin x cos y = sin ( x + y ) + sin ( x − y )
Uvodimo smenu α = x + y \alpha = x + y α = x + y β = x − y \beta = x - y β = x − y
Sabiranjem dobijamo α + β = 2 x \alpha + \beta = 2x α + β = 2 x x = α + β 2 x = \dfrac{\alpha+\beta}{2} x = 2 α + β
Oduzimanjem dobijamo α − β = 2 y \alpha - \beta = 2y α − β = 2 y y = α − β 2 y = \dfrac{\alpha-\beta}{2} y = 2 α − β
sin α + sin β = 2 sin α + β 2 cos α − β 2 \sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} sin α + sin β = 2 sin 2 α + β cos 2 α − β
Formula za sin α − sin β \sin\alpha - \sin\beta sin α − sin β
Polazimo od:
2 cos x sin y = sin ( x + y ) − sin ( x − y ) 2\cos x\sin y = \sin(x+y) - \sin(x-y) 2 cos x sin y = sin ( x + y ) − sin ( x − y )
Istom smenom α = x + y \alpha = x+y α = x + y β = x − y \beta = x-y β = x − y
sin α − sin β = 2 cos α + β 2 sin α − β 2 \sin\alpha - \sin\beta = 2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} sin α − sin β = 2 cos 2 α + β sin 2 α − β
Formula za cos α + cos β \cos\alpha + \cos\beta cos α + cos β
Polazimo od:
2 cos x cos y = cos ( x − y ) + cos ( x + y ) 2\cos x\cos y = \cos(x-y) + \cos(x+y) 2 cos x cos y = cos ( x − y ) + cos ( x + y )
Istom smenom:
cos α + cos β = 2 cos α + β 2 cos α − β 2 \cos\alpha + \cos\beta = 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} cos α + cos β = 2 cos 2 α + β cos 2 α − β
Formula za cos α − cos β \cos\alpha - \cos\beta cos α − cos β
Polazimo od:
2 sin x sin y = cos ( x − y ) − cos ( x + y ) 2\sin x\sin y = \cos(x-y) - \cos(x+y) 2 sin x sin y = cos ( x − y ) − cos ( x + y )
Odatle cos ( x + y ) − cos ( x − y ) = − 2 sin x sin y \cos(x+y) - \cos(x-y) = -2\sin x\sin y cos ( x + y ) − cos ( x − y ) = − 2 sin x sin y
cos α − cos β = − 2 sin α + β 2 sin α − β 2 \cos\alpha - \cos\beta = -2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} cos α − cos β = − 2 sin 2 α + β sin 2 α − β
Ove četiri formule nije potrebno posebno pamtiti ako se dobro znaju formule za pretvaranje proizvoda u zbir. Smena α = x + y \alpha = x+y α = x + y β = x − y \beta = x-y β = x − y
Primeri
Zadatak. Transformisati u proizvod: sin 20 ° + cos 50 ° \sin 20° + \cos 50° sin 20° + cos 50°
Da bismo primenili formule, oba člana moraju biti iste trigonometrijske funkcije. Koristimo cos 50 ° = sin ( 90 ° − 50 ° ) = sin 40 ° \cos 50° = \sin(90° - 50°) = \sin 40° cos 50° = sin ( 90° − 50° ) = sin 40°
sin 20 ° + sin 40 ° = 2 sin 20 ° + 40 ° 2 cos 20 ° − 40 ° 2 = 2 sin 30 ° cos ( − 10 ° ) \sin 20° + \sin 40° = 2\sin\frac{20°+40°}{2}\cos\frac{20°-40°}{2} = 2\sin 30°\cos(-10°) sin 20° + sin 40° = 2 sin 2 20° + 40° cos 2 20° − 40° = 2 sin 30° cos ( − 10° )
Pošto je cos ( − 10 ° ) = cos 10 ° \cos(-10°) = \cos 10° cos ( − 10° ) = cos 10° sin 30 ° = 1 2 \sin 30° = \dfrac{1}{2} sin 30° = 2 1
= 2 ⋅ 1 2 ⋅ cos 10 ° = cos 10 ° = 2 \cdot \frac{1}{2} \cdot \cos 10° = \cos 10° = 2 ⋅ 2 1 ⋅ cos 10° = cos 10°
Zadatak. Transformisati u proizvod: 1 − sin α + cos α 1 - \sin\alpha + \cos\alpha 1 − sin α + cos α
Grupišemo ( 1 + cos α ) − sin α (1 + \cos\alpha) - \sin\alpha ( 1 + cos α ) − sin α 1 + cos α = 2 cos 2 α 2 1 + \cos\alpha = 2\cos^2\dfrac{\alpha}{2} 1 + cos α = 2 cos 2 2 α sin α = 2 sin α 2 cos α 2 \sin\alpha = 2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2} sin α = 2 sin 2 α cos 2 α
2 cos 2 α 2 − 2 sin α 2 cos α 2 = 2 cos α 2 ( cos α 2 − sin α 2 ) 2\cos^2\frac{\alpha}{2} - 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2} = 2\cos\frac{\alpha}{2}\left(\cos\frac{\alpha}{2} - \sin\frac{\alpha}{2}\right) 2 cos 2 2 α − 2 sin 2 α cos 2 α = 2 cos 2 α ( cos 2 α − sin 2 α )
Zadatak. Transformisati u proizvod: sin x + 3 cos x \sin x + \sqrt{3}\cos x sin x + 3 cos x
Zapisujemo 3 = tg π 3 = sin π 3 cos π 3 \sqrt{3} = \operatorname{tg}\dfrac{\pi}{3} = \dfrac{\sin\frac{\pi}{3}}{\cos\frac{\pi}{3}} 3 = tg 3 π = cos 3 π sin 3 π cos π 3 \cos\dfrac{\pi}{3} cos 3 π
sin x cos π 3 + cos x sin π 3 cos π 3 \frac{\sin x\cos\frac{\pi}{3} + \cos x\sin\frac{\pi}{3}}{\cos\frac{\pi}{3}} c o s 3 π s i n x c o s 3 π + c o s x s i n 3 π
Brojilac prepoznajemo kao adicionu formulu za sinus zbira:
sin ( x + π 3 ) 1 2 = 2 sin ( x + π 3 ) \frac{\sin\left(x + \frac{\pi}{3}\right)}{\frac{1}{2}} = 2\sin\left(x + \frac{\pi}{3}\right) 2 1 s i n ( x + 3 π ) = 2 sin ( x + 3 π )
Dokazivanje identiteta
Zadatak. Dokazati: sin α − 2 sin 2 α + sin 3 α cos α − 2 cos 2 α + cos 3 α = tg 2 α \dfrac{\sin\alpha - 2\sin 2\alpha + \sin 3\alpha}{\cos\alpha - 2\cos 2\alpha + \cos 3\alpha} = \operatorname{tg} 2\alpha cos α − 2 cos 2 α + cos 3 α sin α − 2 sin 2 α + sin 3 α = tg 2 α
Korak 1. Grupisemo prvi i treći član posebno u brojiocu i imeniocu:
L = ( sin 3 α + sin α ) − 2 sin 2 α ( cos 3 α + cos α ) − 2 cos 2 α L = \frac{(\sin 3\alpha + \sin\alpha) - 2\sin 2\alpha}{(\cos 3\alpha + \cos\alpha) - 2\cos 2\alpha} L = ( c o s 3 α + c o s α ) − 2 c o s 2 α ( s i n 3 α + s i n α ) − 2 s i n 2 α
Korak 2. Primenjujemo formule za zbir sinusa i kosinusa:
sin 3 α + sin α = 2 sin 2 α cos α \sin 3\alpha + \sin\alpha = 2\sin 2\alpha\cos\alpha sin 3 α + sin α = 2 sin 2 α cos α
cos 3 α + cos α = 2 cos 2 α cos α \cos 3\alpha + \cos\alpha = 2\cos 2\alpha\cos\alpha cos 3 α + cos α = 2 cos 2 α cos α
Korak 3. Uvrštavamo i izvlačimo zajedničke faktore:
L = 2 sin 2 α cos α − 2 sin 2 α 2 cos 2 α cos α − 2 cos 2 α = 2 sin 2 α ( cos α − 1 ) 2 cos 2 α ( cos α − 1 ) = sin 2 α cos 2 α = tg 2 α ✓ L = \frac{2\sin 2\alpha\cos\alpha - 2\sin 2\alpha}{2\cos 2\alpha\cos\alpha - 2\cos 2\alpha} = \frac{2\sin 2\alpha(\cos\alpha - 1)}{2\cos 2\alpha(\cos\alpha - 1)} = \frac{\sin 2\alpha}{\cos 2\alpha} = \operatorname{tg} 2\alpha \checkmark L = 2 c o s 2 α c o s α − 2 c o s 2 α 2 s i n 2 α c o s α − 2 s i n 2 α = 2 c o s 2 α ( c o s α − 1 ) 2 s i n 2 α ( c o s α − 1 ) = c o s 2 α s i n 2 α = tg 2 α ✓
Zadatak. Dokazati: 1 + cos α + cos 2 α + cos 3 α cos α + cos 2 α = 2 cos α \dfrac{1 + \cos\alpha + \cos 2\alpha + \cos 3\alpha}{\cos\alpha + \cos 2\alpha} = 2\cos\alpha cos α + cos 2 α 1 + cos α + cos 2 α + cos 3 α = 2 cos α
Korak 1. Grupišemo u brojiocu ( 1 + cos 2 α ) + ( cos α + cos 3 α ) (1 + \cos 2\alpha) + (\cos\alpha + \cos 3\alpha) ( 1 + cos 2 α ) + ( cos α + cos 3 α )
Korak 2. Na prvi par primenjujemo 1 + cos 2 α = 2 cos 2 α 1 + \cos 2\alpha = 2\cos^2\alpha 1 + cos 2 α = 2 cos 2 α
cos α + cos 3 α = 2 cos 2 α cos α \cos\alpha + \cos 3\alpha = 2\cos 2\alpha\cos\alpha cos α + cos 3 α = 2 cos 2 α cos α
Korak 3. Brojioc postaje 2 cos 2 α + 2 cos 2 α cos α = 2 cos α ( cos α + cos 2 α ) 2\cos^2\alpha + 2\cos 2\alpha\cos\alpha = 2\cos\alpha(\cos\alpha + \cos 2\alpha) 2 cos 2 α + 2 cos 2 α cos α = 2 cos α ( cos α + cos 2 α )
L = 2 cos α ( cos α + cos 2 α ) cos α + cos 2 α = 2 cos α ✓ L = \frac{2\cos\alpha(\cos\alpha + \cos 2\alpha)}{\cos\alpha + \cos 2\alpha} = 2\cos\alpha \checkmark L = c o s α + c o s 2 α 2 c o s α ( c o s α + c o s 2 α ) = 2 cos α ✓
Zadatak. Dokazati: 1 − cos x = 2 sin 2 x 2 1 - \cos x = 2\sin^2\dfrac{x}{2} 1 − cos x = 2 sin 2 2 x
Zapisujemo 1 = cos 0 1 = \cos 0 1 = cos 0 cos α − cos β = − 2 sin α + β 2 sin α − β 2 \cos\alpha - \cos\beta = -2\sin\dfrac{\alpha+\beta}{2}\sin\dfrac{\alpha-\beta}{2} cos α − cos β = − 2 sin 2 α + β sin 2 α − β α = 0 \alpha = 0 α = 0 β = x \beta = x β = x
cos 0 − cos x = − 2 sin x 2 sin ( − x 2 ) = − 2 sin x 2 ⋅ ( − sin x 2 ) = 2 sin 2 x 2 ✓ \cos 0 - \cos x = -2\sin\frac{x}{2}\sin\left(-\frac{x}{2}\right) = -2\sin\frac{x}{2}\cdot\left(-\sin\frac{x}{2}\right) = 2\sin^2\frac{x}{2} \checkmark cos 0 − cos x = − 2 sin 2 x sin ( − 2 x ) = − 2 sin 2 x ⋅ ( − sin 2 x ) = 2 sin 2 2 x ✓
Izračunavanje pri poznatim vrednostima
Zadatak. Odrediti tg α ⋅ tg β \operatorname{tg}\alpha\cdot\operatorname{tg}\beta tg α ⋅ tg β cos ( α + β ) = 1 3 \cos(\alpha+\beta) = \dfrac{1}{3} cos ( α + β ) = 3 1 cos ( α − β ) = 1 5 \cos(\alpha-\beta) = \dfrac{1}{5} cos ( α − β ) = 5 1
Korak 1. Izražavamo proizvod tangensa preko sinusa i kosinusa i primenjujemo formule za pretvaranje proizvoda u zbir:
tg α ⋅ tg β = sin α sin β cos α cos β = 1 2 [ cos ( α − β ) − cos ( α + β ) ] 1 2 [ cos ( α − β ) + cos ( α + β ) ] \operatorname{tg}\alpha\cdot\operatorname{tg}\beta = \frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta} = \frac{\frac{1}{2}[\cos(\alpha-\beta) - \cos(\alpha+\beta)]}{\frac{1}{2}[\cos(\alpha-\beta) + \cos(\alpha+\beta)]} tg α ⋅ tg β = c o s α c o s β s i n α s i n β = 2 1 [ c o s ( α − β ) + c o s ( α + β )] 2 1 [ c o s ( α − β ) − c o s ( α + β )]
Korak 2. Skraćujemo 1 2 \dfrac{1}{2} 2 1
= 1 5 − 1 3 1 5 + 1 3 = 3 − 5 15 3 + 5 15 = − 2 8 = − 1 4 = \frac{\frac{1}{5} - \frac{1}{3}}{\frac{1}{5} + \frac{1}{3}} = \frac{\frac{3-5}{15}}{\frac{3+5}{15}} = \frac{-2}{8} = -\frac{1}{4} = 5 1 + 3 1 5 1 − 3 1 = 15 3 + 5 15 3 − 5 = 8 − 2 = − 4 1
Zadatak. Uprostiti: tg 40 ° + ctg 40 ° \operatorname{tg} 40° + \operatorname{ctg} 40° tg 40° + ctg 40°
Korak 1. Zapisujemo preko sinusa i kosinusa i svodimo na zajednički imenilac:
sin 40 ° cos 40 ° + cos 40 ° sin 40 ° = sin 2 40 ° + cos 2 40 ° sin 40 ° cos 40 ° = 1 sin 40 ° cos 40 ° \frac{\sin 40°}{\cos 40°} + \frac{\cos 40°}{\sin 40°} = \frac{\sin^2 40° + \cos^2 40°}{\sin 40°\cos 40°} = \frac{1}{\sin 40°\cos 40°} c o s 40° s i n 40° + s i n 40° c o s 40° = s i n 40° c o s 40° s i n 2 40° + c o s 2 40° = s i n 40° c o s 40° 1
Korak 2. Na imenilac primenjujemo formulu sin α cos β = 1 2 [ sin ( α + β ) + sin ( α − β ) ] \sin\alpha\cos\beta = \frac{1}{2}[\sin(\alpha+\beta) + \sin(\alpha-\beta)] sin α cos β = 2 1 [ sin ( α + β ) + sin ( α − β )] α = β = 40 ° \alpha = \beta = 40° α = β = 40°
sin 40 ° cos 40 ° = 1 2 ( sin 80 ° + sin 0 ° ) = 1 2 sin 80 ° \sin 40°\cos 40° = \frac{1}{2}(\sin 80° + \sin 0°) = \frac{1}{2}\sin 80° sin 40° cos 40° = 2 1 ( sin 80° + sin 0° ) = 2 1 sin 80°
Korak 3. Rezultat:
1 1 2 sin 80 ° = 2 sin 80 ° \frac{1}{\frac{1}{2}\sin 80°} = \frac{2}{\sin 80°} 2 1 s i n 80° 1 = s i n 80° 2
Relacije između uglova trougla
Kada su α \alpha α β \beta β γ \gamma γ α + β + γ = π \alpha + \beta + \gamma = \pi α + β + γ = π
Zadatak. Dokazati da za uglove trougla α \alpha α β \beta β γ \gamma γ
sin α 2 = cos β + γ 2 cos α 2 = sin β + γ 2 tg α 2 = ctg β + γ 2 \sin\frac{\alpha}{2} = \cos\frac{\beta+\gamma}{2} \qquad \cos\frac{\alpha}{2} = \sin\frac{\beta+\gamma}{2} \qquad \operatorname{tg}\frac{\alpha}{2} = \operatorname{ctg}\frac{\beta+\gamma}{2} sin 2 α = cos 2 β + γ cos 2 α = sin 2 β + γ tg 2 α = ctg 2 β + γ
Iz α + β + γ = π \alpha + \beta + \gamma = \pi α + β + γ = π β + γ = π − α \beta + \gamma = \pi - \alpha β + γ = π − α 2 2 2
β + γ 2 = π 2 − α 2 \frac{\beta+\gamma}{2} = \frac{\pi}{2} - \frac{\alpha}{2} 2 β + γ = 2 π − 2 α
Sada primenjujemo svođenje na oštar ugao:
cos β + γ 2 = cos ( π 2 − α 2 ) = sin α 2 ✓ \cos\frac{\beta+\gamma}{2} = \cos\!\left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = \sin\frac{\alpha}{2} \checkmark cos 2 β + γ = cos ( 2 π − 2 α ) = sin 2 α ✓
sin β + γ 2 = sin ( π 2 − α 2 ) = cos α 2 ✓ \sin\frac{\beta+\gamma}{2} = \sin\!\left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = \cos\frac{\alpha}{2} \checkmark sin 2 β + γ = sin ( 2 π − 2 α ) = cos 2 α ✓
ctg β + γ 2 = ctg ( π 2 − α 2 ) = tg α 2 ✓ \operatorname{ctg}\frac{\beta+\gamma}{2} = \operatorname{ctg}\!\left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = \operatorname{tg}\frac{\alpha}{2} \checkmark ctg 2 β + γ = ctg ( 2 π − 2 α ) = tg 2 α ✓