Kombinovanjem adicionih formula za sinus i kosinus dobijaju se formule koje omogućavaju pretvaranje proizvoda trigonometrijskih funkcija u zbir ili razliku, kao i obrnuto.
Njihova primena olakšava transformaciju i sređivanje trigonometrijskih izraza.
Proizvod u zbir:
sin α ⋅ sin β = 1 2 [ cos ( α − β ) − cos ( α + β ) ] \sin\alpha \cdot \sin\beta = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)] sin α ⋅ sin β = 2 1 [ cos ( α − β ) − cos ( α + β )]
cos α ⋅ cos β = 1 2 [ cos ( α − β ) + cos ( α + β ) ] \cos\alpha \cdot \cos\beta = \frac{1}{2}[\cos(\alpha - \beta) + \cos(\alpha + \beta)] cos α ⋅ cos β = 2 1 [ cos ( α − β ) + cos ( α + β )]
sin α ⋅ cos β = 1 2 [ sin ( α + β ) + sin ( α − β ) ] \sin\alpha \cdot \cos\beta = \frac{1}{2}[\sin(\alpha + \beta) + \sin(\alpha - \beta)] sin α ⋅ cos β = 2 1 [ sin ( α + β ) + sin ( α − β )]
cos α ⋅ sin β = 1 2 [ sin ( α + β ) − sin ( α − β ) ] \cos\alpha \cdot \sin\beta = \frac{1}{2}[\sin(\alpha + \beta) - \sin(\alpha - \beta)] cos α ⋅ sin β = 2 1 [ sin ( α + β ) − sin ( α − β )]
Sve četiri formule dobijaju se oduzimanjem ili sabiranjem adicionih formula za sinus i kosinus.
Formula za sin α sin β \sin\alpha\sin\beta sin α sin β
Polazimo od adicionih formula za kosinus:
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
Oduzimamo drugu od prve:
cos ( α − β ) − cos ( α + β ) = 2 sin α sin β \cos(\alpha - \beta) - \cos(\alpha + \beta) = 2\sin\alpha\sin\beta cos ( α − β ) − cos ( α + β ) = 2 sin α sin β
Delimo sa 2 2 2
sin α sin β = 1 2 [ cos ( α − β ) − cos ( α + β ) ] \sin\alpha\sin\beta = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)] sin α sin β = 2 1 [ cos ( α − β ) − cos ( α + β )]
Formula za cos α cos β \cos\alpha\cos\beta cos α cos β
Polazimo od istih adicionih formula za kosinus i ovaj put sabiramo:
cos ( α − β ) + cos ( α + β ) = 2 cos α cos β \cos(\alpha - \beta) + \cos(\alpha + \beta) = 2\cos\alpha\cos\beta cos ( α − β ) + cos ( α + β ) = 2 cos α cos β
Delimo sa 2 2 2
cos α cos β = 1 2 [ cos ( α − β ) + cos ( α + β ) ] \cos\alpha\cos\beta = \frac{1}{2}[\cos(\alpha - \beta) + \cos(\alpha + \beta)] cos α cos β = 2 1 [ cos ( α − β ) + cos ( α + β )]
Formula za sin α cos β \sin\alpha\cos\beta sin α cos β
Polazimo od adicionih formula za sinus:
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
Sabiramo obe jednačine:
sin ( α + β ) + sin ( α − β ) = 2 sin α cos β \sin(\alpha + \beta) + \sin(\alpha - \beta) = 2\sin\alpha\cos\beta sin ( α + β ) + sin ( α − β ) = 2 sin α cos β
Delimo sa 2 2 2
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 )]
Formula za cos α sin β \cos\alpha\sin\beta cos α sin β
Polazimo od istih adicionih formula za sinus i oduzimamo drugu od prve:
sin ( α + β ) − sin ( α − β ) = 2 cos α sin β \sin(\alpha + \beta) - \sin(\alpha - \beta) = 2\cos\alpha\sin\beta sin ( α + β ) − sin ( α − β ) = 2 cos α sin β
Delimo sa 2 2 2
cos α sin β = 1 2 [ sin ( α + β ) − sin ( α − β ) ] \cos\alpha\sin\beta = \frac{1}{2}[\sin(\alpha + \beta) - \sin(\alpha - \beta)] cos α sin β = 2 1 [ sin ( α + β ) − sin ( α − β )]
Primeri
Zadatak. Uprostiti: cos 10 ° cos 50 ° cos 70 ° \cos 10° \cos 50° \cos 70° cos 10° cos 50° cos 70°
Korak 1. Primenjujemo formulu cos α cos β = 1 2 [ cos ( α − β ) + cos ( α + β ) ] \cos\alpha\cos\beta = \frac{1}{2}[\cos(\alpha-\beta)+\cos(\alpha+\beta)] cos α cos β = 2 1 [ cos ( α − β ) + cos ( α + β )] cos 50 ° cos 70 ° \cos 50°\cos 70° cos 50° cos 70°
cos 50 ° cos 70 ° = 1 2 ( cos 20 ° + cos 120 ° ) = 1 2 ( cos 20 ° − 1 2 ) \cos 50°\cos 70° = \frac{1}{2}(\cos 20° + \cos 120°) = \frac{1}{2}\left(\cos 20° - \frac{1}{2}\right) cos 50° cos 70° = 2 1 ( cos 20° + cos 120° ) = 2 1 ( cos 20° − 2 1 )
Korak 2. Množimo sa cos 10 ° \cos 10° cos 10°
cos 10 ° ⋅ 1 2 ( cos 20 ° − 1 2 ) = 1 2 cos 10 ° cos 20 ° − 1 4 cos 10 ° \cos 10° \cdot \frac{1}{2}\left(\cos 20° - \frac{1}{2}\right) = \frac{1}{2}\cos 10°\cos 20° - \frac{1}{4}\cos 10° cos 10° ⋅ 2 1 ( cos 20° − 2 1 ) = 2 1 cos 10° cos 20° − 4 1 cos 10°
Korak 3. Primenjujemo istu formulu na cos 10 ° cos 20 ° \cos 10°\cos 20° cos 10° cos 20°
cos 10 ° cos 20 ° = 1 2 ( cos 10 ° + cos 30 ° ) \cos 10°\cos 20° = \frac{1}{2}(\cos 10° + \cos 30°) cos 10° cos 20° = 2 1 ( cos 10° + cos 30° )
Korak 4. Uvrštavamo i sređujemo:
1 2 ⋅ 1 2 ( cos 10 ° + cos 30 ° ) − 1 4 cos 10 ° = 1 4 cos 10 ° + 1 4 cos 30 ° − 1 4 cos 10 ° = 1 4 ⋅ 3 2 = 3 8 \frac{1}{2} \cdot \frac{1}{2}(\cos 10° + \cos 30°) - \frac{1}{4}\cos 10° = \frac{1}{4}\cos 10° + \frac{1}{4}\cos 30° - \frac{1}{4}\cos 10° = \frac{1}{4} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8} 2 1 ⋅ 2 1 ( cos 10° + cos 30° ) − 4 1 cos 10° = 4 1 cos 10° + 4 1 cos 30° − 4 1 cos 10° = 4 1 ⋅ 2 3 = 8 3
Dokazivanje identiteta
Zadatak. Dokazati: 4 sin α sin 2 α sin 3 α = sin 2 α + sin 4 α − sin 6 α 4\sin\alpha\sin 2\alpha\sin 3\alpha = \sin 2\alpha + \sin 4\alpha - \sin 6\alpha 4 sin α sin 2 α sin 3 α = sin 2 α + sin 4 α − sin 6 α
Korak 1. Grupišemo prvi i treći faktor i primenjujemo sin α sin β = 1 2 [ cos ( α − β ) − cos ( α + β ) ] \sin\alpha\sin\beta = \frac{1}{2}[\cos(\alpha-\beta)-\cos(\alpha+\beta)] sin α sin β = 2 1 [ cos ( α − β ) − cos ( α + β )]
sin α sin 3 α = 1 2 ( cos 2 α − cos 4 α ) \sin\alpha\sin 3\alpha = \frac{1}{2}(\cos 2\alpha - \cos 4\alpha) sin α sin 3 α = 2 1 ( cos 2 α − cos 4 α )
Korak 2. Uvrštavamo nazad:
4 sin 2 α ⋅ 1 2 ( cos 2 α − cos 4 α ) = 2 sin 2 α cos 2 α − 2 sin 2 α cos 4 α 4\sin 2\alpha \cdot \frac{1}{2}(\cos 2\alpha - \cos 4\alpha) = 2\sin 2\alpha\cos 2\alpha - 2\sin 2\alpha\cos 4\alpha 4 sin 2 α ⋅ 2 1 ( cos 2 α − cos 4 α ) = 2 sin 2 α cos 2 α − 2 sin 2 α cos 4 α
Korak 3. Na prvi član primenjujemo 2 sin x cos x = sin 2 x 2\sin x\cos x = \sin 2x 2 sin x cos x = sin 2 x 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 )]
sin 4 α − 2 ⋅ 1 2 ( sin 6 α + sin ( − 2 α ) ) = sin 4 α − sin 6 α + sin 2 α \sin 4\alpha - 2 \cdot \frac{1}{2}(\sin 6\alpha + \sin(-2\alpha)) = \sin 4\alpha - \sin 6\alpha + \sin 2\alpha sin 4 α − 2 ⋅ 2 1 ( sin 6 α + sin ( − 2 α )) = sin 4 α − sin 6 α + sin 2 α
= sin 2 α + sin 4 α − sin 6 α ✓ = \sin 2\alpha + \sin 4\alpha - \sin 6\alpha \checkmark = sin 2 α + sin 4 α − sin 6 α ✓
Zadatak. Dokazati: 4 sin α sin β cos ( α + β ) = cos 2 α + cos 2 β − cos 2 ( α + β ) − 1 4\sin\alpha\sin\beta\cos(\alpha+\beta) = \cos 2\alpha + \cos 2\beta - \cos 2(\alpha+\beta) - 1 4 sin α sin β cos ( α + β ) = cos 2 α + cos 2 β − cos 2 ( α + β ) − 1
Korak 1. Primenjujemo sin α sin β = 1 2 [ cos ( α − β ) − cos ( α + β ) ] \sin\alpha\sin\beta = \frac{1}{2}[\cos(\alpha-\beta)-\cos(\alpha+\beta)] sin α sin β = 2 1 [ cos ( α − β ) − cos ( α + β )]
4 ⋅ 1 2 [ cos ( α − β ) − cos ( α + β ) ] cos ( α + β ) 4 \cdot \frac{1}{2}[\cos(\alpha-\beta) - \cos(\alpha+\beta)]\cos(\alpha+\beta) 4 ⋅ 2 1 [ cos ( α − β ) − cos ( α + β )] cos ( α + β )
= 2 cos ( α − β ) cos ( α + β ) − 2 cos 2 ( α + β ) = 2\cos(\alpha-\beta)\cos(\alpha+\beta) - 2\cos^2(\alpha+\beta) = 2 cos ( α − β ) cos ( α + β ) − 2 cos 2 ( α + β )
Korak 2. Na prvi sabirak primenjujemo cos α cos β = 1 2 [ cos ( α − β ) + cos ( α + β ) ] \cos\alpha\cos\beta = \frac{1}{2}[\cos(\alpha-\beta)+\cos(\alpha+\beta)] cos α cos β = 2 1 [ cos ( α − β ) + cos ( α + β )]
2 cos ( α − β ) cos ( α + β ) = cos 2 α + cos ( − 2 β ) = cos 2 α + cos 2 β 2\cos(\alpha-\beta)\cos(\alpha+\beta) = \cos 2\alpha + \cos(-2\beta) = \cos 2\alpha + \cos 2\beta 2 cos ( α − β ) cos ( α + β ) = cos 2 α + cos ( − 2 β ) = cos 2 α + cos 2 β
Korak 3. Drugi sabirak svodimo koristeci 2 cos 2 x = cos 2 x + 1 2\cos^2 x = \cos 2x + 1 2 cos 2 x = cos 2 x + 1 cos 2 x = 2 cos 2 x − 1 \cos 2x = 2\cos^2 x - 1 cos 2 x = 2 cos 2 x − 1
2 cos 2 ( α + β ) = cos 2 ( α + β ) + 1 2\cos^2(\alpha+\beta) = \cos 2(\alpha+\beta) + 1 2 cos 2 ( α + β ) = cos 2 ( α + β ) + 1
Korak 4. Spajamo:
( cos 2 α + cos 2 β ) − ( cos 2 ( α + β ) + 1 ) = cos 2 α + cos 2 β − cos 2 ( α + β ) − 1 ✓ (\cos 2\alpha + \cos 2\beta) - (\cos 2(\alpha+\beta) + 1) = \cos 2\alpha + \cos 2\beta - \cos 2(\alpha+\beta) - 1 \checkmark ( cos 2 α + cos 2 β ) − ( cos 2 ( α + β ) + 1 ) = cos 2 α + cos 2 β − cos 2 ( α + β ) − 1 ✓
Stepenovanje i dokazivanje sa višestrukim uglovima
Zadatak. Dokazati: sin 3 α cos α = 1 8 ( 2 sin 2 α − sin 4 α ) \sin^3\alpha\cos\alpha = \dfrac{1}{8}(2\sin 2\alpha - \sin 4\alpha) sin 3 α cos α = 8 1 ( 2 sin 2 α − sin 4 α )
Korak 1. Pišemo sin 3 α cos α = sin 2 α ⋅ ( sin α cos α ) \sin^3\alpha\cos\alpha = \sin^2\alpha \cdot (\sin\alpha\cos\alpha) sin 3 α cos α = sin 2 α ⋅ ( sin α cos α ) sin α cos α = 1 2 sin 2 α \sin\alpha\cos\alpha = \frac{1}{2}\sin 2\alpha sin α cos α = 2 1 sin 2 α sin 2 α = 1 − cos 2 α 2 \sin^2\alpha = \frac{1-\cos 2\alpha}{2} sin 2 α = 2 1 − c o s 2 α
L = 1 − cos 2 α 2 ⋅ sin 2 α 2 = 1 4 ( 1 − cos 2 α ) sin 2 α L = \frac{1-\cos 2\alpha}{2} \cdot \frac{\sin 2\alpha}{2} = \frac{1}{4}(1-\cos 2\alpha)\sin 2\alpha L = 2 1 − c o s 2 α ⋅ 2 s i n 2 α = 4 1 ( 1 − cos 2 α ) sin 2 α
Korak 2. Razvijamo i na sin 2 α cos 2 α \sin 2\alpha\cos 2\alpha sin 2 α cos 2 α 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 )]
1 4 ( sin 2 α − sin 2 α cos 2 α ) = 1 4 ( sin 2 α − 1 2 sin 4 α ) \frac{1}{4}\left(\sin 2\alpha - \sin 2\alpha\cos 2\alpha\right) = \frac{1}{4}\left(\sin 2\alpha - \frac{1}{2}\sin 4\alpha\right) 4 1 ( sin 2 α − sin 2 α cos 2 α ) = 4 1 ( sin 2 α − 2 1 sin 4 α )
Korak 3. Izvlačimo 1 2 \frac{1}{2} 2 1
= 1 8 ( 2 sin 2 α − sin 4 α ) ✓ = \frac{1}{8}(2\sin 2\alpha - \sin 4\alpha) \checkmark = 8 1 ( 2 sin 2 α − sin 4 α ) ✓
Zadatak. Dokazati: sin 2 α cos 3 α = 1 16 ( 2 cos α − cos 3 α − cos 5 α ) \sin^2\alpha\cos^3\alpha = \dfrac{1}{16}(2\cos\alpha - \cos 3\alpha - \cos 5\alpha) sin 2 α cos 3 α = 16 1 ( 2 cos α − cos 3 α − cos 5 α )
Korak 1. Koristimo sin 2 α = 1 − cos 2 α 2 \sin^2\alpha = \dfrac{1-\cos 2\alpha}{2} sin 2 α = 2 1 − cos 2 α cos 3 α = 3 cos α + cos 3 α 4 \cos^3\alpha = \dfrac{3\cos\alpha + \cos 3\alpha}{4} cos 3 α = 4 3 cos α + cos 3 α
L = 1 − cos 2 α 2 ⋅ 3 cos α + cos 3 α 4 = 1 8 ( 3 cos α + cos 3 α − cos 2 α ( 3 cos α + cos 3 α ) ) L = \frac{1-\cos 2\alpha}{2} \cdot \frac{3\cos\alpha + \cos 3\alpha}{4} = \frac{1}{8}(3\cos\alpha + \cos 3\alpha - \cos 2\alpha(3\cos\alpha + \cos 3\alpha)) L = 2 1 − c o s 2 α ⋅ 4 3 c o s α + c o s 3 α = 8 1 ( 3 cos α + cos 3 α − cos 2 α ( 3 cos α + cos 3 α ))
Korak 2. Primenjujemo cos α cos β = 1 2 [ cos ( α − β ) + cos ( α + β ) ] \cos\alpha\cos\beta = \frac{1}{2}[\cos(\alpha-\beta)+\cos(\alpha+\beta)] cos α cos β = 2 1 [ cos ( α − β ) + cos ( α + β )]
cos 2 α cos α = 1 2 ( cos α + cos 3 α ) \cos 2\alpha\cos\alpha = \frac{1}{2}(\cos\alpha + \cos 3\alpha) cos 2 α cos α = 2 1 ( cos α + cos 3 α )
cos 2 α cos 3 α = 1 2 ( cos α + cos 5 α ) \cos 2\alpha\cos 3\alpha = \frac{1}{2}(\cos\alpha + \cos 5\alpha) cos 2 α cos 3 α = 2 1 ( cos α + cos 5 α )
Korak 3. Uvrštavamo i sređujemo:
L = 1 8 ( 3 cos α + cos 3 α − 3 2 cos α − 3 2 cos 3 α − 1 2 cos α − 1 2 cos 5 α ) L = \frac{1}{8}\left(3\cos\alpha + \cos 3\alpha - \frac{3}{2}\cos\alpha - \frac{3}{2}\cos 3\alpha - \frac{1}{2}\cos\alpha - \frac{1}{2}\cos 5\alpha\right) L = 8 1 ( 3 cos α + cos 3 α − 2 3 cos α − 2 3 cos 3 α − 2 1 cos α − 2 1 cos 5 α )
= 1 8 ⋅ 1 2 ( 4 cos α − cos 3 α − cos 5 α ) = \frac{1}{8} \cdot \frac{1}{2}(4\cos\alpha - \cos 3\alpha - \cos 5\alpha) = 8 1 ⋅ 2 1 ( 4 cos α − cos 3 α − cos 5 α )
= 1 16 ( 2 cos α − cos 3 α − cos 5 α ) ✓ = \frac{1}{16}(2\cos\alpha - \cos 3\alpha - \cos 5\alpha) \checkmark = 16 1 ( 2 cos α − cos 3 α − cos 5 α ) ✓