Trigonometrijske funkcije dvostrukog ugla
Formule za dvostruki ugao su poseban slučaj adicionih formula kada su oba ugla jednaka: α + α = 2 α \alpha + \alpha = 2\alpha α + α = 2 α β = α \beta = \alpha β = α
Sinus dvostrukog ugla:
sin 2 α = 2 sin α cos α \sin 2\alpha = 2\sin\alpha\cos\alpha sin 2 α = 2 sin α cos α
Kosinus dvostrukog ugla (tri ekvivalentna oblika):
cos 2 α = cos 2 α − sin 2 α \cos 2\alpha = \cos^2\alpha - \sin^2\alpha cos 2 α = cos 2 α − sin 2 α
Tangens i kotangens dvostrukog ugla:
tg 2 α = 2 tg α 1 − tg 2 α ctg 2 α = ctg 2 α − 1 2 ctg α \operatorname{tg} 2\alpha = \frac{2\operatorname{tg}\alpha}{1 - \operatorname{tg}^2\alpha} \qquad \operatorname{ctg} 2\alpha = \frac{\operatorname{ctg}^2\alpha - 1}{2\operatorname{ctg}\alpha} tg 2 α = 1 − tg 2 α 2 tg α ctg 2 α = 2 ctg α ctg 2 α − 1
Alternativni oblici sa tangensom:
sin 2 α = 2 tg α 1 + tg 2 α cos 2 α = 1 − tg 2 α 1 + tg 2 α \sin 2\alpha = \frac{2\operatorname{tg}\alpha}{1 + \operatorname{tg}^2\alpha} \qquad \cos 2\alpha = \frac{1 - \operatorname{tg}^2\alpha}{1 + \operatorname{tg}^2\alpha} sin 2 α = 1 + tg 2 α 2 tg α cos 2 α = 1 + tg 2 α 1 − tg 2 α
Sve formule slede direktno iz adicionih formula zamenom β = α \beta = \alpha β = α
Sinus: iz 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 β = α \beta = \alpha β = α
sin 2 α = sin α cos α + cos α sin α = 2 sin α cos α \sin 2\alpha = \sin\alpha\cos\alpha + \cos\alpha\sin\alpha = 2\sin\alpha\cos\alpha sin 2 α = sin α cos α + cos α sin α = 2 sin α cos α
Kosinus: iz 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 β = α \beta = \alpha β = α
cos 2 α = cos 2 α − sin 2 α \cos 2\alpha = \cos^2\alpha - \sin^2\alpha cos 2 α = cos 2 α − sin 2 α
Tangens: iz tg ( α + β ) = tg α + tg β 1 − tg α tg β \operatorname{tg}(\alpha + \beta) = \dfrac{\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 β = α \beta = \alpha β = α
tg 2 α = 2 tg α 1 − tg 2 α \operatorname{tg} 2\alpha = \frac{2\operatorname{tg}\alpha}{1 - \operatorname{tg}^2\alpha} tg 2 α = 1 − tg 2 α 2 tg α
Kotangens: iz ctg ( α + β ) = ctg α ctg β − 1 ctg α + ctg β \operatorname{ctg}(\alpha + \beta) = \dfrac{\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 β = α \beta = \alpha β = α
ctg 2 α = ctg 2 α − 1 2 ctg α \operatorname{ctg} 2\alpha = \frac{\operatorname{ctg}^2\alpha - 1}{2\operatorname{ctg}\alpha} ctg 2 α = 2 ctg α ctg 2 α − 1
Alternativni oblici sa tangensom dobijaju se iz sin 2 α = 2 sin α cos α \sin 2\alpha = 2\sin\alpha\cos\alpha sin 2 α = 2 sin α cos α cos 2 α = cos 2 α − sin 2 α \cos 2\alpha = \cos^2\alpha - \sin^2\alpha cos 2 α = cos 2 α − sin 2 α cos 2 α \cos^2\alpha cos 2 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
sin 2 α = 2 sin α cos α sin 2 α + cos 2 α = 2 sin α cos α 1 + sin 2 α cos 2 α = 2 tg α 1 + tg 2 α \sin 2\alpha = \frac{2\sin\alpha\cos\alpha}{\sin^2\alpha + \cos^2\alpha} = \frac{\frac{2\sin\alpha}{\cos\alpha}}{1 + \frac{\sin^2\alpha}{\cos^2\alpha}} = \frac{2\operatorname{tg}\alpha}{1 + \operatorname{tg}^2\alpha} sin 2 α = s i n 2 α + c o s 2 α 2 s i n α c o s α = 1 + c o s 2 α s i n 2 α c o s α 2 s i n α = 1 + tg 2 α 2 tg α
cos 2 α = cos 2 α − sin 2 α sin 2 α + cos 2 α = 1 − sin 2 α cos 2 α 1 + sin 2 α cos 2 α = 1 − tg 2 α 1 + tg 2 α \cos 2\alpha = \frac{\cos^2\alpha - \sin^2\alpha}{\sin^2\alpha + \cos^2\alpha} = \frac{1 - \frac{\sin^2\alpha}{\cos^2\alpha}}{1 + \frac{\sin^2\alpha}{\cos^2\alpha}} = \frac{1 - \operatorname{tg}^2\alpha}{1 + \operatorname{tg}^2\alpha} cos 2 α = s i n 2 α + c o s 2 α c o s 2 α − s i n 2 α = 1 + c o s 2 α s i n 2 α 1 − c o s 2 α s i n 2 α = 1 + tg 2 α 1 − tg 2 α
3. Primeri
3.1 Izračunavanje funkcija dvostrukog ugla iz poznatih vrednosti
Najpre odredimo nepoznatu funkciju iz kvadranta i osnovnog identiteta, pa primenimo formule.
Zadatak. Izračunati sin 2 α \sin 2\alpha sin 2 α cos 2 α \cos 2\alpha cos 2 α tg 2 α \operatorname{tg} 2\alpha tg 2 α sin α = 3 5 \sin\alpha = \dfrac{3}{5} sin α = 5 3 α ∈ ( π 2 , π ) \alpha \in \left(\dfrac{\pi}{2}, \pi\right) α ∈ ( 2 π , π )
Korak 1. Određujemo cos α \cos\alpha cos α cos 2 α = 1 − 9 25 = 16 25 ⟹ cos α = − 4 5 \cos^2\alpha = 1 - \frac{9}{25} = \frac{16}{25} \implies \cos\alpha = -\frac{4}{5} cos 2 α = 1 − 25 9 = 25 16 ⟹ cos α = − 5 4
Korak 2. Primenjujemo formule:
sin 2 α = 2 ⋅ 3 5 ⋅ ( − 4 5 ) = − 24 25 \sin 2\alpha = 2 \cdot \frac{3}{5} \cdot \left(-\frac{4}{5}\right) = -\frac{24}{25} sin 2 α = 2 ⋅ 5 3 ⋅ ( − 5 4 ) = − 25 24
cos 2 α = cos 2 α − sin 2 α = 16 25 − 9 25 = 7 25 \cos 2\alpha = \cos^2\alpha - \sin^2\alpha = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} cos 2 α = cos 2 α − sin 2 α = 25 16 − 25 9 = 25 7
Korak 3. Tangens dvostrukog ugla najlakše dobijamo iz već izračunatih vrednosti:
tg 2 α = sin 2 α cos 2 α = − 24 25 7 25 = − 24 7 \operatorname{tg} 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} = \frac{-\frac{24}{25}}{\frac{7}{25}} = -\frac{24}{7} tg 2 α = c o s 2 α s i n 2 α = 25 7 − 25 24 = − 7 24
Zadatak. Izračunati sin 2 α \sin 2\alpha sin 2 α cos 2 α \cos 2\alpha cos 2 α tg 2 α \operatorname{tg} 2\alpha tg 2 α cos α = − 5 13 \cos\alpha = -\dfrac{5}{13} cos α = − 13 5 sin α > 0 \sin\alpha > 0 sin α > 0
Pošto je cos α < 0 \cos\alpha < 0 cos α < 0 sin α > 0 \sin\alpha > 0 sin α > 0
sin α = 1 − 25 169 = 12 13 \sin\alpha = \sqrt{1 - \frac{25}{169}} = \frac{12}{13} sin α = 1 − 169 25 = 13 12
sin 2 α = 2 ⋅ 12 13 ⋅ ( − 5 13 ) = − 120 169 \sin 2\alpha = 2 \cdot \frac{12}{13} \cdot \left(-\frac{5}{13}\right) = -\frac{120}{169} sin 2 α = 2 ⋅ 13 12 ⋅ ( − 13 5 ) = − 169 120
cos 2 α = 25 169 − 144 169 = − 119 169 \cos 2\alpha = \frac{25}{169} - \frac{144}{169} = -\frac{119}{169} cos 2 α = 169 25 − 169 144 = − 169 119
tg 2 α = − 120 169 − 119 169 = 120 119 \operatorname{tg} 2\alpha = \frac{-\frac{120}{169}}{-\frac{119}{169}} = \frac{120}{119} tg 2 α = − 169 119 − 169 120 = 119 120
Tangens dvostrukog ugla je najlakše računati kao sin 2 α cos 2 α \dfrac{\sin 2\alpha}{\cos 2\alpha} cos 2 α sin 2 α 2 tg α 1 − tg 2 α \dfrac{2\operatorname{tg}\alpha}{1 - \operatorname{tg}^2\alpha} 1 − tg 2 α 2 tg α tg α \operatorname{tg}\alpha tg α
Kada je trigonometrijska vrednost izražena parametrom, postupak je isti, ali rezultat sadrži parametar.
Zadatak. Odrediti sin 2 x \sin 2x sin 2 x cos 2 x \cos 2x cos 2 x sin x = m − n m + n \sin x = \dfrac{m - n}{m + n} sin x = m + n m − n m + n ≠ 0 m + n \neq 0 m + n = 0 m n > 0 mn > 0 mn > 0
Korak 1. Računamo cos 2 x \cos^2 x cos 2 x cos 2 x = 1 − ( m − n m + n ) 2 = ( m + n ) 2 − ( m − n ) 2 ( m + n ) 2 = 4 m n ( m + n ) 2 \cos^2 x = 1 - \left(\frac{m-n}{m+n}\right)^2 = \frac{(m+n)^2 - (m-n)^2}{(m+n)^2} = \frac{4mn}{(m+n)^2} cos 2 x = 1 − ( m + n m − n ) 2 = ( m + n ) 2 ( m + n ) 2 − ( m − n ) 2 = ( m + n ) 2 4 mn
Pošto m n > 0 mn > 0 mn > 0 ∣ cos x ∣ = 2 m n m + n |\cos x| = \dfrac{2\sqrt{mn}}{m+n} ∣ cos x ∣ = m + n 2 mn cos x = ± 2 m n m + n \cos x = \pm\dfrac{2\sqrt{mn}}{m+n} cos x = ± m + n 2 mn
Korak 2. Sinus dvostrukog ugla (znak zavisi od kvadranta):
sin 2 x = 2 ⋅ m − n m + n ⋅ ( ± 2 m n m + n ) = ± 4 ( m − n ) m n ( m + n ) 2 \sin 2x = 2 \cdot \frac{m-n}{m+n} \cdot \left(\pm\frac{2\sqrt{mn}}{m+n}\right) = \pm\frac{4(m-n)\sqrt{mn}}{(m+n)^2} sin 2 x = 2 ⋅ m + n m − n ⋅ ( ± m + n 2 mn ) = ± ( m + n ) 2 4 ( m − n ) mn
Korak 3. Kosinus dvostrukog ugla ne zavisi od znaka cos x \cos x cos x cos 2 x = cos 2 x − sin 2 x = 4 m n ( m + n ) 2 − ( m − n ) 2 ( m + n ) 2 = 4 m n − m 2 + 2 m n − n 2 ( m + n ) 2 = − m 2 − 6 m n + n 2 ( m + n ) 2 \cos 2x = \cos^2 x - \sin^2 x = \frac{4mn}{(m+n)^2} - \frac{(m-n)^2}{(m+n)^2} = \frac{4mn - m^2 + 2mn - n^2}{(m+n)^2} = -\frac{m^2 - 6mn + n^2}{(m+n)^2} cos 2 x = cos 2 x − sin 2 x = ( m + n ) 2 4 mn − ( m + n ) 2 ( m − n ) 2 = ( m + n ) 2 4 mn − m 2 + 2 mn − n 2 = − ( m + n ) 2 m 2 − 6 mn + n 2
3.3 Uprošćavanje prepoznavanjem obrasca
Kada izraz sadrži sin α cos α \sin\alpha\cos\alpha sin α cos α cos 2 α − sin 2 α \cos^2\alpha - \sin^2\alpha cos 2 α − sin 2 α
Zadatak. Uprostiti: ( cos 2 α + 2 sin α cos α − sin 2 α ) 2 (\cos^2\alpha + 2\sin\alpha\cos\alpha - \sin^2\alpha)^2 ( cos 2 α + 2 sin α cos α − sin 2 α ) 2
Prepoznajemo unutar zagrade cos 2 α = cos 2 α − sin 2 α \cos 2\alpha = \cos^2\alpha - \sin^2\alpha cos 2 α = cos 2 α − sin 2 α sin 2 α = 2 sin α cos α \sin 2\alpha = 2\sin\alpha\cos\alpha sin 2 α = 2 sin α cos α
( cos 2 α + sin 2 α ) 2 = cos 2 2 α + 2 sin 2 α cos 2 α + sin 2 2 α (\cos 2\alpha + \sin 2\alpha)^2 = \cos^2 2\alpha + 2\sin 2\alpha\cos 2\alpha + \sin^2 2\alpha ( cos 2 α + sin 2 α ) 2 = cos 2 2 α + 2 sin 2 α cos 2 α + sin 2 2 α
= ( sin 2 2 α + cos 2 2 α ) + 2 sin 2 α cos 2 α = 1 + sin 4 α = (\sin^2 2\alpha + \cos^2 2\alpha) + 2\sin 2\alpha\cos 2\alpha = 1 + \sin 4\alpha = ( sin 2 2 α + cos 2 2 α ) + 2 sin 2 α cos 2 α = 1 + sin 4 α
Zadatak. Uprostiti: sin 3 x sin x − cos 3 x cos x \dfrac{\sin 3x}{\sin x} - \dfrac{\cos 3x}{\cos x} sin x sin 3 x − cos x cos 3 x
Svodimo na zajednički imenilac:
sin 3 x cos x − cos 3 x sin x sin x cos x \frac{\sin 3x\cos x - \cos 3x\sin x}{\sin x\cos x} s i n x c o s x s i n 3 x c o s x − c o s 3 x s i n x
Brojilac je obrazac za sin ( 3 x − x ) = sin 2 x \sin(3x - x) = \sin 2x sin ( 3 x − x ) = sin 2 x 1 2 sin 2 x \frac{1}{2}\sin 2x 2 1 sin 2 x
sin 2 x 1 2 sin 2 x = 2 \frac{\sin 2x}{\frac{1}{2}\sin 2x} = 2 2 1 s i n 2 x s i n 2 x = 2
Vrednost izraza je konstantna i ne zavisi od x x x
Zadatak. Dokazati: sin 15 ° cos 15 ° = 1 4 \sin 15°\cos 15° = \dfrac{1}{4} sin 15° cos 15° = 4 1
Množimo i delimo sa 2 2 2 sin 2 α \sin 2\alpha sin 2 α
sin 15 ° cos 15 ° = 2 sin 15 ° cos 15 ° 2 = sin 30 ° 2 = 1 2 2 = 1 4 ✓ \sin 15°\cos 15° = \frac{2\sin 15°\cos 15°}{2} = \frac{\sin 30°}{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4} \checkmark sin 15° cos 15° = 2 2 s i n 15° c o s 15° = 2 s i n 30° = 2 2 1 = 4 1 ✓
3.4 Dokazivanje identiteta
Identiteti se dokazuju razvijanjem jedne strane formulama za dvostruki ugao i algebarskim sređivanjem.
Zadatak. Dokazati: 1 − cos 2 α + sin 2 α 1 + cos 2 α + sin 2 α = tg α \dfrac{1 - \cos 2\alpha + \sin 2\alpha}{1 + \cos 2\alpha + \sin 2\alpha} = \operatorname{tg}\alpha 1 + cos 2 α + sin 2 α 1 − cos 2 α + sin 2 α = tg α
Zamenjujemo cos 2 α = cos 2 α − sin 2 α \cos 2\alpha = \cos^2\alpha - \sin^2\alpha cos 2 α = cos 2 α − sin 2 α sin 2 α = 2 sin α cos α \sin 2\alpha = 2\sin\alpha\cos\alpha sin 2 α = 2 sin α cos α 1 = sin 2 α + cos 2 α 1 = \sin^2\alpha + \cos^2\alpha 1 = sin 2 α + cos 2 α
sin 2 α + cos 2 α − cos 2 α + sin 2 α + 2 sin α cos α sin 2 α + cos 2 α + cos 2 α − sin 2 α + 2 sin α cos α = 2 sin 2 α + 2 sin α cos α 2 cos 2 α + 2 sin α cos α \frac{\sin^2\alpha + \cos^2\alpha - \cos^2\alpha + \sin^2\alpha + 2\sin\alpha\cos\alpha}{\sin^2\alpha + \cos^2\alpha + \cos^2\alpha - \sin^2\alpha + 2\sin\alpha\cos\alpha} = \frac{2\sin^2\alpha + 2\sin\alpha\cos\alpha}{2\cos^2\alpha + 2\sin\alpha\cos\alpha} s i n 2 α + c o s 2 α + c o s 2 α − s i n 2 α + 2 s i n α c o s α s i n 2 α + c o s 2 α − c o s 2 α + s i n 2 α + 2 s i n α c o s α = 2 c o s 2 α + 2 s i n α c o s α 2 s i n 2 α + 2 s i n α c o s α
Izvlačimo zajedničke faktore:
= 2 sin α ( sin α + cos α ) 2 cos α ( cos α + sin α ) = sin α cos α = tg α ✓ = \frac{2\sin\alpha(\sin\alpha + \cos\alpha)}{2\cos\alpha(\cos\alpha + \sin\alpha)} = \frac{\sin\alpha}{\cos\alpha} = \operatorname{tg}\alpha \checkmark = 2 c o s α ( c o s α + s i n α ) 2 s i n α ( s i n α + c o s α ) = c o s α s i n α = tg α ✓
Zadatak. Dokazati: cos 4 α + sin 4 α = 1 − 1 2 sin 2 2 α \cos^4\alpha + \sin^4\alpha = 1 - \dfrac{1}{2}\sin^2 2\alpha cos 4 α + sin 4 α = 1 − 2 1 sin 2 2 α
Koristimo a 4 + b 4 = ( a 2 + b 2 ) 2 − 2 a 2 b 2 a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 a 4 + b 4 = ( a 2 + b 2 ) 2 − 2 a 2 b 2 a = cos α a = \cos\alpha a = cos α b = sin α b = \sin\alpha b = sin α
cos 4 α + sin 4 α = ( cos 2 α + sin 2 α ) 2 − 2 sin 2 α cos 2 α = 1 − 2 sin 2 α cos 2 α \cos^4\alpha + \sin^4\alpha = (\cos^2\alpha + \sin^2\alpha)^2 - 2\sin^2\alpha\cos^2\alpha = 1 - 2\sin^2\alpha\cos^2\alpha cos 4 α + sin 4 α = ( cos 2 α + sin 2 α ) 2 − 2 sin 2 α cos 2 α = 1 − 2 sin 2 α cos 2 α
Pošto sin 2 α = 2 sin α cos α \sin 2\alpha = 2\sin\alpha\cos\alpha sin 2 α = 2 sin α cos α sin 2 2 α = 4 sin 2 α cos 2 α \sin^2 2\alpha = 4\sin^2\alpha\cos^2\alpha sin 2 2 α = 4 sin 2 α cos 2 α 2 sin 2 α cos 2 α = 1 2 sin 2 2 α 2\sin^2\alpha\cos^2\alpha = \dfrac{1}{2}\sin^2 2\alpha 2 sin 2 α cos 2 α = 2 1 sin 2 2 α
= 1 − 1 2 sin 2 2 α ✓ = 1 - \frac{1}{2}\sin^2 2\alpha \checkmark = 1 − 2 1 sin 2 2 α ✓
Zadatak. Dokazati: 1 sin 10 ° − 3 cos 10 ° = 4 \dfrac{1}{\sin 10°} - \dfrac{\sqrt{3}}{\cos 10°} = 4 sin 10° 1 − cos 10° 3 = 4
Svodimo na zajednički imenilac:
cos 10 ° − 3 sin 10 ° sin 10 ° cos 10 ° \frac{\cos 10° - \sqrt{3}\sin 10°}{\sin 10°\cos 10°} s i n 10° c o s 10° c o s 10° − 3 s i n 10°
U brojiocu izvlačimo faktor 2 2 2 2 ( 1 2 cos 10 ° − 3 2 sin 10 ° ) sin 10 ° cos 10 ° = 2 ( sin 30 ° cos 10 ° − cos 30 ° sin 10 ° ) sin 10 ° cos 10 ° \frac{2\left(\frac{1}{2}\cos 10° - \frac{\sqrt{3}}{2}\sin 10°\right)}{\sin 10°\cos 10°} = \frac{2(\sin 30°\cos 10° - \cos 30°\sin 10°)}{\sin 10°\cos 10°} s i n 10° c o s 10° 2 ( 2 1 c o s 10° − 2 3 s i n 10° ) = s i n 10° c o s 10° 2 ( s i n 30° c o s 10° − c o s 30° s i n 10° )
Prepoznajemo sin ( 30 ° − 10 ° ) = sin 20 ° \sin(30° - 10°) = \sin 20° sin ( 30° − 10° ) = sin 20° sin 10 ° cos 10 ° = 1 2 sin 20 ° \sin 10°\cos 10° = \dfrac{1}{2}\sin 20° sin 10° cos 10° = 2 1 sin 20°
= 2 sin 20 ° 1 2 sin 20 ° = 4 ✓ = \frac{2\sin 20°}{\frac{1}{2}\sin 20°} = 4 \checkmark = 2 1 s i n 20° 2 s i n 20° = 4 ✓
Formule za sin 3 α \sin 3\alpha sin 3 α cos 3 α \cos 3\alpha cos 3 α tg 3 α \operatorname{tg} 3\alpha tg 3 α ctg 3 α \operatorname{ctg} 3\alpha ctg 3 α
Zadatak. Dokazati: cos 3 α = 4 cos 3 α − 3 cos α \cos 3\alpha = 4\cos^3\alpha - 3\cos\alpha cos 3 α = 4 cos 3 α − 3 cos α
Pišemo cos 3 α = cos ( 2 α + α ) \cos 3\alpha = \cos(2\alpha + \alpha) cos 3 α = cos ( 2 α + α )
cos 2 α cos α − sin 2 α sin α = ( cos 2 α − sin 2 α ) cos α − 2 sin α cos α ⋅ sin α \cos 2\alpha\cos\alpha - \sin 2\alpha\sin\alpha = (\cos^2\alpha - \sin^2\alpha)\cos\alpha - 2\sin\alpha\cos\alpha\cdot\sin\alpha cos 2 α cos α − sin 2 α sin α = ( cos 2 α − sin 2 α ) cos α − 2 sin α cos α ⋅ sin α
= cos 3 α − sin 2 α cos α − 2 sin 2 α cos α = cos 3 α − 3 sin 2 α cos α = \cos^3\alpha - \sin^2\alpha\cos\alpha - 2\sin^2\alpha\cos\alpha = \cos^3\alpha - 3\sin^2\alpha\cos\alpha = cos 3 α − sin 2 α cos α − 2 sin 2 α cos α = cos 3 α − 3 sin 2 α cos α
Zamenjujemo sin 2 α = 1 − cos 2 α \sin^2\alpha = 1 - \cos^2\alpha sin 2 α = 1 − cos 2 α
= cos 3 α − 3 ( 1 − cos 2 α ) cos α = 4 cos 3 α − 3 cos α ✓ = \cos^3\alpha - 3(1 - \cos^2\alpha)\cos\alpha = 4\cos^3\alpha - 3\cos\alpha \checkmark = cos 3 α − 3 ( 1 − cos 2 α ) cos α = 4 cos 3 α − 3 cos α ✓
Zadatak. Dokazati: tg 3 α = 3 tg α − tg 3 α 1 − 3 tg 2 α \operatorname{tg} 3\alpha = \dfrac{3\operatorname{tg}\alpha - \operatorname{tg}^3\alpha}{1 - 3\operatorname{tg}^2\alpha} tg 3 α = 1 − 3 tg 2 α 3 tg α − tg 3 α
Pišemo tg 3 α = tg ( 2 α + α ) \operatorname{tg} 3\alpha = \operatorname{tg}(2\alpha + \alpha) tg 3 α = tg ( 2 α + α )
tg 2 α + tg α 1 − tg 2 α ⋅ tg α \frac{\operatorname{tg} 2\alpha + \operatorname{tg}\alpha}{1 - \operatorname{tg} 2\alpha\cdot\operatorname{tg}\alpha} 1 − tg 2 α ⋅ tg α tg 2 α + tg α
Uvrštavamo tg 2 α = 2 tg α 1 − tg 2 α \operatorname{tg} 2\alpha = \dfrac{2\operatorname{tg}\alpha}{1 - \operatorname{tg}^2\alpha} tg 2 α = 1 − tg 2 α 2 tg α ( 1 − tg 2 α ) (1 - \operatorname{tg}^2\alpha) ( 1 − tg 2 α )
= 2 tg α + tg α ( 1 − tg 2 α ) 1 − tg 2 α 1 − tg 2 α − 2 tg 2 α 1 − tg 2 α = 3 tg α − tg 3 α 1 − 3 tg 2 α ✓ = \frac{\dfrac{2\operatorname{tg}\alpha + \operatorname{tg}\alpha(1 - \operatorname{tg}^2\alpha)}{1 - \operatorname{tg}^2\alpha}}{\dfrac{1 - \operatorname{tg}^2\alpha - 2\operatorname{tg}^2\alpha}{1 - \operatorname{tg}^2\alpha}} = \frac{3\operatorname{tg}\alpha - \operatorname{tg}^3\alpha}{1 - 3\operatorname{tg}^2\alpha} \checkmark = 1 − tg 2 α 1 − tg 2 α − 2 tg 2 α 1 − tg 2 α 2 tg α + tg α ( 1 − tg 2 α ) = 1 − 3 tg 2 α 3 tg α − tg 3 α ✓