Ushbu darsda trigonometrik funksiyalar yig‘indisini ko‘paytmaga va aksincha, ko‘paytmani yig‘indiga aylantiruvchi formulalar haqida so‘z yuritamiz. Ular tenglama yechish va ifoda soddalashtirishda juda qo‘l keladi.
§1. Ko‘paytmani yig‘indiga aylantirish
Qo‘shish formulalarini qo‘shib yoki ayirib, ko‘paytmalarni yig‘indiga aylantiruvchi formulalarni hosil qilamiz.
Ta’rif. Ikki trigonometrik funksiya ko‘paytmasini ularning yig‘indisiga (yoki aksincha) aylantiruvchi tengliklar yig‘indi-ko‘paytma formulalari deyiladi.
Qoida. $$ \cos\alpha\cos\beta=\tfrac{1}{2}\big[\cos(\alpha-\beta)+\cos(\alpha+\beta)\big], $$
sin α sin β = 1 2 [ cos ( α − β ) − cos ( α + β ) ] , \sin\alpha\sin\beta=\tfrac{1}{2}\big[\cos(\alpha-\beta)-\cos(\alpha+\beta)\big],
sin α sin β = 2 1 [ cos ( α − β ) − cos ( α + β ) ] , sin α cos β = 1 2 [ sin ( α + β ) + sin ( α − β ) ] . \sin\alpha\cos\beta=\tfrac{1}{2}\big[\sin(\alpha+\beta)+\sin(\alpha-\beta)\big].
sin α cos β = 2 1 [ sin ( α + β ) + sin ( α − β ) ] .
Savol. Nega bu foydali? Ko‘paytmani qo‘shishga aylantirsak, integrallash yoki hisoblash osonlashadi — qo‘shiluvchilarni alohida-alohida qo‘llash mumkin.
Misol. sin 75 ∘ cos 15 ∘ \sin75^\circ\cos15^\circ sin 7 5 ∘ cos 1 5 ∘ Yechish. sin α cos β \sin\alpha\cos\beta sin α cos β α = 75 ∘ \alpha=75^\circ α = 7 5 ∘ β = 15 ∘ \beta=15^\circ β = 1 5 ∘
sin 75 ∘ cos 15 ∘ = 1 2 [ sin ( 75 ∘ + 15 ∘ ) + sin ( 75 ∘ − 15 ∘ ) ] . \sin75^\circ\cos15^\circ=\tfrac{1}{2}\big[\sin(75^\circ+15^\circ)+\sin(75^\circ-15^\circ)\big].
sin 7 5 ∘ cos 1 5 ∘ = 2 1 [ sin ( 7 5 ∘ + 1 5 ∘ ) + sin ( 7 5 ∘ − 1 5 ∘ ) ] . = 1 2 [ sin 90 ∘ + sin 60 ∘ ] = 1 2 [ 1 + 3 2 ] = 2 + 3 4 . =\tfrac{1}{2}\big[\sin90^\circ+\sin60^\circ\big]=\tfrac{1}{2}\left[1+\frac{\sqrt3}{2}\right]=\frac{2+\sqrt3}{4}.
= 2 1 [ sin 9 0 ∘ + sin 6 0 ∘ ] = 2 1 [ 1 + 2 3 ] = 4 2 + 3 . Javob. 2 + 3 4 \dfrac{2+\sqrt3}{4} 4 2 + 3
§2. Yig‘indini ko‘paytmaga aylantirish
Teskari yo‘nalishdagi formulalar (ko‘pincha "yig‘indi-ayirma" formulalari deb ataladi):
Qoida. $$ \sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}, $$
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 .
Eslatma. cos α − cos β \cos\alpha-\cos\beta cos α − cos β
Misol. sin 40 ∘ + sin 20 ∘ \sin40^\circ+\sin20^\circ sin 4 0 ∘ + sin 2 0 ∘ Yechish. α + β 2 = 40 ∘ + 20 ∘ 2 = 30 ∘ \frac{\alpha+\beta}{2}=\frac{40^\circ+20^\circ}{2}=30^\circ 2 α + β = 2 4 0 ∘ + 2 0 ∘ = 3 0 ∘ α − β 2 = 40 ∘ − 20 ∘ 2 = 10 ∘ \frac{\alpha-\beta}{2}=\frac{40^\circ-20^\circ}{2}=10^\circ 2 α − β = 2 4 0 ∘ − 2 0 ∘ = 1 0 ∘
sin 40 ∘ + sin 20 ∘ = 2 sin 30 ∘ cos 10 ∘ = 2 ⋅ 1 2 ⋅ cos 10 ∘ = cos 10 ∘ . \sin40^\circ+\sin20^\circ=2\sin30^\circ\cos10^\circ=2\cdot\frac{1}{2}\cdot\cos10^\circ=\cos10^\circ.
sin 4 0 ∘ + sin 2 0 ∘ = 2 sin 3 0 ∘ cos 1 0 ∘ = 2 ⋅ 2 1 ⋅ cos 1 0 ∘ = cos 1 0 ∘ . Javob. cos 10 ∘ \cos10^\circ cos 1 0 ∘
§3. Formulalarni birlashtirib qo‘llash
Ko‘pincha ikkala guruh formulalari birga ishlatiladi.
yig‘indi
ko‘paytma
Misol. sin 3 α + sin α cos 3 α + cos α \dfrac{\sin3\alpha+\sin\alpha}{\cos3\alpha+\cos\alpha} cos 3 α + cos α sin 3 α + sin α Yechish. Surat va maxrajni yig‘indi-ko‘paytma formulalari bilan yozamiz. 3 α + α 2 = 2 α \frac{3\alpha+\alpha}{2}=2\alpha 2 3 α + α = 2 α 3 α − α 2 = α \frac{3\alpha-\alpha}{2}=\alpha 2 3 α − α = α
2 sin 2 α cos α 2 cos 2 α cos α = sin 2 α cos 2 α = tan 2 α . \frac{2\sin2\alpha\cos\alpha}{2\cos2\alpha\cos\alpha}=\frac{\sin2\alpha}{\cos2\alpha}=\tan2\alpha.
2 cos 2 α cos α 2 sin 2 α cos α = cos 2 α sin 2 α = tan 2 α . Javob. tan 2 α \tan2\alpha tan 2 α
Misol. cos 40 ∘ − cos 80 ∘ \cos40^\circ-\cos80^\circ cos 4 0 ∘ − cos 8 0 ∘ Yechish. 40 ∘ + 80 ∘ 2 = 60 ∘ \frac{40^\circ+80^\circ}{2}=60^\circ 2 4 0 ∘ + 8 0 ∘ = 6 0 ∘ 40 ∘ − 80 ∘ 2 = − 20 ∘ \frac{40^\circ-80^\circ}{2}=-20^\circ 2 4 0 ∘ − 8 0 ∘ = − 2 0 ∘
cos 40 ∘ − cos 80 ∘ = − 2 sin 60 ∘ sin ( − 20 ∘ ) = 2 sin 60 ∘ sin 20 ∘ = 3 sin 20 ∘ . \cos40^\circ-\cos80^\circ=-2\sin60^\circ\sin(-20^\circ)=2\sin60^\circ\sin20^\circ=\sqrt3\sin20^\circ.
cos 4 0 ∘ − cos 8 0 ∘ = − 2 sin 6 0 ∘ sin ( − 2 0 ∘ ) = 2 sin 6 0 ∘ sin 2 0 ∘ = 3 sin 2 0 ∘ . Javob. 3 sin 20 ∘ \sqrt3\sin20^\circ 3 sin 2 0 ∘
§ Lug‘at
ko‘paytma — product