Matematika

10-sinf · Algebra · 30 daqiqa

Keltirish formulalari

Avval shuni biling:Trigonometrik funksiyalar va ularning xossalari
Trigonometriya: mavzular

Bu nima va nega kerak

Trigonometrik jadvallar asosan 0∘0^\circ dan 90∘90^\circ gacha bo'lgan o'tkir burchaklar uchun tuziladi. Biroq 120∘,225∘,300∘120^\circ, 225^\circ, 300^\circ yoki 7π6\dfrac{7\pi}{6} kabi burchaklarning qiymatlarini ham tez va aniq hisoblash lozim.

Ixtiyoriy chorakdagi burchakni o'tkir burchak trigonometrik funksiyasi orqali ifodalash formulalari keltirish formulalari deyiladi. Jami 3232 ta alohida formula bo'lib, ularni yodlamasdan ikkita oddiy mnemonik qoida bilan xatosiz yozish mumkin.

Asosiy g'oya

Keltirish formulalari π2⋅k±α\dfrac{\pi}{2} \cdot k \pm \alpha yoki 90∘⋅k±α90^\circ \cdot k \pm \alpha (k∈Zk \in \mathbb{Z}) ko'rinishidagi burchaklar uchun qo'llanadi. Bunda α\alpha har doim o'tkir burchak deb qaraladi.

1-qoida: Nom o'zgarishi

  1. Gorizontal o'q (OxOx): π±α\pi \pm \alpha, 2π±α2\pi \pm \alpha (180∘±α180^\circ \pm \alpha, 360∘±α360^\circ \pm \alpha).

    • Funksiya nomi o'zgarmaydi: sin⁡→sin⁡\sin \to \sin, cos⁡→cos⁡\cos \to \cos, tg⁡→tg⁡\operatorname{tg} \to \operatorname{tg}, ctg⁡→ctg⁡\operatorname{ctg} \to \operatorname{ctg}.

  2. Vertikal o'q (OyOy): π2±α\dfrac{\pi}{2} \pm \alpha, 3π2±α\dfrac{3\pi}{2} \pm \alpha (90∘±α90^\circ \pm \alpha, 270∘±α270^\circ \pm \alpha).

    • Funksiya nomi kofunksiyasiga o'zgaradi: sin⁡↔cos⁡\sin \leftrightarrow \cos, tg⁡↔ctg⁡\operatorname{tg} \leftrightarrow \operatorname{ctg}.

2-qoida: Ishora tanlash

Natija oldidagi ishora (++ yoki −-) dastlabki (berilgan) funksiyaning qaralayotgan chorakdagi ishorasi bilan aniqlanadi.

Muhim eslatma: Ishora hosil bo'lgan yangi funksiyaga qarab emas, balki boshlang'ich funksiyaga qarab qo'yiladi!

Asosiy holatlar jadvali

Argumentsin⁡\sincos⁡\costg⁡\operatorname{tg}ctg⁡\operatorname{ctg}
π2−α\dfrac{\pi}{2} - \alphacos⁡α\cos \alphasin⁡α\sin \alphactg⁡α\operatorname{ctg} \alphatg⁡α\operatorname{tg} \alpha
π2+α\dfrac{\pi}{2} + \alphacos⁡α\cos \alpha−sin⁡α-\sin \alpha−ctg⁡α-\operatorname{ctg} \alpha−tg⁡α-\operatorname{tg} \alpha
π−α\pi - \alphasin⁡α\sin \alpha−cos⁡α-\cos \alpha−tg⁡α-\operatorname{tg} \alpha−ctg⁡α-\operatorname{ctg} \alpha
π+α\pi + \alpha−sin⁡α-\sin \alpha−cos⁡α-\cos \alphatg⁡α\operatorname{tg} \alphactg⁡α\operatorname{ctg} \alpha
3π2−α\dfrac{3\pi}{2} - \alpha−cos⁡α-\cos \alpha−sin⁡α-\sin \alphactg⁡α\operatorname{ctg} \alphatg⁡α\operatorname{tg} \alpha
3π2+α\dfrac{3\pi}{2} + \alpha−cos⁡α-\cos \alphasin⁡α\sin \alpha−ctg⁡α-\operatorname{ctg} \alpha−tg⁡α-\operatorname{tg} \alpha
2π−α2\pi - \alpha−sin⁡α-\sin \alphacos⁡α\cos \alpha−tg⁡α-\operatorname{tg} \alpha−ctg⁡α-\operatorname{ctg} \alpha

Davriylikdan foydalanish

Agar burchak 2π2\pi (360∘360^\circ) dan katta bo'lsa, avval to'liq davrlar tashlab yuboriladi: sin⁡(α+2πk)=sin⁡α,cos⁡(α+2πk)=cos⁡α\sin(\alpha + 2\pi k) = \sin \alpha, \qquad \cos(\alpha + 2\pi k) = \cos \alpha tg⁡(α+πk)=tg⁡α,ctg⁡(α+πk)=ctg⁡α\operatorname{tg}(\alpha + \pi k) = \operatorname{tg} \alpha, \qquad \operatorname{ctg}(\alpha + \pi k) = \operatorname{ctg} \alpha

Yechilgan misollar

1-misol. Katta burchaklarning aniq qiymatlarini topish

Hisoblang: a) cos⁡120∘\cos 120^\circ; b) sin⁡225∘\sin 225^\circ; c) tg⁡300∘\operatorname{tg} 300^\circ.

Yechilishi:

a) 120∘=180∘−60∘120^\circ = 180^\circ - 60^\circ (II chorak):

  • II chorakda kosinus manfiy (−-).
  • 180∘180^\circ gorizontal o'qda, nom o'zgarmaydi. cos⁡120∘=cos⁡(180∘−60∘)=−cos⁡60∘=−12.\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}.

b) 225∘=180∘+45∘225^\circ = 180^\circ + 45^\circ (III chorak):

  • III chorakda sinus manfiy (−-).
  • 180∘180^\circ da nom o'zgarmaydi: sin⁡225∘=sin⁡(180∘+45∘)=−sin⁡45∘=−22.\sin 225^\circ = \sin(180^\circ + 45^\circ) = -\sin 45^\circ = -\frac{\sqrt{2}}{2}.

c) 300∘=360∘−60∘300^\circ = 360^\circ - 60^\circ (IV chorak):

  • IV chorakda tangens manfiy (−-).
  • 360∘360^\circ da nom o'zgarmaydi: tg⁡300∘=tg⁡(360∘−60∘)=−tg⁡60∘=−3.\operatorname{tg} 300^\circ = \operatorname{tg}(360^\circ - 60^\circ) = -\operatorname{tg} 60^\circ = -\sqrt{3}.

Javob: a) −12-\dfrac{1}{2}; b) −22-\dfrac{\sqrt{2}}{2}; c) −3-\sqrt{3}.

2-misol. Radianda berilgan burchaklarni hisoblash

Hisoblang: a) sin⁡(7π6)\sin\left(\dfrac{7\pi}{6}\right); b) cos⁡(5π3)\cos\left(\dfrac{5\pi}{3}\right); c) ctg⁡(3π4)\operatorname{ctg}\left(\dfrac{3\pi}{4}\right).

Yechilishi:

a) 7π6=π+π6\dfrac{7\pi}{6} = \pi + \dfrac{\pi}{6} (III chorak): sin⁡(π+π6)=−sin⁡(π6)=−12.\sin\left(\pi + \frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}.

b) 5π3=2π−π3\dfrac{5\pi}{3} = 2\pi - \dfrac{\pi}{3} (IV chorak, kosinus musbat): cos⁡(2π−π3)=cos⁡(π3)=12.\cos\left(2\pi - \frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}.

c) 3π4=π−π4\dfrac{3\pi}{4} = \pi - \dfrac{\pi}{4} (II chorak, kotangens manfiy): ctg⁡(π−π4)=−ctg⁡(π4)=−1.\operatorname{ctg}\left(\pi - \frac{\pi}{4}\right) = -\operatorname{ctg}\left(\frac{\pi}{4}\right) = -1.

Javob: a) −12-\dfrac{1}{2}; b) 12\dfrac{1}{2}; c) −1-1.

3-misol. Harfli ifodani soddalashtirish

Ifodani soddalashtiring: A=sin⁡(π−α)⋅cos⁡(π2+α)cos⁡(2π−α)⋅sin⁡(3π2−α).A = \frac{\sin(\pi - \alpha) \cdot \cos\left(\frac{\pi}{2} + \alpha\right)}{\cos(2\pi - \alpha) \cdot \sin\left(\frac{3\pi}{2} - \alpha\right)}.

Yechilishi: Har bir ko'paytuvchini alohida soddalashtiramiz:

  1. sin⁡(π−α)=sin⁡α\sin(\pi - \alpha) = \sin \alpha (II chorak, sin⁡>0\sin > 0, nom o'zgarmaydi);
  2. cos⁡(π2+α)=−sin⁡α\cos\left(\dfrac{\pi}{2} + \alpha\right) = -\sin \alpha (II chorak, cos⁡<0\cos < 0, nom o'zgaradi);
  3. cos⁡(2π−α)=cos⁡α\cos(2\pi - \alpha) = \cos \alpha (IV chorak, cos⁡>0\cos > 0, nom o'zgarmaydi);
  4. sin⁡(3π2−α)=−cos⁡α\sin\left(\dfrac{3\pi}{2} - \alpha\right) = -\cos \alpha (III chorak, sin⁡<0\sin < 0, nom o'zgaradi).

O'rniga qo'yamiz: A=sin⁡α⋅(−sin⁡α)cos⁡α⋅(−cos⁡α)=−sin⁡2α−cos⁡2α=sin⁡2αcos⁡2α=tg⁡2α.A = \frac{\sin \alpha \cdot (-\sin \alpha)}{\cos \alpha \cdot (-\cos \alpha)} = \frac{-\sin^2 \alpha}{-\cos^2 \alpha} = \frac{\sin^2 \alpha}{\cos^2 \alpha} = \operatorname{tg}^2 \alpha.

Javob: tg⁡2α\operatorname{tg}^2 \alpha.

4-misol. DTM uslubidagi hisoblash masalasi

Ifodaning qiymatini hisoblang: B=cos⁡10∘−sin⁡80∘tg⁡40∘⋅tg⁡50∘+sin⁡390∘.B = \frac{\cos 10^\circ - \sin 80^\circ}{\operatorname{tg} 40^\circ \cdot \operatorname{tg} 50^\circ} + \sin 390^\circ.

Yechilishi:

  1. Keltirish formulasi: sin⁡80∘=sin⁡(90∘−10∘)=cos⁡10∘\sin 80^\circ = \sin(90^\circ - 10^\circ) = \cos 10^\circ. Demak, surat: cos⁡10∘−cos⁡10∘=0\cos 10^\circ - \cos 10^\circ = 0.
  2. Maxraj: tg⁡50∘=tg⁡(90∘−40∘)=ctg⁡40∘\operatorname{tg} 50^\circ = \operatorname{tg}(90^\circ - 40^\circ) = \operatorname{ctg} 40^\circ. tg⁡40∘⋅ctg⁡40∘=1≠0\operatorname{tg} 40^\circ \cdot \operatorname{ctg} 40^\circ = 1 \neq 0. Demak, kasrning qiymati: 01=0\dfrac{0}{1} = 0.
  3. Davriylik: sin⁡390∘=sin⁡(360∘+30∘)=sin⁡30∘=12\sin 390^\circ = \sin(360^\circ + 30^\circ) = \sin 30^\circ = \dfrac{1}{2}.

Natija: B=0+12=0,5.B = 0 + \frac{1}{2} = 0{,}5.

Javob: 0,50{,}5 (yoki 12\dfrac{1}{2}).

Mashq

Quyidagi masalalarni mustaqil ishlang va javoblarni tekshiring:

  1. Hisoblang: a) sin⁡135∘\sin 135^\circ; b) cos⁡210∘\cos 210^\circ; c) tg⁡330∘\operatorname{tg} 330^\circ.

  2. Radiandagi burchaklarni hisoblang: a) cos⁡(4π3)\cos\left(\dfrac{4\pi}{3}\right); b) sin⁡(11π6)\sin\left(\dfrac{11\pi}{6}\right); c) tg⁡(3π4)\operatorname{tg}\left(\dfrac{3\pi}{4}\right).

  3. Ifodani soddalashtiring: sin⁡(π2−α)−cos⁡(π−α)+sin⁡(π+α).\sin\left(\frac{\pi}{2} - \alpha\right) - \cos(\pi - \alpha) + \sin(\pi + \alpha).

  4. Ifodani soddalashtiring: tg⁡(π−α)⋅cos⁡(3π2+α)sin⁡(π+α).\frac{\operatorname{tg}(\pi - \alpha) \cdot \cos\left(\frac{3\pi}{2} + \alpha\right)}{\sin(\pi + \alpha)}.

  5. Hisoblang: sin⁡(−330∘)+cos⁡(−420∘)\sin(-330^\circ) + \cos(-420^\circ).

  6. Soddalashtiring: sin⁡2(π2+α)+sin⁡2(π+α)\sin^2\left(\dfrac{\pi}{2} + \alpha\right) + \sin^2(\pi + \alpha).

  7. Ifodaning qiymatini toping: sin⁡130∘cos⁡40∘+tg⁡25∘⋅tg⁡65∘.\frac{\sin 130^\circ}{\cos 40^\circ} + \operatorname{tg} 25^\circ \cdot \operatorname{tg} 65^\circ.

  8. Agar α,β,γ\alpha, \beta, \gamma uchburchakning ichki burchaklari bo'lsa, sin⁡(α+β)−sin⁡γ\sin(\alpha + \beta) - \sin \gamma ayirmani hisoblang.

Javoblar va yechimlar

  1. Yechilishi: a) sin⁡135∘=sin⁡(180∘−45∘)=sin⁡45∘=22\sin 135^\circ = \sin(180^\circ - 45^\circ) = \sin 45^\circ = \dfrac{\sqrt{2}}{2}. b) cos⁡210∘=cos⁡(180∘+30∘)=−cos⁡30∘=−32\cos 210^\circ = \cos(180^\circ + 30^\circ) = -\cos 30^\circ = -\dfrac{\sqrt{3}}{2}. c) tg⁡330∘=tg⁡(360∘−30∘)=−tg⁡30∘=−33\operatorname{tg} 330^\circ = \operatorname{tg}(360^\circ - 30^\circ) = -\operatorname{tg} 30^\circ = -\dfrac{\sqrt{3}}{3}.

  2. Yechilishi: a) cos⁡(π+π3)=−cos⁡(π3)=−12\cos\left(\pi + \dfrac{\pi}{3}\right) = -\cos\left(\dfrac{\pi}{3}\right) = -\dfrac{1}{2}. b) sin⁡(2π−π6)=−sin⁡(π6)=−12\sin\left(2\pi - \dfrac{\pi}{6}\right) = -\sin\left(\dfrac{\pi}{6}\right) = -\dfrac{1}{2}. c) tg⁡(π−π4)=−tg⁡(π4)=−1\operatorname{tg}\left(\pi - \dfrac{\pi}{4}\right) = -\operatorname{tg}\left(\dfrac{\pi}{4}\right) = -1.

  3. Yechilishi: sin⁡(π/2−α)=cos⁡α\sin(\pi/2 - \alpha) = \cos \alpha, cos⁡(π−α)=−cos⁡α\cos(\pi - \alpha) = -\cos \alpha, sin⁡(π+α)=−sin⁡α\sin(\pi + \alpha) = -\sin \alpha. Ifoda: cos⁡α−(−cos⁡α)+(−sin⁡α)=2cos⁡α−sin⁡α\cos \alpha - (-\cos \alpha) + (-\sin \alpha) = 2\cos \alpha - \sin \alpha.

  4. Yechilishi: tg⁡(π−α)=−tg⁡α=−sin⁡αcos⁡α\operatorname{tg}(\pi - \alpha) = -\operatorname{tg} \alpha = -\dfrac{\sin \alpha}{\cos \alpha}. cos⁡(3π/2+α)=sin⁡α\cos(3\pi/2 + \alpha) = \sin \alpha, sin⁡(π+α)=−sin⁡α\sin(\pi + \alpha) = -\sin \alpha. −tg⁡α⋅sin⁡α−sin⁡α=tg⁡α.\frac{-\operatorname{tg} \alpha \cdot \sin \alpha}{-\sin \alpha} = \operatorname{tg} \alpha.

  5. Yechilishi: sin⁡(−330∘)=−sin⁡(330∘)=−(−sin⁡30∘)=12\sin(-330^\circ) = -\sin(330^\circ) = -(-\sin 30^\circ) = \dfrac{1}{2}. cos⁡(−420∘)=cos⁡(420∘)=cos⁡(360∘+60∘)=cos⁡60∘=12\cos(-420^\circ) = \cos(420^\circ) = \cos(360^\circ + 60^\circ) = \cos 60^\circ = \dfrac{1}{2}. Yig'indi: 12+12=1\dfrac{1}{2} + \dfrac{1}{2} = 1.

  6. Yechilishi: sin⁡(π/2+α)=cos⁡α  ⟹  sin⁡2(π/2+α)=cos⁡2α\sin(\pi/2 + \alpha) = \cos \alpha \implies \sin^2(\pi/2 + \alpha) = \cos^2 \alpha. sin⁡(π+α)=−sin⁡α  ⟹  sin⁡2(π+α)=sin⁡2α\sin(\pi + \alpha) = -\sin \alpha \implies \sin^2(\pi + \alpha) = \sin^2 \alpha. Yig'indi: cos⁡2α+sin⁡2α=1\cos^2 \alpha + \sin^2 \alpha = 1.

  7. Yechilishi: sin⁡130∘=sin⁡(90∘+40∘)=cos⁡40∘  ⟹  cos⁡40∘cos⁡40∘=1\sin 130^\circ = \sin(90^\circ + 40^\circ) = \cos 40^\circ \implies \dfrac{\cos 40^\circ}{\cos 40^\circ} = 1. tg⁡65∘=tg⁡(90∘−25∘)=ctg⁡25∘  ⟹  tg⁡25∘⋅ctg⁡25∘=1\operatorname{tg} 65^\circ = \operatorname{tg}(90^\circ - 25^\circ) = \operatorname{ctg} 25^\circ \implies \operatorname{tg} 25^\circ \cdot \operatorname{ctg} 25^\circ = 1. Natija: 1+1=21 + 1 = 2.

  8. Yechilishi: α+β+γ=180∘  ⟹  α+β=180∘−γ\alpha + \beta + \gamma = 180^\circ \implies \alpha + \beta = 180^\circ - \gamma. sin⁡(α+β)=sin⁡(180∘−γ)=sin⁡γ\sin(\alpha + \beta) = \sin(180^\circ - \gamma) = \sin \gamma. Demak: sin⁡γ−sin⁡γ=0\sin \gamma - \sin \gamma = 0.

Xatolar ro'yxati

Keltirish formulalarini qo'llashda eng ko'p uchraydigan xatolar:

  1. Ishorani yangi funksiya bo'yicha tanlash:

    • ✗ Noto'g'ri: cos⁡(π2+α)\cos\left(\dfrac{\pi}{2} + \alpha\right) da natija sinus bo'lgani uchun II chorakda sinus musbat deb +sin⁡α+\sin \alpha yozish.
    • ✓ To'g'ri: Ishora dastlabki funksiya (cos⁡\cos) bo'yicha aniqlanadi. II chorakda kosinus manfiy, shuning uchun −sin⁡α-\sin \alpha.

  2. O'qni adashtirib, nomni noto'g'ri o'zgartirish:

    • ✗ Noto'g'ri: sin⁡(π+α)=−cos⁡α\sin(\pi + \alpha) = -\cos \alpha.
    • ✓ To'g'ri: π\pi gorizontal o'qda, shuning uchun funksiya nomi o'zgarmaydi: sin⁡(π+α)=−sin⁡α\sin(\pi + \alpha) = -\sin \alpha.

  3. α\alpha burchak o'tmas bo'lsa, formula ishlamaydi deb o'ylash:

    • ✗ Noto'g'ri: Agar α>90∘\alpha > 90^\circ bo'lsa, keltirish formulasi xato natija beradi deb hisoblash.
    • ✓ To'g'ri: Qoida mantiqiy jihatdan α\alpha o'tkir deb qabul qilinganda ham, ixtiyoriy haqiqiy α\alpha uchun ayniyat sifatida to'g'ri qolaveradi.

  4. Kvadrat darajada manfiy ishorani qavsdan tashqariga chiqarish:

    • ✗ Noto'g'ri: sin⁡2(π+α)=−sin⁡2α\sin^2(\pi + \alpha) = -\sin^2 \alpha.
    • ✓ To'g'ri: sin⁡2(π+α)=(−sin⁡α)2=sin⁡2α\sin^2(\pi + \alpha) = (-\sin \alpha)^2 = \sin^2 \alpha. Kvadrat har doim musbat natija beradi.

DTM uslubidagi test

Yuqoridagi frontmatter bo'limida keltirilgan 5 ta test savoli orqali mavzuni o'zlashtirganlik darajangizni sinab ko'ring. Har bir savol DTM standartlariga moslab tuzilgan.

Test

0 / 5 javob berildi

  1. 1.sin⁡150∘\sin 150^\circ ning qiymatini toping.

  2. 2.Ifodani soddalashtiring: cos⁡(π2+α)+sin⁡(π−α)\cos\left(\dfrac{\pi}{2} + \alpha\right) + \sin(\pi - \alpha).

  3. 3.tg⁡225∘\operatorname{tg} 225^\circ ning qiymatini toping.

  4. 4.Ifodaning qiymatini hisoblash: sin⁡20∘cos⁡70∘+tg⁡10∘⋅tg⁡80∘\dfrac{\sin 20^\circ}{\cos 70^\circ} + \operatorname{tg} 10^\circ \cdot \operatorname{tg} 80^\circ.

  5. 5.Ifodani soddalashtiring: sin⁡(3π2−α)cos⁡(π+α)\dfrac{\sin\left(\dfrac{3\pi}{2} - \alpha\right)}{\cos(\pi + \alpha)}.