Qo'shish va ikkilangan burchak formulalari

Bu nima va nega kerak

Trigonometrik jadvallarda asosan standart burchaklarning ($0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$) qiymatlari beriladi. Ammo amaliyotda ko'pincha nostandart burchaklar, masalan $15^\circ$ yoki $75^\circ$ kabi burchaklar uchraydi.

Tabiiy savol tug'iladi: $\sin 75^\circ$ ni qanday hisoblash mumkin? $75^\circ = 45^\circ + 30^\circ$ ekanini bilamiz. Lekin shunchaki $\sin(45^\circ + 30^\circ) = \sin 45^\circ + \sin 30^\circ$ deb yozish mumkinmi? Keling tekshirib ko'ramiz: $\sin 45^\circ + \sin 30^\circ = \frac{\sqrt{2}}{2} + \frac{1}{2} \approx 0{,}71 + 0{,}5 = 1{,}21.$ Biroq sinusning qiymati hech qachon $1$ dan oshishi mumkin emas! Demak, trigonometrik funksiyalar burchaklar yig'indisiga oddiy chiziqli taqsimlanmaydi.

Ikki burchak yig'indisi va ayirmasining trigonometrik funksiyalarini to'g'ri topish uchun qo'shish formulalari xizmat qiladi. Ushbu formulalar to'lqinlar fizikasi, optika, akustika, tebranishlar interferensiyasi hamda elektrotexnikada o'zgaruvchan tok zanjirlarini hisoblashda eng muhim matematik poydevor hisoblanadi.

Asosiy g'oya

1. Qo'shish formulalari

Har qanday $\alpha$ va $\beta$ burchaklar uchun quyidagi ayniyatlar o'rinli:

Kosinus uchun qo'shish formulalari: $\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$ $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$

E'tibor bering: kosinus formulasida argumentdagi ishora teskarisiga o'zgaradi — burchaklar ayirmasida o'rtada plyus, yig'indisida esa minus bo'ladi.

Sinus uchun qo'shish formulalari: $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$ $\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$

Sinus formulasida funksiyalar almashinib keladi ($\sin\cdot\cos$) va o'rtadagi ishora saqlanadi.

Tangens uchun qo'shish formulalari: $\operatorname{tg}(\alpha + \beta) = \frac{\operatorname{tg}\alpha + \operatorname{tg}\beta}{1 - \operatorname{tg}\alpha\operatorname{tg}\beta}$ $\operatorname{tg}(\alpha - \beta) = \frac{\operatorname{tg}\alpha - \operatorname{tg}\beta}{1 + \operatorname{tg}\alpha\operatorname{tg}\beta}$

(Bu formulalar $\cos(\alpha \pm \beta) \neq 0$, $\cos\alpha \neq 0$ va $\cos\beta \neq 0$ bo'lganda o'rinli).

2. Ikkilangan burchak formulalari

Qo'shish formulalarida $\beta = \alpha$ deb olsak, burchakni ikkilantirish formulalari hosil bo'ladi:

Sinus uchun: $\sin 2\alpha = 2\sin\alpha\cos\alpha$

Kosinus uchun: $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha$

Asosiy trigonometrik ayniyat $\sin^2\alpha + \cos^2\alpha = 1$ dan foydalanib, $\cos 2\alpha$ ni faqat bitta funksiya orqali ham yozish mumkin: $\cos 2\alpha = 2\cos^2\alpha - 1$ $\cos 2\alpha = 1 - 2\sin^2\alpha$

Tangens uchun: $\operatorname{tg} 2\alpha = \frac{2\operatorname{tg}\alpha}{1 - \operatorname{tg}^2\alpha}$

3. Darajani pasaytirish formulalari

Kosinusning ikkilangan burchak formulalaridan darajani pasaytirish formulalari kelib chiqadi. Ular integrallash va murakkab ifodalarni soddalashtirishda juda keng qo'llanadi: $\sin^2\alpha = \frac{1 - \cos 2\alpha}{2}$ $\cos^2\alpha = \frac{1 + \cos 2\alpha}{2}$

4. Yig'indi va ayirmani ko'paytmaga keltirish formulalari

Ifodalarni ko'paytuvchilarga ajratish va qisqartirish uchun quyidagi munosabatlardan foydalaniladi: $\sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ $\sin\alpha - \sin\beta = 2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}$ $\cos\alpha + \cos\beta = 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ $\cos\alpha - \cos\beta = -2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$

Yechilgan misollar

1-misol. Nostandart burchak qiymatini hisoblash

$\cos 15^\circ$ va $\sin 75^\circ$ ning aniq qiymatlarini toping.

• $15^\circ = 45^\circ - 30^\circ$ deb yozib, kosinuslar ayirmasi formulasini qo'llaymiz: $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ.$
• Standart qiymatlarni qo'yamiz: $\cos 15^\circ = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}.$
• Keltirish formulasiga ko'ra $\sin 75^\circ = \cos(90^\circ - 75^\circ) = \cos 15^\circ$. Demak, $\sin 75^\circ$ ham aynan shu qiymatga teng: $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}.$

Javob: $\cos 15^\circ = \sin 75^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}$.

2-misol. Ikkilangan burchak funksiyalarini hisoblash

Agar $\sin\alpha = \dfrac{3}{5}$ va $\alpha \in \left(\dfrac{\pi}{2}; \pi\right)$ (ikkinchi chorak) bo'lsa, $\sin 2\alpha$, $\cos 2\alpha$ va $\operatorname{tg} 2\alpha$ ni hisoblang.

• Avval $\cos\alpha$ ni topamiz. Ikkinchi chorakda kosinus manfiy bo'ladi: $\cos\alpha = -\sqrt{1 - \sin^2\alpha} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}.$
• Ikkilangan burchak formulalariga qo'yamiz: $\sin 2\alpha = 2\sin\alpha\cos\alpha = 2\cdot\left(\frac{3}{5}\right)\cdot\left(-\frac{4}{5}\right) = -\frac{24}{25}.$ $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = \left(-\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}.$ $\operatorname{tg} 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} = \frac{-24/25}{7/25} = -\frac{24}{7}.$

Javob: $\sin 2\alpha = -\dfrac{24}{25}$, $\cos 2\alpha = \dfrac{7}{25}$, $\operatorname{tg} 2\alpha = -\dfrac{24}{7}$.

3-misol. Kasr ko'rinishidagi ifodani soddalashtirish

Kasrni qisqartiring va soddalashtiring: $\frac{\sin 3x - \sin x}{\cos 3x + \cos x}.$

• Suratga sinuslar ayirmasi, maxrajga kosinuslar yig'indisi formulasini qo'llaymiz: $\sin 3x - \sin x = 2\sin\frac{3x - x}{2}\cos\frac{3x + x}{2} = 2\sin x\cos 2x.$ $\cos 3x + \cos x = 2\cos\frac{3x + x}{2}\cos\frac{3x - x}{2} = 2\cos 2x\cos x.$
• Surat va maxrajni kasrga qo'yib, umumiy ko'paytuvchi $2\cos 2x$ ga qisqartiramiz: $\frac{2\sin x\cos 2x}{2\cos 2x\cos x} = \frac{\sin x}{\cos x} = \operatorname{tg} x.$

Javob: $\operatorname{tg} x$.

4-misol. Yordamchi burchak kiritish usuli

$f(x) = \sqrt{3}\sin x + \cos x$ funksiyaning eng katta qiymatini toping.

• Ifodaning umumiy ko'paytuvchisini qavsdan tashqariga chiqaramiz. Buning uchun koeffitsiyentlar kvadratlari yig'indisining ildizini hisoblaymiz: $R = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2.$
• Ifodani $2$ ga ko'paytirib va bo'lamiz: $f(x) = 2\left(\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x\right).$
• $\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$ va $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$ ekanini bilamiz. Qo'shish formulasidan foydalanamiz: $f(x) = 2\left(\sin x\cos\frac{\pi}{6} + \cos x\sin\frac{\pi}{6}\right) = 2\sin\left(x + \frac{\pi}{6}\right).$
• Sinus funksiyasining eng katta qiymati $1$ bo'lgani sababli, $f(x)$ ning eng katta qiymati $2\cdot 1 = 2$ bo'ladi.

Javob: $2$.

Mashq

Quyidagi topshiriqlarni mustaqil bajaring va javoblaringizni tekshiring:

1. $\sin 20^\circ\cos 40^\circ + \cos 20^\circ\sin 40^\circ$ ifodaning qiymatini toping.
2. $\cos 70^\circ\cos 10^\circ + \sin 70^\circ\sin 10^\circ$ ifodaning qiymatini hisoblang.
3. $\sin 15^\circ$ ning aniq qiymatini qo'shish formulasi yordamida toping.
4. $\cos^2 15^\circ - \sin^2 15^\circ$ ning qiymatini toping.
5. $\dfrac{2\operatorname{tg} 15^\circ}{1 - \operatorname{tg}^2 15^\circ}$ ifodani hisoblang.
6. Agar $\cos\alpha = \dfrac{5}{13}$ va $\alpha \in \left(\dfrac{3\pi}{2}; 2\pi\right)$ bo'lsa, $\sin 2\alpha$ ni hisoblang.
7. $\dfrac{\cos 2\alpha}{\cos\alpha - \sin\alpha} - \cos\alpha$ ifodani soddalashtiring.
8. $\dfrac{\sin 50^\circ + \sin 10^\circ}{\cos 20^\circ}$ ifodaning son qiymatini toping.
9. $\cos 105^\circ + \cos 15^\circ$ yig'indini ko'paytmaga keltirib hisoblang.
10. $y = 3\sin x - 4\cos x$ funksiyaning eng katta qiymatini toping.

Javoblar va yechim yo'llari

1. Sinuslar yig'indisi formulasi: $\sin(20^\circ + 40^\circ) = \sin 60^\circ = \dfrac{\sqrt{3}}{2}$. Javob: $\dfrac{\sqrt{3}}{2}$.

2. Kosinuslar ayirmasi formulasi: $\cos(70^\circ - 10^\circ) = \cos 60^\circ = \dfrac{1}{2}$. Javob: $\dfrac{1}{2}$.

3. $\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ\cos 30^\circ - \cos 45^\circ\sin 30^\circ = \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2} = \dfrac{\sqrt{6} - \sqrt{2}}{4}$. Javob: $\dfrac{\sqrt{6} - \sqrt{2}}{4}$.

4. Ikkilangan burchak kosinusi: $\cos^2 15^\circ - \sin^2 15^\circ = \cos(2\cdot 15^\circ) = \cos 30^\circ = \dfrac{\sqrt{3}}{2}$. Javob: $\dfrac{\sqrt{3}}{2}$.

5. Ikkilangan burchak tangensi: $\dfrac{2\operatorname{tg} 15^\circ}{1 - \operatorname{tg}^2 15^\circ} = \operatorname{tg}(2\cdot 15^\circ) = \operatorname{tg} 30^\circ = \dfrac{\sqrt{3}}{3}$. Javob: $\dfrac{\sqrt{3}}{3}$.

6. $\alpha \in \left(\dfrac{3\pi}{2}; 2\pi\right)$ to'rtinchi chorak bo'lgani uchun $\sin\alpha < 0$: $\sin\alpha = -\sqrt{1 - \left(\dfrac{5}{13}\right)^2} = -\sqrt{\dfrac{144}{169}} = -\dfrac{12}{13}$. $\sin 2\alpha = 2\sin\alpha\cos\alpha = 2\cdot\left(-\dfrac{12}{13}\right)\cdot\left(\dfrac{5}{13}\right) = -\dfrac{120}{169}$. Javob: $-\dfrac{120}{169}$.

7. $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = (\cos\alpha - \sin\alpha)(\cos\alpha + \sin\alpha)$. Kasrni qisqartiramiz: $\dfrac{(\cos\alpha - \sin\alpha)(\cos\alpha + \sin\alpha)}{\cos\alpha - \sin\alpha} = \cos\alpha + \sin\alpha$. Ayiramiz: $(\cos\alpha + \sin\alpha) - \cos\alpha = \sin\alpha$. Javob: $\sin\alpha$.

8. Yig'indini ko'paytmaga keltiramiz: $\sin 50^\circ + \sin 10^\circ = 2\sin\dfrac{50^\circ + 10^\circ}{2}\cos\dfrac{50^\circ - 10^\circ}{2} = 2\sin 30^\circ\cos 20^\circ$. $\sin 30^\circ = \dfrac{1}{2}$ bo'lgani uchun: $\dfrac{2\cdot\dfrac{1}{2}\cdot\cos 20^\circ}{\cos 20^\circ} = \dfrac{\cos 20^\circ}{\cos 20^\circ} = 1$. Javob: $1$.

9. $\cos 105^\circ + \cos 15^\circ = 2\cos\dfrac{105^\circ + 15^\circ}{2}\cos\dfrac{105^\circ - 15^\circ}{2} = 2\cos 60^\circ\cos 45^\circ = 2\cdot\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{2}}{2}$. Javob: $\dfrac{\sqrt{2}}{2}$.

10. Yordamchi burchak usuliga ko'ra eng katta qiymat: $R = \sqrt{a^2 + b^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. Javob: $5$.

Keng tarqalgan xatolar

1. Burchaklar yig'indisini funksiyalar yig'indisi deb hisoblash. ✗ $\sin(\alpha + \beta) = \sin\alpha + \sin\beta$ yoki $\cos(\alpha - \beta) = \cos\alpha - \cos\beta$. ✓ Trigonometrik funksiyalar chiziqli emas! Doimo to'liq formuladan foydalaning: $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$.

2. Kosinus qo'shish formulasidagi ishorada adashish. ✗ $\cos(\alpha + \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$ deb plyus yozish. ✓ Kosinus formulasida burchaklar orasidagi ishora teskarisiga aylanadi: $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$.

3. Ikkilangan burchakni koeffitsiyent bilan chalkashtirish. ✗ $\cos 2\alpha = 2\cos\alpha$ deb yozish. ✓ Argument $2$ marta kattalashganda funksiya qiymati $2$ ga ko'paymaydi: $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha$.

4. Darajani pasaytirish formulasida ishorani chalkashtirish. ✗ $\sin^2\alpha = \dfrac{1 + \cos 2\alpha}{2}$ deb plyus qo'yish. ✓ Sinus kvadratida minus, kosinus kvadratida esa plyus bo'ladi: $\sin^2\alpha = \dfrac{1 - \cos 2\alpha}{2}$ va $\cos^2\alpha = \dfrac{1 + \cos 2\alpha}{2}$.

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Keyingi qadam

Qo'shish va ikkilangan burchak formulalari yordamida har qanday murakkab ifodalarni soddalashtirishni o'rgandik. Endi ushbu formulalarni noma'lum burchak qatnashgan tenglamalarga tatbiq etamiz — navbatdagi mavzu Trigonometrik tenglamalar.