Sinus, kosinus, tangens

Bu nima va nega kerak

Baland bino yoki tog' cho'qqisining balandligini to'g'ridan-to'g'ri o'lchash qiyin. Biroq yerdan turib burchakni va masofani o'lchash orqali balandlikni hisoblash mumkin.

To'g'ri burchakli uchburchakda biror burchak ma'lum bo'lsa, uchburchak o'lchamidan qat'i nazar, uning mos tomonlari nisbati o'zgarmasdir, chunki burchaklari teng bo'lgan to'g'ri burchakli uchburchaklar o'zaro o'xshash bo'ladi. Ushbu nisbatlar trigonometrik funksiyalar deyiladi. 9-sinfda biz o'tkir burchaklardan tashqari, $0^\circ$ dan $180^\circ$ gacha bo'lgan burchaklar trigonometriyasini o'rganamiz.

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Asosiy g'oya va ta'riflar

1. To'g'ri burchakli uchburchakda trigonometrik nisbatlar

Gipotenuzasi $c$, katetlari $a$ va $b$ bo'lgan to'g'ri burchakli $ABC$ uchburchakda ($\angle C = 90^\circ$):

          B
         /|
      c / | a (qarshisidagi katet)
       /  |
      /___|
     A  b C
 (yopishgan katet)

$\alpha = \angle A$ o'tkir burchak uchun:

[!IMPORTANT]

• Sinus: $\sin \alpha = \dfrac{a}{c} = \dfrac{\text{qarshisidagi katet}}{\text{gipotenuza}}$
• Kosinus: $\cos \alpha = \dfrac{b}{c} = \dfrac{\text{yopishgan katet}}{\text{gipotenuza}}$
• Tangens: $\operatorname{tg} \alpha = \dfrac{a}{b} = \dfrac{\text{qarshisidagi katet}}{\text{yopishgan katet}}$
• Kotangens: $\operatorname{ctg} \alpha = \dfrac{b}{a} = \dfrac{\text{yopishgan katet}}{\text{qarshisidagi katet}}$

Har qanday o'tkir burchak uchun $0 < \sin \alpha < 1$ va $0 < \cos \alpha < 1$.

2. Asosiy trigonometrik ayniyatlar

Pifagor teoremasidan ($a^2 + b^2 = c^2$) quyidagi tenglik kelib chiqadi: $\sin^2 \alpha + \cos^2 \alpha = 1$

Shuningdek:

• $\operatorname{tg} \alpha = \dfrac{\sin \alpha}{\cos \alpha}, \quad \operatorname{ctg} \alpha = \dfrac{\cos \alpha}{\sin \alpha}, \quad \operatorname{tg} \alpha \cdot \operatorname{ctg} \alpha = 1$
• $1 + \operatorname{tg}^2 \alpha = \dfrac{1}{\cos^2 \alpha}, \quad 1 + \operatorname{ctg}^2 \alpha = \dfrac{1}{\sin^2 \alpha}$

To'ldiruvchi burchaklar uchun: $\sin(90^\circ - \alpha) = \cos \alpha, \quad \cos(90^\circ - \alpha) = \sin \alpha$

3. $0^\circ \le \alpha \le 180^\circ$ oralig'i (birlik yarim aylana)

Koordinata tekisligida radiusi $R = 1$ bo'lgan birlik yarim aylanada musbat $Ox$ nuri bilan $\alpha$ burchak hosil qiluvchi radius uchining koordinatalari $(x, y) = (\cos \alpha, \sin \alpha)$ bo'ladi.

• $0^\circ \le \alpha \le 180^\circ$ oraliqda $\sin \alpha \ge 0$;
• O'tkir burchakda $\cos \alpha > 0$, to'g'ri burchakda $\cos 90^\circ = 0$, o'tmas burchakda $\cos \alpha < 0$.

[!IMPORTANT] Keltirish formulalari: $\sin(180^\circ - \alpha) = \sin \alpha$ $\cos(180^\circ - \alpha) = -\cos \alpha$ $\operatorname{tg}(180^\circ - \alpha) = -\operatorname{tg} \alpha \quad (\alpha \ne 90^\circ)$

4. Asosiy burchaklar jadvali

$\alpha$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$135^\circ$	$150^\circ$	$180^\circ$
$\sin \alpha$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$0$
$\cos \alpha$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$0$	$-\frac{1}{2}$	$-\frac{\sqrt{2}}{2}$	$-\frac{\sqrt{3}}{2}$	$-1$
$\operatorname{tg} \alpha$	$0$	$\frac{\sqrt{3}}{3}$	$1$	$\sqrt{3}$	—	$-\sqrt{3}$	$-1$	$-\frac{\sqrt{3}}{3}$	$0$

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Yechilgan misollar

1-misol. Katetlar bo'yicha trigonometrik nisbatlarni topish

To'g'ri burchakli uchburchakning katetlari $6\text{ sm}$ va $8\text{ sm}$. Kichik burchakning sinusi, kosinusi va tangensini hisoblang.

• Gipotenuzani topamiz: $c = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ sm}$.
• Kichik burchak kichik katet ($6\text{ sm}$) qarshisida yotadi:
  • $\sin \alpha = \dfrac{6}{10} = 0{,}6$
  • $\cos \alpha = \dfrac{8}{10} = 0{,}8$
  • $\operatorname{tg} \alpha = \dfrac{6}{8} = 0{,}75$

Javob: $\sin \alpha = 0{,}6$, $\cos \alpha = 0{,}8$, $\operatorname{tg} \alpha = 0{,}75$.

2-misol. Asosiy ayniyat orqali noma'lumni hisoblash

O'tkir burchak $\alpha$ uchun $\cos \alpha = \dfrac{5}{13}$ bo'lsa, $\sin \alpha$ va $\operatorname{tg} \alpha$ ni toping.

• $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(\dfrac{5}{13}\right)^2 = 1 - \dfrac{25}{169} = \dfrac{144}{169}$.
• O'tkir burchak uchun sinus musbat: $\sin \alpha = \dfrac{12}{13}$.
• Tangens: $\operatorname{tg} \alpha = \dfrac{\sin \alpha}{\cos \alpha} = \dfrac{12/13}{5/13} = \dfrac{12}{5} = 2{,}4$.

Javob: $\sin \alpha = \dfrac{12}{13}$, $\operatorname{tg} \alpha = 2{,}4$.

3-misol. Qiymatlarni hisoblash

Ifodaning qiymatini hisoblang: $E = 2 \sin 30^\circ + 4 \cos 60^\circ - \sqrt{3} \operatorname{tg} 30^\circ + 2 \cos 180^\circ + \sin 150^\circ$

• $\sin 30^\circ = 0{,}5$, $\cos 60^\circ = 0{,}5$, $\operatorname{tg} 30^\circ = \dfrac{\sqrt{3}}{3}$.
• $\cos 180^\circ = -1$, $\sin 150^\circ = \sin 30^\circ = 0{,}5$.
• $E = 2(0{,}5) + 4(0{,}5) - \sqrt{3}\left(\dfrac{\sqrt{3}}{3}\right) + 2(-1) + 0{,}5 = 1 + 2 - 1 - 2 + 0{,}5 = 0{,}5$.

Javob: $0{,}5$.

4-misol. Teng yonli uchburchak elementlari

Teng yonli uchburchakning asosi $16\text{ sm}$, asosidagi burchagi $\alpha$ bo'lib, $\cos \alpha = 0{,}8$. Uchburchakning yon tomoni va balandligini toping.

• Asosga tushirilgan balandlik asosni teng ikkiga bo'ladi: yarim asos $8\text{ sm}$.
• Hosil bo'lgan to'g'ri burchakli uchburchakdan: $\cos \alpha = \dfrac{8}{b} \implies 0{,}8 = \dfrac{8}{b} \implies b = 10\text{ sm}$.
• Balandlik: $h = \sqrt{10^2 - 8^2} = \sqrt{36} = 6\text{ sm}$.

Javob: Yon tomoni $10\text{ sm}$, balandligi $6\text{ sm}$.

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Mashq

1. Katetlari $5\text{ sm}$ va $12\text{ sm}$ bo'lgan to'g'ri burchakli uchburchakning kichik burchagi sinusi va kosinusini toping.
2. O'tkir burchak $\alpha$ uchun $\sin \alpha = 0{,}6$ bo'lsa, $\cos \alpha$ va $\operatorname{tg} \alpha$ ni hisoblang.
3. Hisoblang: $2 \cos 30^\circ \cdot \operatorname{tg} 60^\circ - \sin 90^\circ$.
4. To'g'ri burchakli uchburchakning gipotenuzasi $10\text{ sm}$, bitta o'tkir burchagi $30^\circ$. Uning katetlarini toping.
5. $\alpha = 120^\circ$ uchun $\sin 120^\circ$, $\cos 120^\circ$ va $\operatorname{tg} 120^\circ$ qiymatlarini yozing.
6. Soddalashtiring: $(1 - \cos \alpha)(1 + \cos \alpha) - \sin^2 \alpha$.
7. Rombning tomoni $10\text{ sm}$, o'tkir burchagi $60^\circ$. Kichik diagonali va yuzini hisoblang.
8. Agar $\operatorname{tg} \alpha = 2$ bo'lsa, $\dfrac{3\sin \alpha - \cos \alpha}{\sin \alpha + 2\cos \alpha}$ kasr qiymatini toping.
9. Teng yonli to'g'ri burchakli uchburchakning kateti $6\text{ sm}$. Gipotenuzasi va o'tkir burchagining tangensini toping.
10. To'g'ri to'rtburchakning diagonali $20\text{ sm}$ bo'lib, u katta tomon bilan $30^\circ$ li burchak hosil qiladi. Uning tomonlari va yuzini toping.

Javoblar va yechim yo'llari

1. $c = \sqrt{5^2 + 12^2} = 13\text{ sm}$. Kichik katet qarshisida: $\sin \alpha = \dfrac{5}{13}$, $\cos \alpha = \dfrac{12}{13}$. Javob: $\sin \alpha = \dfrac{5}{13}, \cos \alpha = \dfrac{12}{13}$.
2. $\cos \alpha = \sqrt{1 - 0{,}6^2} = 0{,}8$; $\operatorname{tg} \alpha = \dfrac{0{,}6}{0{,}8} = 0{,}75$. Javob: $\cos \alpha = 0{,}8, \operatorname{tg} \alpha = 0{,}75$.
3. $2 \cdot \dfrac{\sqrt{3}}{2} \cdot \sqrt{3} - 1 = 3 - 1 = 2$. Javob: 2.
4. $a = 10 \sin 30^\circ = 5\text{ sm}$, $b = 10 \cos 30^\circ = 5\sqrt{3}\text{ sm}$. Javob: $5\text{ sm}$ va $5\sqrt{3}\text{ sm}$.
5. $\sin 120^\circ = \sin 60^\circ = \dfrac{\sqrt{3}}{2}$, $\cos 120^\circ = -\cos 60^\circ = -0{,}5$, $\operatorname{tg} 120^\circ = -\sqrt{3}$. Javob: $\dfrac{\sqrt{3}}{2}, -0{,}5, -\sqrt{3}$.
6. $1 - \cos^2 \alpha - \sin^2 \alpha = \sin^2 \alpha - \sin^2 \alpha = 0$. Javob: 0.
7. Kichik diagonal $d = 10\text{ sm}$. Yuza: $S = 10^2 \cdot \sin 60^\circ = 50\sqrt{3}\text{ sm}^2$. Javob: $10\text{ sm}, 50\sqrt{3}\text{ sm}^2$.
8. Surat va maxrajni $\cos \alpha$ ga bo'lsak: $\dfrac{3(2) - 1}{2 + 2} = \dfrac{5}{4} = 1{,}25$. Javob: $1{,}25$.
9. $c = \sqrt{6^2 + 6^2} = 6\sqrt{2}\text{ sm}$, $\operatorname{tg} 45^\circ = 1$. Javob: $6\sqrt{2}\text{ sm}, 1$.
10. $a = 20 \cos 30^\circ = 10\sqrt{3}\text{ sm}$, $b = 20 \sin 30^\circ = 10\text{ sm}$, $S = 100\sqrt{3}\text{ sm}^2$. Javob: $10\text{ sm}$ va $10\sqrt{3}\text{ sm}, 100\sqrt{3}\text{ sm}^2$.

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Keng tarqalgan xatolar

1. Sinus bilan kosinus nisbatlarini almashtirib qo'yish. ✗ $\sin \alpha = \dfrac{\text{yopishgan katet}}{\text{gipotenuza}}$ deb hisoblash. ✓ Sinus — burchak qarshisidagi katetning, kosinus esa yopishgan katetning gipotenuzaga nisbatidir.

2. O'tmas burchak kosinusi ishorasini musbat deb olish. ✗ $\cos 120^\circ = \dfrac{1}{2}$ deb yozish. ✓ Ikkinchi chorakda absissalar manfiy bo'lgani sababli, o'tmas burchakning kosinusi manfiydir: $\cos 120^\circ = -0{,}5$.

3. Trigonometrik funksiyani burchak qo'shiluvchilariga tarqatish. ✗ $\sin(\alpha + \beta) = \sin \alpha + \sin \beta$ deb yozish. ✓ $\sin(30^\circ + 60^\circ) = \sin 90^\circ = 1$, biroq $\sin 30^\circ + \sin 60^\circ = \dfrac{1 + \sqrt{3}}{2} \ne 1$. Funksiya ko'paytuvchi emas.

4. Argumentni qisqartirishga urinish. ✗ $\dfrac{\sin 2x}{2} = \sin x$ deb hisoblash. ✓ Burchak argumenti funksiya ichida joylashgan bo'lib, uni tashqi songa bo'lib bo'lmaydi.

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Keyingi qadam

To'g'ri burchakli uchburchakda trigonometrik nisbatlarni va asosiy ayniyatlarni o'rgandik. Biroq amaliyotda ixtiyoriy burchakli uchburchaklar uchraydi. Keyingi darsda ixtiyoriy uchburchak tomonlari va burchaklarini bog'lovchi asosiy qonunlar — Sinuslar va kosinuslar teoremasi bilan tanishamiz.