Trigonometrik funksiyalar va ularning xossalari

Bu nima va nega kerak

Geometriyada sinus, kosinus va tangens to'g'ri burchakli uchburchak tomonlari nisbati sifatida faqat o'tkir burchaklar ($0^\circ < \alpha < 90^\circ$) uchun ta'riflangan edi.

Biroq tebranishlar, to'lqinlar va aylanma harakatni o'rganishda har qanday burchak (manfiy yoki $360^\circ$ dan katta) talab qilinadi. Shu sababli trigonometriya uchburchak doirasidan chiqib, birlik aylana yordamida barcha haqiqiy sonlar uchun umumlashtiriladi.

Asosiy g'oya

1. Birlik aylana va trigonometrik nuqta

Koordinatalar boshida markazlashgan va radiusi $R = 1$ bo'lgan aylana birlik aylana deyiladi.

Boshlang'ich $P_0(1; 0)$ nuqtani $\alpha$ burchakka burishdan hosil bo'lgan nuqta $P_\alpha(x; y)$ bo'lsin (soat miliga qarshi — musbat, soat mili bo'yicha — manfiy yo'nalish).

Ta'riflar:

• $\sin \alpha = y$ ($P_\alpha$ nuqtaning ordinatasi);
• $\cos \alpha = x$ ($P_\alpha$ nuqtaning abssissasi);
• $\operatorname{tg} \alpha = \dfrac{y}{x} = \dfrac{\sin \alpha}{\cos \alpha}$ ($x \neq 0$);
• $\operatorname{ctg} \alpha = \dfrac{x}{y} = \dfrac{\cos \alpha}{\sin \alpha}$ ($y \neq 0$).

Demak, aylanadagi har bir nuqta $P_\alpha = (\cos \alpha; \sin \alpha)$ koordinataga ega.

2. Aniqlanish va qiymatlar sohasi

• $y = \sin x$ va $y = \cos x$: $D(f) = \mathbb{R}$, $E(f) = [-1; 1]$.
• $y = \operatorname{tg} x$: $x \neq \dfrac{\pi}{2} + \pi k$, $E(f) = \mathbb{R}$.
• $y = \operatorname{ctg} x$: $x \neq \pi k$, $E(f) = \mathbb{R}$.

Har qanday $x$ uchun: $-1 \le \sin x \le 1$ va $-1 \le \cos x \le 1$.

3. Funksiyalarning ishoralari

Chorak	Oraliq	$\sin \alpha$	$\cos \alpha$	$\operatorname{tg} \alpha, \operatorname{ctg} \alpha$
I	$\left(0; \dfrac{\pi}{2}\right)$	$+$	$+$	$+$
II	$\left(\dfrac{\pi}{2}; \pi\right)$	$+$	$-$	$-$
III	$\left(\pi; \dfrac{3\pi}{2}\right)$	$-$	$-$	$+$
IV	$\left(\dfrac{3\pi}{2}; 2\pi\right)$	$-$	$+$	$-$

4. Juft va toqligi

• $\cos(-\alpha) = \cos \alpha$ — juft funksiya;
• $\sin(-\alpha) = -\sin \alpha$, $\operatorname{tg}(-\alpha) = -\operatorname{tg} \alpha$, $\operatorname{ctg}(-\alpha) = -\operatorname{ctg} \alpha$ — toq funksiyalar.

5. Asosiy trigonometrik ayniyatlar

Birlik aylanada $x^2 + y^2 = 1$ bo'lgani sababli:

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

Yechilgan misollar

1-misol. Bitta funksiya bo'yicha qolganlarini topish

Agar $\cos \alpha = -\dfrac{5}{13}$ va $\alpha \in \left(\dfrac{\pi}{2}; \pi\right)$ (II chorak) bo'lsa, $\sin \alpha$, $\operatorname{tg} \alpha$ va $\operatorname{ctg} \alpha$ larni toping.

Yechilishi:

1. Asosiy ayniyatdan $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \dfrac{25}{169} = \dfrac{144}{169}$.
2. II chorakda $\sin \alpha > 0$, demak $\sin \alpha = \dfrac{12}{13}$.
3. $\operatorname{tg} \alpha = \dfrac{\sin \alpha}{\cos \alpha} = \dfrac{12/13}{-5/13} = -\dfrac{12}{5} = -2{,}4$.
4. $\operatorname{ctg} \alpha = \dfrac{1}{\operatorname{tg} \alpha} = -\dfrac{5}{12}$.

Javob: $\sin \alpha = \dfrac{12}{13}$; $\operatorname{tg} \alpha = -2{,}4$; $\operatorname{ctg} \alpha = -\dfrac{5}{12}$.

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2-misol. Ayniyatlar yordamida soddalashtirish

Quyidagi ifodalarni soddalashtiring: a) $\dfrac{\sin^4 \alpha - \cos^4 \alpha}{\sin^2 \alpha - \cos^2 \alpha}$; b) $(\sin \alpha + \cos \alpha)^2 - 2\sin \alpha \cos \alpha$.

Yechilishi: a) $\sin^4 \alpha - \cos^4 \alpha = (\sin^2 \alpha - \cos^2 \alpha)(\sin^2 \alpha + \cos^2 \alpha) = (\sin^2 \alpha - \cos^2 \alpha) \cdot 1$. Kasr qisqarib, natija $1$ bo'ladi. b) Qavsni ochsak: $\sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha - 2\sin \alpha \cos \alpha = \sin^2 \alpha + \cos^2 \alpha = 1$.

Javob: a) $1$; b) $1$.

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3-misol. Sonli ifodaning qiymatini hisoblash

Hisoblang: $A = \sin\left(-\dfrac{\pi}{6}\right) + 2\cos\left(\dfrac{\pi}{3}\right) - \operatorname{tg}\left(-\dfrac{\pi}{4}\right)$.

Yechilishi: $\sin(-\pi/6) = -1/2$, $\cos(\pi/3) = 1/2$, $\operatorname{tg}(-\pi/4) = -1$. $A = -\frac{1}{2} + 2 \cdot \frac{1}{2} - (-1) = -\frac{1}{2} + 1 + 1 = 1{,}5.$

Javob: $1{,}5$.

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4-misol. Funksiyaning qiymatlar sohasini topish

Funksiyalarning eng katta va eng kichik qiymatlarini toping: a) $y = 2\sin x - 7$; b) $y = \cos^2 x - 2\cos x + 3$.

Yechilishi: a) $\sin x \in [-1; 1] \implies 2\sin x \in [-2; 2] \implies 2\sin x - 7 \in [-9; -5]$. Demak, $y_{\min} = -9$, $y_{\max} = -5$.

b) To'la kvadrat ajratamiz: $y = (\cos x - 1)^2 + 2$. $t = \cos x \in [-1; 1]$ bo'lgani uchun:

• $t = 1$ da: $y_{\min} = (1 - 1)^2 + 2 = 2$.
• $t = -1$ da: $y_{\max} = (-1 - 1)^2 + 2 = 6$. Demak, $E(y) = [2; 6]$.

Javob: a) $y_{\min} = -9$, $y_{\max} = -5$; b) $y_{\min} = 2$, $y_{\max} = 6$.

Mashq

Quyidagi masalalarni mustaqil ishlang va javoblarni tekshiring:

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

2. Agar $\operatorname{tg} \alpha = 2$ bo'lsa, $\dfrac{3\sin \alpha - \cos \alpha}{\sin \alpha + 2\cos \alpha}$ kasrning qiymatini hisoblang.

3. Ifodani soddalashtiring: $(1 - \cos \alpha)(1 + \cos \alpha) \cdot (1 + \operatorname{ctg}^2 \alpha)$.

4. Hisoblang: $\cos(-60^\circ) + \sin(-30^\circ) - \operatorname{ctg}(-45^\circ)$.

5. $y = 4 - 3\sin^2 x$ funksiyaning eng katta va eng kichik qiymatlarini toping.

6. Agar $\sin \alpha + \cos \alpha = 0{,}8$ bo'lsa, $\sin \alpha \cdot \cos \alpha$ ko'paytmasini toping.

7. Birlik aylanada $\alpha = \dfrac{4\pi}{3}$ burchakka mos keluvchi nuqtaning koordinatalarini toping.

8. $y = \sin^2 x + 4\sin x + 1$ funksiyaning qiymatlar sohasini toping.

Javoblar va yechimlar

1. Yechilishi: II chorakda kosinus manfiy: $\cos \alpha = -\sqrt{1 - (3/5)^2} = -\dfrac{4}{5}$. $\operatorname{tg} \alpha = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}$, $\operatorname{ctg} \alpha = -\dfrac{4}{3}$.

2. Yechilishi: Kasrning surat va maxrajini $\cos \alpha \neq 0$ ga bo'lamiz: $\frac{3\frac{\sin \alpha}{\cos \alpha} - 1}{\frac{\sin \alpha}{\cos \alpha} + 2} = \frac{3\operatorname{tg} \alpha - 1}{\operatorname{tg} \alpha + 2} = \frac{3(2) - 1}{2 + 2} = \frac{5}{4} = 1{,}25.$

3. Yechilishi: $(1 - \cos^2 \alpha) \cdot (1 + \operatorname{ctg}^2 \alpha) = \sin^2 \alpha \cdot \dfrac{1}{\sin^2 \alpha} = 1$.

4. Yechilishi: $\cos(60^\circ) - \sin(30^\circ) + \operatorname{ctg}(45^\circ) = \dfrac{1}{2} - \dfrac{1}{2} + 1 = 1$.

5. Yechilishi: $\sin^2 x \in [0; 1]$. $\sin^2 x = 0 \implies y_{\max} = 4$. $\sin^2 x = 1 \implies y_{\min} = 4 - 3(1) = 1$. $E(y) = [1; 4]$.

6. Yechilishi: Tenglikning ikkala tomonini kvadratga ko'taramiz: $(\sin \alpha + \cos \alpha)^2 = 0{,}8^2 \implies \sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha = 0{,}64.$ $1 + 2\sin \alpha \cos \alpha = 0{,}64 \implies 2\sin \alpha \cos \alpha = -0{,}36 \implies \sin \alpha \cos \alpha = -0{,}18$.

7. Yechilishi: $\cos\left(\dfrac{4\pi}{3}\right) = \cos(240^\circ) = -\dfrac{1}{2}$. $\sin\left(\dfrac{4\pi}{3}\right) = \sin(240^\circ) = -\dfrac{\sqrt{3}}{2}$. Nuqta koordinatalari: $\left(-\dfrac{1}{2}; -\dfrac{\sqrt{3}}{2}\right)$.

8. Yechilishi: $t = \sin x \in [-1; 1]$. Funksiya: $f(t) = t^2 + 4t + 1 = (t + 2)^2 - 3$. Parabola uchi $t = -2$ oraliqdan tashqarida. $[-1; 1]$ kesmada funksiya o'suvchi: $t = -1 \implies f(-1) = 1 - 4 + 1 = -2$. $t = 1 \implies f(1) = 1 + 4 + 1 = 6$. Demak, $E(y) = [-2; 6]$.

Xatolar ro'yxati

O'quvchilar bu mavzuda ko'p uchraydigan odatiy xatolar:

1. Kvadrat ildiz chiqarishda burchak choragini inobatga olmaslik:

   • ✗ Noto'g'ri: $\cos \alpha = \sqrt{1 - \sin^2 \alpha}$ deb har doim musbat ishora olish.
   • ✓ To'g'ri: $\cos \alpha = \pm\sqrt{1 - \sin^2 \alpha}$. Ishora $\alpha$ burchak qaysi chorakka tegishli ekaniga qarab tanlanadi (II va III chorakda manfiy).

2. Kosinus manfiy ishorasini oldinga chiqarish:

   • ✗ Noto'g'ri: $\cos(-x) = -\cos x$.
   • ✓ To'g'ri: Kosinus — yagona juft asosiy trigonometrik funksiyadir: $\cos(-x) = \cos x$. Qolganlari esa toq: $\sin(-x) = -\sin x$, $\operatorname{tg}(-x) = -\operatorname{tg} x$.

3. Yozuvdagi qavslarni e'tiborsiz qoldirish:

   • ✗ Noto'g'ri: $\sin^2 x$ bilan $\sin x^2$ ni bir xil deb tushunish.
   • ✓ To'g'ri: $\sin^2 x = (\sin x)^2$ (funksiya qiymatining kvadrati), $\sin x^2$ esa $x^2$ burchakning sinusidir.

4. Tangens va kotangens aniqlanish sohasini unutish:

   • ✗ Noto'g'ri: $\operatorname{tg} \alpha \cdot \operatorname{ctg} \alpha = 1$ barcha sonlar uchun to'g'ri deyish.
   • ✓ To'g'ri: $\alpha = \dfrac{\pi}{2}k$ ($k \in \mathbb{Z}$) nuqtalarda tangens yoki kotangens mavjud emas. Tenglik faqat ikkala funksiya ham aniqlanganda o'rinli.

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