Trigonometriya (DTM)

Bu mavzuda DTM nimani so'raydi

DTM imtihonlarida trigonometriyadan har yili 2–4 ta savol tushadi:

1. Ayniyatlar va soddalashtirish: asosiy ayniyatlar, darajani pasaytirish, ikkilangan burchak hamda yig'indini ko'paytmaga keltirish.
2. Qiymatlar sohasi va ekstremumlar: $a\sin x + b\cos x$ ko'rinishidagi ifodalar, kvadratik shakllar.
3. Tenglamalar: berilgan oraliqdagi ildizlar soni yoki ildizlar yig'indisi.
4. Teskari trigonometrik funksiyalar: $\arcsin$, $\arccos$, $\operatorname{arctg}$ qiymatlari va ta'rif sohalari.

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Asosiy formulalar va usullar

1. Muhim ayniyatlar

• $\sin^2\alpha + \cos^2\alpha = 1$, $\operatorname{tg}\alpha \cdot \operatorname{ctg}\alpha = 1$
• $\sin 2\alpha = 2\sin\alpha\cos\alpha$, $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha - 1 = 1 - 2\sin^2\alpha$
• Darajani pasaytirish: $\sin^2\alpha = \dfrac{1 - \cos 2\alpha}{2}$, $\cos^2\alpha = \dfrac{1 + \cos 2\alpha}{2}$

2. Yordamchi burchak usuli

$a\sin x + b\cos x = \sqrt{a^2 + b^2} \sin(x + \varphi)$ Qiymatlar sohasi: $[-\sqrt{a^2 + b^2}; \sqrt{a^2 + b^2}]$.

3. Teskari trigonometrik funksiyalar

• $\arcsin x \in \left[-\dfrac{\pi}{2}; \dfrac{\pi}{2}\right]$, $\arccos x \in [0; \pi]$
• $\arcsin(-x) = -\arcsin x$, ammo $\arccos(-x) = \pi - \arccos x$

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Yechilgan misollar

1-misol

Agar $\operatorname{tg}\alpha + \operatorname{ctg}\alpha = 3$ bo'lsa, $\operatorname{tg}^3\alpha + \operatorname{ctg}^3\alpha$ ning qiymatini toping.

A) $18$
B) $21$
C) $27$
D) $15$

Yechimi:
Kublar yig'indisi formulasidan: $\operatorname{tg}^3\alpha + \operatorname{ctg}^3\alpha = (\operatorname{tg}\alpha + \operatorname{ctg}\alpha)\left((\operatorname{tg}\alpha + \operatorname{ctg}\alpha)^2 - 3\operatorname{tg}\alpha\operatorname{ctg}\alpha\right)$ $\operatorname{tg}\alpha\operatorname{ctg}\alpha = 1$ va $\operatorname{tg}\alpha + \operatorname{ctg}\alpha = 3$: $3 \cdot (3^2 - 3 \cdot 1) = 3 \cdot (9 - 3) = 18$

To'g'ri javob: A.

Tezroq yo'l: $u + \dfrac{1}{u} = p$ bo'lsa, $u^3 + \dfrac{1}{u^3} = p^3 - 3p = 27 - 9 = 18$.

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2-misol

$f(x) = 5\sin x - 12\cos x + 4$ funksiyaning eng katta qiymatini toping.

A) $13$
B) $17$
C) $21$
D) $19$

Yechimi:
$5\sin x - 12\cos x$ ifodaning eng katta qiymati $\sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = 13$.
Funksiyaning eng katta qiymati: $13 + 4 = 17$.

To'g'ri javob: B.

Tezroq yo'l: $5$ va $12$ sonlari Pifagor uchligi ($5, 12, 13$). Eng katta qiymat darhol $13 + 4 = 17$.

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3-misol

$\cos 20^\circ \cdot \cos 40^\circ \cdot \cos 80^\circ$ ko'paytmaning qiymatini hisoblang.

A) $\dfrac{1}{4}$
B) $\dfrac{1}{8}$
C) $\dfrac{1}{16}$
D) $\dfrac{\sqrt{3}}{8}$

Yechimi:
Ifodani $2\sin 20^\circ$ ga ko'paytirib bo'lamiz: $\frac{2\sin 20^\circ \cos 20^\circ \cos 40^\circ \cos 80^\circ}{2\sin 20^\circ} = \frac{\sin 40^\circ \cos 40^\circ \cos 80^\circ}{2\sin 20^\circ} = \frac{\sin 80^\circ \cos 80^\circ}{4\sin 20^\circ} = \frac{\sin 160^\circ}{8\sin 20^\circ}$ $\sin 160^\circ = \sin(180^\circ - 20^\circ) = \sin 20^\circ$ bo'lgani uchun kasr $\dfrac{1}{8}$ ga teng.

To'g'ri javob: B.

Tezroq yo'l: $\cos\alpha \cos(60^\circ - \alpha)\cos(60^\circ + \alpha) = \dfrac{1}{4}\cos 3\alpha$ formulasida $\alpha = 20^\circ$ qo'yilsa: $\dfrac{1}{4}\cos 60^\circ = \dfrac{1}{8}$.

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4-misol

$\sin 2x = \cos x$ tenglamaning $[0; \pi]$ kesmadagi ildizlari yig'indisini toping.

A) $\pi$
B) $\dfrac{3\pi}{2}$
C) $2\pi$
D) $\dfrac{5\pi}{6}$

Yechimi:
$2\sin x \cos x - \cos x = 0 \iff \cos x(2\sin x - 1) = 0$

1. $\cos x = 0 \implies x_1 = \dfrac{\pi}{2} \in [0; \pi]$.
2. $\sin x = \dfrac{1}{2} \implies x_2 = \dfrac{\pi}{6}, x_3 = \dfrac{5\pi}{6} \in [0; \pi]$.
   Yig'indi: $\dfrac{\pi}{2} + \dfrac{\pi}{6} + \dfrac{5\pi}{6} = \dfrac{3\pi}{2}$.

To'g'ri javob: B.

Tezroq yo'l: $\sin x = \dfrac{1}{2}$ ildizlari $\dfrac{\pi}{2}$ ga nisbatan simmetrik: ularning yig'indisi $\pi$. Uchinchi ildiz $\dfrac{\pi}{2}$. Jami: $\pi + \dfrac{\pi}{2} = \dfrac{3\pi}{2}$.

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5-misol

$\arccos\left(\cos\dfrac{7\pi}{5}\right)$ ifodaning qiymatini hisoblang.

A) $\dfrac{7\pi}{5}$
B) $\dfrac{2\pi}{5}$
C) $\dfrac{3\pi}{5}$
D) $\dfrac{4\pi}{5}$

Yechimi:
$\arccos$ qiymatlar sohasi $[0; \pi]$. $\dfrac{7\pi}{5} > \pi$ bo'lgani uchun to'g'ridan-to'g'ri tenglik bajarilmaydi.
Kosinusning davriyligidan: $\cos\dfrac{7\pi}{5} = \cos\left(2\pi - \dfrac{7\pi}{5}\right) = \cos\dfrac{3\pi}{5}$.
$\dfrac{3\pi}{5} \in [0; \pi]$ bo'lgani sababli: $\arccos\left(\cos\dfrac{3\pi}{5}\right) = \dfrac{3\pi}{5}$.

To'g'ri javob: C.

Tezroq yo'l: Burchak $252^\circ$ (3-chorak). 2-chorakdagi unga teng kosinusli burchak: $360^\circ - 252^\circ = 108^\circ = \dfrac{3\pi}{5}$.

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Mashq

1. Kasrni soddalashtiring: $\dfrac{\sin 2\alpha}{1 + \cos 2\alpha}$.

Yechimi va javobi

$\frac{\sin 2\alpha}{1 + \cos 2\alpha} = \frac{2\sin\alpha\cos\alpha}{2\cos^2\alpha} = \frac{\sin\alpha}{\cos\alpha} = \operatorname{tg}\alpha$

Javob: $\operatorname{tg}\alpha$

2. Agar $\sin\alpha - \cos\alpha = 0{,}6$ bo'lsa, $\sin 2\alpha$ ning qiymatini toping.

Yechimi va javobi

Ikkala tomonni kvadratga ko'taramiz: $(\sin\alpha - \cos\alpha)^2 = 1 - \sin 2\alpha = 0{,}36 \implies \sin 2\alpha = 1 - 0{,}36 = 0{,}64$.

Javob: $0{,}64$

3. $y = 3\cos^2 x - 4\sin^2 x$ funksiyaning eng kichik qiymatini toping.

Yechimi va javobi

$\cos^2 x = 1 - \sin^2 x$ orqali: $y = 3(1 - \sin^2 x) - 4\sin^2 x = 3 - 7\sin^2 x$. $\sin^2 x = 1$ da eng kichik qiymat: $3 - 7(1) = -4$.

Javob: $-4$

4. $\operatorname{tg} x = \sqrt{3}$ tenglamaning $(-90^\circ; 270^\circ)$ oraliqdagi ildizlari sonini toping.

Yechimi va javobi

$x = 60^\circ + 180^\circ k$. Berilgan oraliqda: $k=0 \implies 60^\circ$, $k=1 \implies 240^\circ$. Jami $2$ ta ildiz.

Javob: $2$ ta

5. Hisoblang: $\sin 15^\circ \cdot \cos 15^\circ$.

Yechimi va javobi

$\sin 15^\circ \cos 15^\circ = \frac{\sin 30^\circ}{2} = \frac{1/2}{2} = \frac{1}{4} = 0{,}25$

Javob: $\dfrac{1}{4}$

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

Yechimi va javobi

Surat va maxrajni $\cos\alpha$ ga bo'lamiz:
$\frac{3\operatorname{tg}\alpha - 1}{\operatorname{tg}\alpha + 2} = \frac{3(2) - 1}{2 + 2} = \frac{5}{4} = 1{,}25$

Javob: $\dfrac{5}{4}$ (yoki $1{,}25$)

7. Hisoblang: $\operatorname{arctg}(1) + \arcsin\left(-\dfrac{1}{2}\right)$.

Yechimi va javobi

$\operatorname{arctg}(1) = \dfrac{\pi}{4}$ va $\arcsin\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}$.
Yig'indi: $\dfrac{\pi}{4} - \dfrac{\pi}{6} = \dfrac{\pi}{12}$.

Javob: $\dfrac{\pi}{12}$

8. $2\sin^2 x - 3\sin x + 1 = 0$ tenglamaning $[0; 2\pi]$ kesmadagi eng kichik musbat ildizini toping.

Yechimi va javobi

$(2\sin x - 1)(\sin x - 1) = 0$. Musbat ildizlar: $\sin x = \dfrac{1}{2} \implies x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$; $\sin x = 1 \implies x = \dfrac{\pi}{2}$. Eng kichigi $\dfrac{\pi}{6}$.

Javob: $\dfrac{\pi}{6}$

9. Hisoblang: $\cos 75^\circ + \cos 15^\circ$.

Yechimi va javobi

$2\cos\frac{75^\circ + 15^\circ}{2}\cos\frac{75^\circ - 15^\circ}{2} = 2\cos 45^\circ \cos 30^\circ = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2}$

Javob: $\dfrac{\sqrt{6}}{2}$

10. $f(x) = 6\sin x + 8\cos x - 5$ funksiyaning eng kichik qiymatini toping.

Yechimi va javobi

$\sqrt{6^2 + 8^2} = 10$. Eng kichik qiymat: $-10 - 5 = -15$.

Javob: $-15$

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Keng tarqalgan tuzoqlar

1. Tenglamaning har ikki tomonini $\cos x$ ga bo'lib yuborish:
   $\sin 2x = \cos x$ tenglamada $\cos x$ ga bo'lish mumkin emas, chunki $\cos x = 0$ bo'lgan ildizlar yo'qoladi. Hammasini bir tomonga o'tkazib, ko'paytuvchilarga ajratish shart.
2. Chorak ishorasiga e'tibor bermaslik:
   $\sin\alpha$ orqali $\cos\alpha$ topilayotganda burchak qaysi chorakdaligiga qarang. 2- va 3-chorakda kosinus manfiydir: $\cos\alpha = -\sqrt{1 - \sin^2\alpha}$.
3. $\arccos(-x)$ formulasidagi adashish:
   $\arcsin(-x) = -\arcsin x$, biroq $\arccos(-x) = \pi - \arccos x$. Buni doimo esda tuting.
4. $\arcsin(\sin x) = x$ ayniyatini har doim qo'llash:
   Bu tenglik faqat $x \in \left[-\dfrac{\pi}{2}; \dfrac{\pi}{2}\right]$ oraliqda to'g'ri. Agar $x$ tashqarida bo'lsa, davriylik va keltirish formulalari bilan asosiy oraliqqa keltiriladi.