Arksinus, arkkosinus, arktangens: matematika darsi

Sinusi $\dfrac{1}{2}$ ga teng burchaklar cheksiz ko'p: $30^\circ$ , $150^\circ$ , $390^\circ$ va hokazo. Teskari funksiya shulardan bittasini, belgilangan oraliqdagisini tanlaydi: $\arcsin\dfrac{1}{2}=\dfrac{\pi}{6}$ . Oraliqni bilmaslik bu mavzudagi eng asosiy xato manbai, masalan, $\arccos\left(-\dfrac{1}{2}\right)$ javobi $-\dfrac{\pi}{3}$ emas, $\dfrac{2\pi}{3}$ .

Ta'riflar

Funksiya	Ma'nosi	$a$ qiymatlari	Burchak oralig'i
Funksiya $\arcsin a$	Ma'nosisinusi $a$ ga teng burchak	$a$ qiymatlari $-1\le a\le 1$	Burchak oralig'i $\left[-\dfrac{\pi}{2};\,\dfrac{\pi}{2}\right]$
Funksiya $\arccos a$	Ma'nosikosinusi $a$ ga teng burchak	$a$ qiymatlari $-1\le a\le 1$	Burchak oralig'i $[0;\,\pi]$
Funksiya $\operatorname{arctg}a$	Ma'nositangensi $a$ ga teng burchak	$a$ qiymatlariistalgan son	Burchak oralig'i $\left(-\dfrac{\pi}{2};\,\dfrac{\pi}{2}\right)$
Funksiya $\operatorname{arcctg}a$	Ma'nosikotangensi $a$ ga teng burchak	$a$ qiymatlariistalgan son	Burchak oralig'i $(0;\,\pi)$

Bu oraliqlarda funksiya har bir qiymatni bir marta qabul qiladi, shuning uchun burchak yagona. Arktangens oralig'ining uchlari kirmaydi, chunki $\operatorname{tg}\left(\pm\dfrac{\pi}{2}\right)$ mavjud emas. Arkkotangens oralig'ining uchlari kirmaydi, chunki $\operatorname{ctg}0$ va $\operatorname{ctg}\pi$ mavjud emas.

$\arcsin a=\varphi$ degani: $\sin\varphi=a$ va $-\dfrac{\pi}{2}\le\varphi\le\dfrac{\pi}{2}$ . $\arccos a=\varphi$ degani: $\cos\varphi=a$ va $0\le\varphi\le\pi$ . $\arcsin 2$ va $\arccos(-1{,}5)$ mavjud emas.

Yaxlit chiziq

y=\arcsin x

, u o'sadi. Punktir chiziq

y=\arccos x

, u kamayadi. Ikkalasi faqat

[-1;\,1]

da mavjud

Jadval qiymatlari

$a$	$\arcsin a$	$\arccos a$
$a$ $-1$	$\arcsin a$ $-\dfrac{\pi}{2}$	$\arccos a$ $\pi$
$a$ $-\dfrac{\sqrt{3}}{2}$	$\arcsin a$ $-\dfrac{\pi}{3}$	$\arccos a$ $\dfrac{5\pi}{6}$
$a$ $-\dfrac{\sqrt{2}}{2}$	$\arcsin a$ $-\dfrac{\pi}{4}$	$\arccos a$ $\dfrac{3\pi}{4}$
$a$ $-\dfrac{1}{2}$	$\arcsin a$ $-\dfrac{\pi}{6}$	$\arccos a$ $\dfrac{2\pi}{3}$
$a$ $0$	$\arcsin a$ $0$	$\arccos a$ $\dfrac{\pi}{2}$
$a$ $\dfrac{1}{2}$	$\arcsin a$ $\dfrac{\pi}{6}$	$\arccos a$ $\dfrac{\pi}{3}$
$a$ $\dfrac{\sqrt{2}}{2}$	$\arcsin a$ $\dfrac{\pi}{4}$	$\arccos a$ $\dfrac{\pi}{4}$
$a$ $\dfrac{\sqrt{3}}{2}$	$\arcsin a$ $\dfrac{\pi}{3}$	$\arccos a$ $\dfrac{\pi}{6}$
$a$ $1$	$\arcsin a$ $\dfrac{\pi}{2}$	$\arccos a$ $0$

$a$	$\operatorname{arctg}a$	$\operatorname{arcctg}a$
$a$ $-\sqrt{3}$	$\operatorname{arctg}a$ $-\dfrac{\pi}{3}$	$\operatorname{arcctg}a$ $\dfrac{5\pi}{6}$
$a$ $-1$	$\operatorname{arctg}a$ $-\dfrac{\pi}{4}$	$\operatorname{arcctg}a$ $\dfrac{3\pi}{4}$
$a$ $-\dfrac{\sqrt{3}}{3}$	$\operatorname{arctg}a$ $-\dfrac{\pi}{6}$	$\operatorname{arcctg}a$ $\dfrac{2\pi}{3}$
$a$ $0$	$\operatorname{arctg}a$ $0$	$\operatorname{arcctg}a$ $\dfrac{\pi}{2}$
$a$ $\dfrac{\sqrt{3}}{3}$	$\operatorname{arctg}a$ $\dfrac{\pi}{6}$	$\operatorname{arcctg}a$ $\dfrac{\pi}{3}$
$a$ $1$	$\operatorname{arctg}a$ $\dfrac{\pi}{4}$	$\operatorname{arcctg}a$ $\dfrac{\pi}{4}$
$a$ $\sqrt{3}$	$\operatorname{arctg}a$ $\dfrac{\pi}{3}$	$\operatorname{arcctg}a$ $\dfrac{\pi}{6}$

Asosiy xossalar

\arcsin(-a)=-\arcsin a

\arccos(-a)=\pi-\arccos a

|a|\le 1

\operatorname{arctg}(-a)=-\operatorname{arctg}a

\operatorname{arcctg}(-a)=\pi-\operatorname{arcctg}a

Istalgan

a

\arcsin a+\arccos a=\frac{\pi}{2}

\operatorname{arctg}a+\operatorname{arcctg}a=\frac{\pi}{2}

Birinchisi

|a|\le 1

da, ikkinchisi istalgan

a

Isbot: $\varphi=\dfrac{\pi}{2}-\arcsin a$ burchak $[0;\,\pi]$ da yotadi va $\cos\varphi=\sin(\arcsin a)=a$ , demak $\varphi=\arccos a$ .

O'sish va kamayish: $\arcsin x$ va $\operatorname{arctg}x$ o'sadi, $\arccos x$ va $\operatorname{arcctg}x$ kamayadi. Bu tengsizliklarda hal qiluvchi: kamayuvchi funksiyadan qutulganda tengsizlik belgisi almashadi.

Funksiya va teskari funksiya

\sin(\arcsin a)=a

\cos(\arccos a)=a

Faqat

|a|\le 1

\operatorname{tg}(\operatorname{arctg}a)=a

\operatorname{ctg}(\operatorname{arcctg}a)=a

Istalgan

a

Teskari tartibda tenglik faqat asosiy oraliqda o'rinli:

Tenglik	Faqat shu $x$ da
$\arcsin(\sin x)=x$	$x\in\left[-\dfrac{\pi}{2};\,\dfrac{\pi}{2}\right]$
$\arccos(\cos x)=x$	$x\in[0;\,\pi]$
$\operatorname{arctg}(\operatorname{tg}x)=x$	$x\in\left(-\dfrac{\pi}{2};\,\dfrac{\pi}{2}\right)$
$\operatorname{arcctg}(\operatorname{ctg}x)=x$	$x\in(0;\,\pi)$

Oraliqdan tashqarida avval keltirish formulasi bilan burchakni asosiy oraliqqa o'tkazamiz:

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