Funksiyalar

Teskari trigonometrik funksiyalar

3 daqiqa o'qish · 10 mashq

Ushbu darsda trigonometrik funksiyalarga teskari bo'lgan teskari trigonometrik funksiyalar — arksinus, arkkosinus, arktangens haqida so'z yuritamiz. Ular "sinusi shu songa teng bo'lgan burchak qaysi?" degan savolga javob beradi.

§1. Nega teskari funksiya kerak?

sinx=12\sin x = \dfrac{1}{2} tenglamaning cheksiz ko'p yechimi bor, chunki sinus davriy. Bittagina, aniq javob olish uchun burchakni ma'lum bir oraliqqa cheklaymiz.

Ta'rif. arcsina\arcsin a — sinusi aa ga teng bo'lgan, [π2; π2]\left[-\dfrac{\pi}{2};\ \dfrac{\pi}{2}\right] oraliqdagi yagona burchak. Bu yerda 1a1-1 \le a \le 1.

Ta'rif. arccosa\arccos a — kosinusi aa ga teng bo'lgan, [0; π][0;\ \pi] oraliqdagi yagona burchak; arctga\operatorname{arctg} a — tangensi aa ga teng bo'lgan, (π2; π2)\left(-\dfrac{\pi}{2};\ \dfrac{\pi}{2}\right) oraliqdagi yagona burchak.

Savol. Nega arcsin\arcsin uchun aynan [π2;π2]\left[-\dfrac{\pi}{2}; \dfrac{\pi}{2}\right] tanlandi? Chunki shu oraliqda sinus qat'iy o'suvchi — har bir qiymatga bitta burchak mos keladi, ya'ni teskari funksiya bir qiymatli bo'ladi.

§2. Xossalari

π/2 −π/2 −1 1 y = arcsin x x

Qoida. y=arcsinxy = \arcsin x: D=[1;1]D = [-1; 1], E=[π2;π2]E = \left[-\dfrac{\pi}{2}; \dfrac{\pi}{2}\right], toq funksiya, o'suvchi. y=arccosxy = \arccos x: D=[1;1]D = [-1; 1], E=[0;π]E = [0; \pi], kamayuvchi. y=arctgxy = \operatorname{arctg} x: D=RD = \mathbb{R}, E=(π2;π2)E = \left(-\dfrac{\pi}{2}; \dfrac{\pi}{2}\right).

Qoida. Foydali ayniyatlar: arcsin(a)=arcsina\arcsin(-a) = -\arcsin a, arccos(a)=πarccosa\arccos(-a) = \pi - \arccos a, hamda

arcsina+arccosa=π2.\arcsin a + \arccos a = \frac{\pi}{2}.

§3. Misollar

Misol. arcsin12\arcsin\dfrac{1}{2} ni toping.
Yechish. [π2;π2]\left[-\dfrac{\pi}{2}; \dfrac{\pi}{2}\right] oralig'ida sinusi 12\dfrac{1}{2} ga teng burchakni izlaymiz:

sinπ6=12    arcsin12=π6.\sin\frac{\pi}{6} = \frac{1}{2} \implies \arcsin\frac{1}{2} = \frac{\pi}{6}.

Javob. π6\dfrac{\pi}{6}.

Misol. arccos(12)\arccos\left(-\dfrac{1}{2}\right) ni toping.
Yechish. Ayniyatdan foydalanamiz:

arccos(12)=πarccos12=ππ3=2π3.\arccos\left(-\frac{1}{2}\right) = \pi - \arccos\frac{1}{2} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}.

Javob. 2π3\dfrac{2\pi}{3}.

Misol. cos(arcsin35)\cos\left(\arcsin\dfrac{3}{5}\right) ni hisoblang.
Yechish. arcsin35=α\arcsin\dfrac{3}{5} = \alpha desak, sinα=35\sin\alpha = \dfrac{3}{5} va α\alpha birinchi chorakda. U holda:

cosα=1sin2α=1925=1625=45.\cos\alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}.

Javob. 45\dfrac{4}{5}.

Misol. arctg1\operatorname{arctg} 1 ni toping.
Yechish. (π2;π2)\left(-\dfrac{\pi}{2}; \dfrac{\pi}{2}\right) da tangensi 11 ga teng burchak:

tgπ4=1    arctg1=π4.\operatorname{tg}\frac{\pi}{4} = 1 \implies \operatorname{arctg} 1 = \frac{\pi}{4}.

Javob. π4\dfrac{\pi}{4}.

Eslatma. Teskari trigonometrik funksiyaning javobi doim o'z qiymatlar sohasida yotishi kerak: arccos\arccos hech qachon manfiy bo'lmaydi (chunki E=[0;π]E=[0;\pi]), arcsin\arcsin esa π2\dfrac{\pi}{2} dan oshmaydi. Manfiy argumentda arccos(a)=πarccosa\arccos(-a) = \pi - \arccos a formulasini ishlating — arccosa-\arccos a deb yozish xato.

§ Lug'at

teskari trigonometrik funksiya — inverse trigonometric function
arksinus — arcsine
arkkosinus — arccosine
arktangens — arctangent
qiymatlar sohasi — range
bir qiymatli — single-valued
qat'iy monoton — strictly monotonic

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