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2.1-bo'lim: Trig funktsiyasini baholash

Ko'pgina trig sinflari bilan bog'liq muammolardan biri shundaki, ular to'g'ri burchakli trigga e'tibor qaratishadi va hamma narsani darajalar bo'yicha bajaradilar. Keyin siz Calculus kursiga o'tasiz, u erda deyarli hamma narsa radyanlarda amalga oshiriladi va birlik doirasi juda foydali vositadir.

Shunday qilib, avval darajalar va radianlarni bog'lash uchun quyidagi jadvalni ko'rib chiqamiz.

Daraja	0	30	45	60	90	180	270	360
Radianlar	0	$\displaystyle \frac{\pi }{6}$	$\displaystyle \frac{\pi }{4}$	$\displaystyle \frac{\pi }{3}$	$\displaystyle \frac{\pi }{2}$	$\displaystyle \pi $	$\displaystyle \frac{{3\pi }}{2}$	$\displaystyle 2\pi $

Ushbu jadvalni biling! Albatta, biz ushbu dars davomida ko'radigan radianlarda boshqa ko'plab burchaklar mavjud, ammo ularning aksariyati bu bir nechta burchaklarga tegishli bo'ladi. Shunday qilib, agar siz ushbu burchaklar bilan shug'ullana olsangiz, boshqalarning ko'pchiligi bilan shug'ullanishingiz mumkin.

Ogoh bo'ling, ko'pchilik hisob darslarida hamma narsa radyanlarda amalga oshiriladi!

Keling, birlik doirasini ko'rib chiqaylik. Quyida faqat birinchi kvadrant to'ldirilgan birlik doirasi mavjud. Birlik doiraning ishlash usuli aylananing markazidan tashqariga berilgan burchakka mos keladigan chiziq chizishdir. Keyin chiziq va aylana kesishgan nuqtaning koordinatalariga qarang. Birinchi koordinata bu burchakning kosinusu, ikkinchi koordinata esa bu burchakning sinusidir. Bir-ikkitasi bor asosiy tez-tez ishlatiladigan burchaklar. Bu $0,\frac{\pi }{6},\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{2},\pi ,\frac{{3\pi }}{2}$ va $2\pi $ va quyida kesishmalar koordinatalari bilan birga ko'rsatilgan. Shunday qilib, pastdagi birlik doirasidan biz $\cos \left( {\frac{\pi }{6}} \right) = \frac{{\sqrt 3 }}{2}$ va $\sin \left( {\frac{\pi }{6}} \right) = \frac{1}{2}$ ni ko'rishimiz mumkin.

$A unit circle with only the first quadrant filled in. Lines are drawn from the origin at the angles $\frac{\pi }{6}$, $\frac{\pi }{4}$ and $\frac{\pi }{3}$ and the point where each meets the circle is labeled with its coordinates, $\left( {\frac{{\sqrt 3 }}{2},\frac{1}{2}} \right)$, $\left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$ and $\left( {\frac{1}{2},\frac{{\sqrt 3 }}{2}} \right)$ respectively. The four quadrantal angles are also labeled with the points $\left( {1,0} \right)$, $\left( {0,1} \right)$, $\left( { - 1,0} \right)$ and $\left( {0, - 1} \right)$.$

Burchak belgilari qanday ishlashini eslang. Agar siz soat yo'nalishi bo'yicha teskari yo'nalishda aylansangiz, burchak musbat bo'ladi va agar siz soat yo'nalishi bo'yicha aylantirsangiz, bu burchak manfiy bo'ladi.

Shuni ham yodda tutingki, bitta to'liq inqilob $2\pi $, shuning uchun musbat $x$ o'qi 0 yoki $2\pi $ (yoki $4\pi $ yoki $6\pi $, $ - 2\pi $ yoki $ - 4\pi $) burchakka mos kelishi mumkin. va hokazo. aylanish yo'nalishiga qarab). Xuddi shunday, $\frac{\pi }{6}$ burchagi (to'liq tasodifiy burchakni tanlash uchun) quyidagi burchaklardan har qanday bo'lishi mumkin:

$\displaystyle \frac{\pi }{6} + 2\pi = \frac{{13\pi }}{6}$ ($\displaystyle \frac{\pi }{6}$ dan boshlang, keyin soat miliga teskari tomonga bir marta aylantiring)

$\displaystyle \frac{\pi }{6} + 4\pi = \frac{{25\pi }}{6}$ ($\displaystyle \frac{\pi }{6}$ da boshlang, keyin soat miliga teskari ikki marta aylantiring)

$\displaystyle \frac{\pi }{6} - 2\pi = - \frac{{11\pi }}{6}$ ($\displaystyle \frac{\pi }{6}$ da boshlang, keyin soat yo'nalishi bo'yicha bir marta aylantiring)

$\displaystyle \frac{\pi }{6} - 4\pi = - \frac{{23\pi }}{6}$ ($\displaystyle \frac{\pi }{6}$ da boshlang, keyin soat yo'nalishi bo'yicha ikki marta aylantiring)

va hokazo.

Aslida, $\frac{\pi }{6}$ quyidagi burchaklardan istalgani bo'lishi mumkin $\frac{\pi }{6} + 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots $ Bu holda $n$ $\frac{\pi }{6}$ dan boshlab birlik doirasi bo'ylab qilgan to'liq aylanishlar soni. $n$ ning musbat qiymatlari soat miliga teskari aylanishlarga, $n$ ning manfiy qiymatlari esa soat miliga teskari aylanishlarga mos keladi.

Xo'sh, nega men faqat birinchi kvadrantga qo'ydim? Javob oddiy. Agar siz birinchi kvadrantni bilsangiz, birinchidan qolgan barcha kvadrantlarni olishingiz mumkin. Buni quyidagi misollarda ko'rasiz.

Quyidagilarning har birining aniq qiymatini toping. Boshqacha qilib aytganda, kalkulyatordan foydalanmang.

$\sin \left( {\frac{{2\pi }}{3}} \right)$ va $\sin \left( { - \frac{{2\pi }}{3}} \right)$

Yechim

Bu erda birinchi baholash $\frac{{2\pi }}{3}$ burchagidan foydalanadi. E'tibor bering, $\frac{{2\pi }}{3} = \pi - \frac{\pi }{3}$. Shunday qilib, $\frac{{2\pi }}{3}$ $\frac{\pi }{3}$ ni manfiy $x$ o'qidan yuqoriga aylantirish orqali topiladi. Bu shuni anglatadiki, $\frac{{2\pi }}{3}$ uchun chiziq faqat ikkinchi kvadrantda $\frac{\pi }{3}$ uchun chiziqning oyna tasviri bo'ladi. $\frac{{2\pi }}{3}$ koordinatalari $\frac{\pi }{3}$ uchun koordinatalar bo'ladi, bundan tashqari $x$ koordinatasi manfiy bo'ladi.

Xuddi shunday, $ - \frac{{2\pi }}{3}$ uchun biz $ - \frac{{2\pi }}{3} = - \pi + \frac{\pi }{3}$ ni payqashimiz mumkin, shuning uchun bu burchakni $\frac{\pi }{3}$ ni salbiy $x$ o'qidan pastga aylantirish orqali topish mumkin. Bu shuni anglatadiki, $ - \frac{{2\pi }}{3}$ uchun chiziq faqat uchinchi kvadrantda $\frac{\pi }{3}$ uchun chiziqning oyna tasviri bo'ladi va koordinatalar $\frac{\pi }{3}$ uchun koordinatalar bilan bir xil bo'ladi, bundan tashqari ikkalasi ham salbiy bo'ladi.

Bu ikkala burchak va ularning koordinatalari quyidagi birlik doirada ko'rsatilgan.

$A unit circle with the first quadrant angles $\frac{\pi }{6}$, $\frac{\pi }{4}$ and $\frac{\pi }{3}$ drawn in and labeled with their coordinates. Also drawn in is the angle $\frac{{2\pi }}{3}$ in the second quadrant, which meets the circle at $\left( { - \frac{1}{2},\frac{{\sqrt 3 }}{2}} \right)$, and the angle $ - \frac{{2\pi }}{3}$ in the third quadrant, which meets the circle at $\left( { - \frac{1}{2}, - \frac{{\sqrt 3 }}{2}} \right)$. The two are mirror images of each other across the x-axis.$

Ushbu birlik doirasidan biz $\sin \left( {\frac{{2\pi }}{3}} \right) = \frac{{\sqrt 3 }}{2}$ va $\sin \left( { - \frac{{2\pi }}{3}} \right) = - \frac{{\sqrt 3 }}{2}$ ekanligini ko'rishimiz mumkin.

Bu sinus funktsiyasi haqida yoqimli haqiqatga olib keladi. Sinus funktsiyasi an deb ataladi g'alati funktsiya va shunga o'xshash har qanday burchak uchun bizda mavjud
\[\sin \left( { - \theta } \right) = - \sin \left( \theta \right)\]
$\cos \left( {\frac{{7\pi }}{6}} \right)$ va $\cos \left( { - \frac{{7\pi }}{6}} \right)$

Yechim

Ushbu misol uchun $\frac{{7\pi }}{6} = \pi + \frac{\pi }{6}$ ni e'tiborga oling, shuning uchun biz $\frac{\pi }{6}$ ni salbiy $x$ o'qidan pastga aylantirib, bu burchakka o'tamiz. Shuningdek, $ - \frac{{7\pi }}{6} = - \pi - \frac{\pi }{6}$ demak, bu burchakka o'tish uchun $\frac{\pi }{6}$ ni salbiy $x$ o'qidan yuqoriga aylantiramiz. Ularning ikkalasi ham quyidagi birlik doiralarida va kesishish nuqtalari uchun tegishli koordinatalar bilan ko'rsatilgan.

$A unit circle with the first quadrant angles $\frac{\pi }{6}$, $\frac{\pi }{4}$ and $\frac{\pi }{3}$ drawn in and labeled with their coordinates. Also drawn in is the angle $\frac{{7\pi }}{6}$ in the third quadrant, which meets the circle at $\left( { - \frac{{\sqrt 3 }}{2}, - \frac{1}{2}} \right)$, and the angle $ - \frac{{7\pi }}{6}$ in the second quadrant, which meets the circle at $\left( { - \frac{{\sqrt 3 }}{2},\frac{1}{2}} \right)$. The two are mirror images of each other across the x-axis and both have the same x-coordinate.$

Ushbu birlik doirasidan biz $\cos \left( {\frac{{7\pi }}{6}} \right) = - \frac{{\sqrt 3 }}{2}$ va $\cos \left( { - \frac{{7\pi }}{6}} \right) = - \frac{{\sqrt 3 }}{2}$ ekanligini ko'rishimiz mumkin. Bu holda kosinus funksiyasi deyiladi hatto funktsiya va shunga o'xshash har qanday burchak uchun bizda mavjud
\[\cos \left( { - \theta } \right) = \cos \left( \theta \right)\]
$\tan \left( { - \frac{\pi }{4}} \right)$ va $\tan \left( {\frac{{7\pi }}{4}} \right)$

Yechim

Bu erda shuni ta'kidlashimiz kerakki, $\frac{{7\pi }}{4} = 2\pi - \frac{\pi }{4}$ shuning uchun $\frac{{7\pi }}{4}$ va $ - \frac{\pi }{4}$ aslida bir xil burchakdir! Bu burchak uchun birlik doirasi

$A unit circle with the first quadrant angles $\frac{\pi }{6}$, $\frac{\pi }{4}$ and $\frac{\pi }{3}$ drawn in and labeled with their coordinates. Also drawn in is a single line into the fourth quadrant labeled with both $ - \frac{\pi }{4}$ and $\frac{{7\pi }}{4}$, showing that these are the same angle, and the point where it meets the circle is labeled $\left( {\frac{{\sqrt 2 }}{2}, - \frac{{\sqrt 2 }}{2}} \right)$.$

Endi, agar biz $\tan \left( x \right) = \frac{{\sin \left( x \right)}}{{\cos \left( x \right)}}$ ni eslasak, tangens funksiyasining qiymatlarini topish uchun birlik doirasidan foydalanishimiz mumkin. Shunday qilib,
\[\tan \left( \frac{7\pi }{4} \right)=\tan \left( -\frac{\pi }{4} \right)=\frac{\sin \left( -{\pi }/{4}\; \right)}{\cos \left( -{\pi }/{4}\; \right)}=\frac{-{\sqrt{2}}/{2}\;}{{\sqrt{2}}/{2}\;}=-1\]
Yon eslatmaga e'tibor bering, $\tan \left( {\frac{\pi }{4}} \right) = 1$ va biz tangens funksiyasi ham deyiladi. g'alati funktsiya va shunga o'xshash har qanday burchak uchun bizda bo'ladi
\[\tan \left( { - \theta } \right) = - \tan \left( \theta \right)\]
$\sin \left( {\frac{{9\pi }}{4}} \right)$

Yechim

Ushbu muammo uchun $\frac{{9\pi }}{4} = 2\pi + \frac{\pi }{4}$ ga e'tibor qaratamiz. Endi eslang, bitta to'liq inqilob $2\pi $. Demak, bu $\frac{{9\pi }}{4}$ va $\frac{\pi }{4}$ birlik aylanasining bir nuqtasida ekanligini anglatadi. Shuning uchun,
\[\sin \left( {\frac{{9\pi }}{4}} \right) = \sin \left( {2\pi + \frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\]
Bu bizni sinus funktsiyasi haqida juda yaxshi haqiqatga olib keladi. Sinus funksiyasi a ga misoldir davriy funktsiyasi. Davriy funktsiya o'zini qayta-qayta takrorlaydigan funktsiyalardir. Funktsiya o'zini takrorlashni boshlashdan oldin o'ngga yoki chapga siljishingiz kerak bo'lgan "masofa" deyiladi davr funksiyasidan.

Sinus holatida davr $2\pi $. Bu sinus funksiyasi har $2\pi $ takrorlanishini bildiradi. Bu sinus funktsiyasi uchun yaxshi formulaga olib keladi.
\[\sin \left( {x + 2\pi n} \right) = \sin \left( x \right)\hspace{0.25in}\,\,\,\,\,\,\,\,n = 0,\, \pm 1,\,\, \pm 2,\, \ldots \]
Bunga ham e'tibor bering, chunki
\[csc \left( x \right) = \frac{1}{{\sin \left( x \right)}}\]
Kosekant haqida ham xuddi shunday deyishimiz mumkin.
\[\csc \left( {x + 2\pi n} \right) = \csc \left( x \right)\hspace{0.25in}\,\,\,\,\,\,\,\,n = 0,\, \pm 1,\,\, \pm 2,\, \ldots \]
Xo'sh, aslida biz bu erda ehtiyot bo'lishimiz kerak. Buni $x \ne n\pi $ bilan ta'minlangan deb aytishimiz mumkin, chunki bu nuqtalarda sinus nolga teng bo'ladi va u erda kosekant mavjud bo'lmaydi!
$\sec \left( {\frac{{25\pi }}{6}} \right)$

Yechim

Bu erda biz $\frac{{25\pi }}{6} = 4\pi + \frac{\pi }{6}$ ekanligini ta'kidlashimiz kerak. Boshqacha qilib aytadigan bo'lsak, biz $\frac{\pi }{6}$ dan boshladik va birlik aylanasining bir nuqtasida zaxira nusxasini tugatish uchun ikki marta aylantirdik. Bu shuni anglatadiki
\[\sec \left( {\frac{{25\pi }}{6}} \right) = \sec \left( {4\pi + \frac{\pi }{6}} \right) = \sec \left( {\frac{\pi }{6}} \right)\]
Keling, bu erda sekant haqida ham hayajonlanmaylik. Shunchaki eslang
\[\sec \left( x \right) = \frac{1}{{\cos \left( x \right)}}\]
va shuning uchun bu erda qilishimiz kerak bo'lgan yagona narsa kosinusni baholashdir! Shuning uchun,
\[\sec \left( \frac{25\pi }{6} \right)=\sec \left( \frac{\pi }{6} \right)=\frac{1}{\cos \left( \frac{\pi }{6} \right)}=\frac{1}{{}^{\sqrt{3}}/{}_{2}}=\frac{2}{\sqrt{3}}\]
Shuni ham ta'kidlash kerakki, kosinus va sekant $2\pi $ davriga ega davriy funksiyalardir. Shunday qilib,
\[\begin{align*}\begin{aligned}\cos \left( {x + 2\pi n} \right) & = \cos \left( x \right)\\ \sec \left( {x + 2\pi n} \right) & = \sec \left( x \right) \end{aligned}& \hspace{0.5in}n = 0, \pm 1, \pm 2,\, \ldots \end{align*}\]
$\tan \left( {\frac{{4\pi }}{3}} \right)$

Yechim

Bu muammoni hal qilish uchun tangens (va shuning uchun kotangent) ham davriy funktsiya ekanligini bilish yordam beradi, lekin sinus va kosinusdan farqli o'laroq, u $\pi $ davriga ega.
\[\begin{align*}\begin{aligned}\tan \left( {x + \pi n} \right) & = \tan \left( x \right)\\ \cot \left( {x + \pi n} \right) & = \cot \left( x \right) \end{aligned}& \hspace{0.5in}n = 0, \pm 1, \pm 2,\, \ldots \end{align*}\]
Shunday qilib, ushbu muammoni hal qilish uchun $\frac{{4\pi }}{3} = \pi + \frac{\pi }{3}$ ekanligini ta'kidlaymiz. Shuning uchun,
\[\tan \left( {\frac{{4\pi }}{3}} \right) = \tan \left( {\pi + \frac{\pi }{3}} \right) = \tan \left( {\frac{\pi }{3}} \right) = \sqrt 3 \]
Trig baholash yakuniy fikrlar

Oldingi misollarda ko'rganimizdek, agar siz birlik doirasining birinchi kvadrantini bilsangiz, birinchi kvadrantda ko'rsatilganlardan biriga bog'liq bo'lishi mumkin bo'lgan HAR QANDAY burchak uchun HAR QANDAY trig funksiyasining qiymatini (nafaqat sinus va kosinus) topishingiz mumkin. Buni eslab qolish yaxshi fikr, chunki bu faqat birinchi kvadrantni yodlash va qolgan uchta kvadrantdagi burchaklarni qanday olish kerakligini anglatadi!

Bu masalalarda men faqat “asosiy” burchaklardan foydalanardim, lekin bu yerdagi ko‘p fikrlarni bu “asosiy” burchaklardan boshqa burchaklarga ham qo‘llash mumkin, chunki biz ko‘rib chiqamiz. Trig tenglamalarni yechish.

Daraja	0	30	45	60	90	180	270	360
Radianlar	0	\(\displaystyle \frac{\pi }{6}\)	\(\displaystyle \frac{\pi }{4}\)	\(\displaystyle \frac{\pi }{3}\)	\(\displaystyle \frac{\pi }{2}\)	\(\displaystyle \pi \)	\(\displaystyle \frac{{3\pi }}{2}\)	\(\displaystyle 2\pi \)

2.1-bo'lim: Trig funktsiyasini baholash

Trig baholash yakuniy fikrlar