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2.5-bo'lim: Teskari trig funktsiyalari

Teskari trig funktsiyalari uchun keng tarqalgan belgilardan biri juda chalkash bo'lishi mumkin. Birinchidan, eksponentatsiya bilan qanday ishlashga odatlanganligingizdan qat'i nazar, biz "-1" ko'rsatkichi bilan teskari trig funktsiyasini bildiramiz. Boshqacha qilib aytganda, teskari kosinus \({\cos ^{ - 1}}\left( x \right)\) sifatida belgilanadi. Shuni ta'kidlash kerakki, bu holda "-1" ko'rsatkich EMAS va shuning uchun,

\[{\cos ^{ - 1}}\left( x \right) \ne \frac{1}{{\cos \left( x \right)}}\]

Teskari trig funksiyalarida “-1” ko‘rsatkichga o‘xshaydi, lekin unday emas, bu shunchaki biz teskari trig funksiyasi bilan ishlayotganimizni bildirish uchun ishlatadigan belgi. Bu biz bu holatda teskari trig funksiyalarini belgilash uchun ishlatadigan belgi. Agar men kosinus ustidan 1 ni ko'rsatishni chindan ham xohlasam, men quyidagilarni ishlatgan bo'lardim.

\[{\left( {\cos \left( x \right)} \right)^{ - 1}} = \frac{1}{{\cos \left( x \right)}}\]

Bu noaniqlikni oldini oladigan teskari trig funktsiyalari uchun yana bir belgi mavjud. Bu quyidagilar.

\[\begin{align*}{\cos ^{ - 1}}\left( x \right) & = \arccos \left( x \right)\\ {\sin ^{ - 1}}\left( x \right) & = \arcsin \left( x \right)\\ {\tan ^{ - 1}}\left( x \right) & = \arctan \left( x \right)\end{align*}\]

Shunday qilib, teskari trig funktsiyalari uchun belgilar bilan ehtiyot bo'ling!

Qolgan uchta trig-funksiya uchun, albatta, shunga o'xshash teskari funktsiyalar mavjud, ammo bular hisob-kitob sinfida ko'radigan asosiy uchtasi, shuning uchun men ularga e'tibor qaratmoqchiman.

Teskari trig funksiyalarini baholash uchun quyidagi iboralar ekvivalent ekanligini unutmang.

\[\begin{align*}\theta & = {\cos ^{ - 1}}\left( x \right) & \hspace{0.5in} & \Leftrightarrow & \hspace{0.5in} x & = \cos \left( \theta \right)\\ \theta & = {\sin ^{ - 1}}\left( x \right) & \hspace{0.5in} & \Leftrightarrow & \hspace{0.5in} x & = \sin \left( \theta \right)\\ \theta & = {\tan ^{ - 1}}\left( x \right) & \hspace{0.5in} & \Leftrightarrow & \hspace{0.5in} x & = \tan \left( \theta \right)\end{align*}\]

Boshqacha qilib aytganda, biz teskari trig funksiyasini baholaganimizda, \(x\) ni olish uchun biz qaysi burchakni, \(\theta \), trig funksiyasiga (muntazam, teskari emas!) ulanganligini so‘raymiz.

Shunday qilib, keling, bularning qanday ishlashini ko'rish uchun bir nechta muammolarni hal qilaylik. Quyidagilarning har birini baholang.

  1. \(\displaystyle {\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)\)
    Yechim

    1-muammoda Trig tenglamalarni yechish bo'limida biz quyidagi tenglamani yechdik.

    \[\cos \left( t \right) = \frac{{\sqrt 3 }}{2}\]

    Boshqacha qilib aytganda, biz \(\frac{{\sqrt 3 }}{2}\) ni olish uchun qanday burchaklarni, \(x\), kosinusga ulanishimiz kerakligini so'radik? Bu bizdan teskari trig funktsiyasini hisoblashni so'rashganda, bu erda so'ragan narsadir.

    \[{\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)\]

    Biroq, bitta juda katta farq bor. 1-masalada biz cheksiz ko'p yechimlarni bergan tenglamani yechdik. Bular edi,

    \[\begin{align*} & \frac{\pi }{6} + 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \\ & \frac{{11\pi }}{6} + 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots \end{align*}\]

    Teskari trig funksiyalarida biz bitta qiymatdan keyin turamiz. Biz cheksiz mumkin bo'lgan javoblardan qaysi birini xohlayotganimizni taxmin qilishni xohlamaymiz. Shunday qilib, teskari trig kosinus funktsiyasidan bitta qiymat olishimizga ishonch hosil qilish uchun biz teskari kosinus bo'yicha quyidagi cheklovlardan foydalanamiz.

    \[\theta = {\cos ^{ - 1}}\left( x \right)\hspace{0.25in}\hspace{0.25in} - 1 \le x \le 1\hspace{0.25in}{\rm{and }}\hspace{0.25in}0 \le \theta \le \pi \]

    \(\theta \)-dagi cheklov biz faqat bitta qiymat burchagini olishimizni kafolatlaydi va biz \(x\) ning 1 dan katta yoki -1 dan kichik bo'lgan kosinus qiymatlarini ololmaganimiz uchun biz bu qiymatlarni teskari trig funktsiyasiga ulay olmaymiz.

    Shunday qilib, 1-muammoni hal qilishda ushbu cheklovlardan foydalanib, biz bu holatda javob ekanligini ko'rishimiz mumkin

    \[{\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{6}\]
  2. \(\displaystyle {\cos ^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right)\)
    Yechim

    Umuman olganda, biz teskari trig funksiyasining qiymatini aniqlash uchun tenglamani echishimiz shart emas. Biz qilishimiz kerak bo'lgan yagona narsa birlik doirasiga qarashdir. Shunday qilib, bu holda biz 0 va \(\pi \) orasidagi burchakdan keyin turibmiz, buning uchun kosinus \( - \frac{{\sqrt 3 }}{2}\) qiymatini oladi. Shunday qilib, quyidagi birlik doirasini tekshiring

    A unit circle with a number of angles drawn in and the coordinates of the point where each angle meets the circle labeled beside it. The angle \(\frac{{5\pi }}{6}\) is drawn into the second quadrant and the point where it meets the circle is labeled \(\left( { - \frac{{\sqrt 3 }}{2},\frac{1}{2}} \right)\), showing that cosine of this angle is \( - \frac{{\sqrt 3 }}{2}\).

    Bundan shuni ko'rishimiz mumkin

    \[{\cos ^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right) = \frac{{5\pi }}{6}\]
  3. \(\displaystyle {\sin ^{ - 1}}\left( { - \frac{1}{2}} \right)\)
    Yechim

    Teskari kosinus funktsiyasi uchun \(\theta \)-ga qo'ygan cheklovlarimiz teskari sinus funksiyasi uchun ishlamaydi. Shunchaki yuqoridagi birlik doirasiga qarang va siz 0 va \(\pi \) orasida ikkita burchak mavjudligini ko'rasiz, ular uchun sinus \(\frac{1}{2}\) bo'ladi va bu biz xohlagan narsa emas. Teskari kosinus funktsiyasida bo'lgani kabi, biz faqat bitta qiymatni xohlaymiz. Shuning uchun teskari sinus funksiyasi uchun quyidagi cheklovlardan foydalanamiz.

    \[\theta = {\sin ^{ - 1}}\left( x \right)\hspace{0.25in}\hspace{0.25in} - 1 \le x \le 1\hspace{0.25in}{\rm{and }}\hspace{0.25in} - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}\]

    Birlik doirasini tekshirish orqali

    A unit circle with a number of angles drawn in and the coordinates of the point where each angle meets the circle labeled beside it. The angle \( - \frac{\pi }{6}\) is drawn into the fourth quadrant and the point where it meets the circle is labeled \(\left( {\frac{{\sqrt 3 }}{2}, - \frac{1}{2}} \right)\), showing that sine of this angle is \( - \frac{1}{2}\).

    ko'ramiz

    \[{\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = - \frac{\pi }{6}\]
  4. \({\tan ^{ - 1}}\left( 1 \right)\)
    Yechim

    Teskari tangens uchun cheklov

    \[\theta = {\tan ^{ - 1}}\left( x \right)\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} - \frac{\pi }{2} < \theta < \frac{\pi }{2}\]

    E'tibor bering, bu safar \(x\)-da hech qanday cheklovlar yo'q. Buning sababi, \(\tan \left( \theta \right)\) har qanday qiymatni manfiy cheksizlikdan musbat cheksizlikgacha olishi mumkin. Agar bu to'g'ri bo'lsa, biz teskari tangens funksiyasiga istalgan qiymatni ham kiritishimiz mumkin. Shuni ham yodda tutingki, biz \(\theta \)-dagi cheklovning ikkita so'nggi nuqtasini o'z ichiga olmaganmiz. Tangens bu ikki nuqtada aniqlanmagan, shuning uchun biz ularni teskari tangens funksiyasiga ulay olmaymiz.

    Ushbu muammoda biz \(\tan \left( \theta \right) = 1\) yoki \(\sin \left( \theta \right) = \cos \left( \theta \right)\) bo'lgan \( - \frac{\pi }{2}\) va \(\frac{\pi }{2}\) o'rtasidagi burchakni qidiramiz. Bu faqat \(\theta = \frac{\pi }{4}\) da sodir bo'lishi mumkin, shuning uchun,

    \[{\tan ^{ - 1}}\left( 1 \right) = \frac{\pi }{4}\]
  5. \(\displaystyle \cos \left( {{{\cos }^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)} \right)\)
    Yechim

    Ushbu bo'limdagi 1-masalaning javobini eslab, bu masalani hal qilish sirtdan ko'rinadiganidan ancha oson.

    \[\cos \left( {{{\cos }^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)} \right) = \cos \left( {\frac{\pi }{6}} \right) = \frac{{\sqrt 3 }}{2}\]

    Bu muammo teskari kosinus haqida bir nechta yoqimli faktlarga olib keladi

    \[\cos \left( {{{\cos }^{ - 1}}\left( x \right)} \right) = x\hspace{0.25in}\hspace{0.25in}{\rm{AND}}\hspace{0.25in}\hspace{0.25in}{\cos ^{ - 1}}\left( {\cos \left( \theta \right)} \right) = \theta \]
  6. \(\displaystyle {\sin ^{ - 1}}\left( {\sin \left( {\frac{\pi }{4}} \right)} \right)\)
    Yechim

    Bu muammo ham unchalik qiyin emas (umid qilamanki...).

    \[{\sin ^{ - 1}}\left( {\sin \left( {\frac{\pi }{4}} \right)} \right) = {\sin ^{ - 1}}\left( {\frac{{\sqrt 2 }}{2}} \right) = \frac{\pi }{4}\]

    Teskari kosinusda bo'lgani kabi bizda ham teskari sinus haqida quyidagi faktlar mavjud.

    \[\sin \left( {{{\sin }^{ - 1}}\left( x \right)} \right) = x\hspace{0.25in}\hspace{0.25in}{\rm{AND}}\hspace{0.25in}\hspace{0.25in}{\sin ^{ - 1}}\left( {\sin \left( \theta \right)} \right) = \theta \]
  7. \(\tan \left( {{{\tan }^{ - 1}}\left( { - 4} \right)} \right)\)
    Yechim

    Teskari kosinus va teskari sinus ular haqida bir nechta yoqimli faktlarga ega bo'lganidek, teskari tangens ham. Mana haqiqat

    \[\tan \left( {{{\tan }^{ - 1}}\left( x \right)} \right) = x\hspace{0.25in}\hspace{0.25in}{\rm{AND}}\hspace{0.25in}\hspace{0.25in}{\tan ^{ - 1}}\left( {\tan \left( \theta \right)} \right) = \theta \]

    Bu faktdan foydalanish buni juda oson muammoga aylantiradi, chunki men \({\tan ^{ - 1}}\left( 4 \right)\) ni qo'lda qila olmadim! Kalkulyator buni osonlikcha bajara olardi, lekin birlik doirasidan aniq javob ololmadim.

    \[\tan \left( {{{\tan }^{ - 1}}\left( { - 4} \right)} \right) = - 4\]