Asosiy tarkibga o'tish

9.6-bo'lim: Qutb koordinatalari

Shu paytgacha biz faqat Dekart (yoki To'rtburchak yoki.) bilan shug'ullanganmiz x-y) koordinatalar tizimi. Biroq, biz ko'rib turganimizdek, bu har doim ham ishlash uchun eng oson koordinatalar tizimi emas. Shunday qilib, ushbu bo'limda biz qutb koordinata tizimini ko'rib chiqamiz.

Koordinata tizimlari haqiqatda kosmosdagi nuqtani aniqlash usulidan boshqa narsa emas. Masalan, Dekart koordinata tizimida nuqtada \(\left( {x,y} \right)\) koordinatalari berilgan va biz bu nuqtani boshlang'ichdan boshlab, so'ngra \(x\) birliklarini gorizontal, keyin esa \(y\) birliklari vertikal ravishda siljitish orqali nuqtani aniqlash uchun foydalanamiz. Bu quyidagi eskizda ko'rsatilgan.

This is a standard xy-axis system with a point (x,y) in the first quadrant. There is a dashed line dropping straight down from the point to the x-axis and is labeled y to indicate the distance from the point to the x-axis. There is also a dashed line moving straight left from the point to the y-axis and is labeled x to indicate the distance from the point to the y-axis.

Biroq, bu ikki o'lchovli fazoda nuqtani aniqlashning yagona yo'li emas. Biz nuqtaga o'tish uchun boshlang'ichdan vertikal va gorizontal harakat qilish o'rniga, biz nuqtaga tegmagunimizcha, to'g'ridan-to'g'ri boshlang'ichdan chiqib ketishimiz va keyin bu chiziqning musbat \(x\) o'qi bilan qiladigan burchagini aniqlashimiz mumkin. Keyin nuqtaning koordinatalari sifatida nuqtaning boshlang'ich nuqtasidan masofasi va musbat \(x\) o'qidan aylanish uchun zarur bo'lgan miqdorni ishlatishimiz mumkin. Bu quyidagi eskizda ko'rsatilgan.

This is the same sketch from the first image except the point is now labeled (r, $\theta$) and there is a line from the origin to the point that is labeled r indicating the distance of the point from the origin. The angle from the positive x-axis and the line for r is labeled $\theta$.

Ushbu shakldagi koordinatalar deyiladi qutb koordinatalari.

Yuqoridagi muhokama \(r\) ijobiy raqam bo'lishi kerak degan fikrga olib kelishi mumkin. Biroq, biz \(r\) ning salbiy bo'lishiga ham ruxsat beramiz. Quyida \(\left( {2,\frac{\pi }{6}} \right)\) va \(\left( { - 2,\frac{\pi }{6}} \right)\) ikkita nuqtaning eskizi keltirilgan.

This is an xy-axis system with the point (2, $\frac{\pi}{6}$) shown in the 1st quadrant. There is a line from the origin to this point and it is shown that the angle from the x-axis to this line is shown to be $\frac{\pi}{6}$. The point (-2, $\frac{\pi}{6}$) is also shown and a dashed line from the origin to this point is included on the sketch.

Ushbu eskizdan biz \(r\) ijobiy bo'lsa, nuqta \(\theta \) bilan bir xil kvadrantda bo'lishini ko'rishimiz mumkin. Boshqa tomondan, agar \(r\) manfiy bo'lsa, nuqta \(\theta \) ga to'liq qarama-qarshi kvadrantda tugaydi. E'tibor bering, \(\left( { - 2,\frac{\pi }{6}} \right)\) koordinatalari \(\left( {2,\frac{{7\pi }}{6}} \right)\) koordinatalari bilan bir xil nuqtani tasvirlaydi. \(\left( {2,\frac{{7\pi }}{6}} \right)\) koordinatalari musbat \(x\) o'qidan \(\frac{{7\pi }}{6}\) burchagini aylantirishimizni aytadi, bu bizni yuqoridagi eskizdagi kesik chiziqqa qo'yadi, va keyin 2 masofaga chiqing.

Bu Kartezian koordinatalari va qutb koordinatalari o'rtasidagi muhim farqga olib keladi. Dekart koordinatalarida har qanday berilgan nuqta uchun aynan bitta koordinatalar to'plami mavjud. Polar koordinatalar bilan bu to'g'ri emas. Qutbli koordinatalarda ma'lum bir nuqta uchun cheksiz koordinatalar mavjud. Misol uchun, quyidagi to'rtta nuqta bir xil nuqta uchun barcha koordinatalardir.

\[\left( {5,\frac{\pi }{3}} \right) = \left( {5, - \frac{{5\pi }}{3}} \right) = \left( { - 5,\frac{{4\pi }}{3}} \right) = \left( { - 5, - \frac{{2\pi }}{3}} \right)\]

Bu to'rtta koordinatalar to'plamida ishlatiladigan burchaklarning eskizi.

This is an xy-axis system with the point (5, $\frac{\pi}{3}$) shown in the 1st quadrant with a solid line from the origin to this point. The point (-5, $\frac{\pi}{3}$) is also shown (but not labeled) and a dashed line from the origin to this point is included on the sketch. The counter clockwise angle from the positive x-axis to the solid line is shown to be $\frac{\pi}{3}$ and the counter clockwise angle from the positive x-axis to the dashed line is shown to be $\frac{4\pi}{3}$. The clockwise angle from the positive x-axis to the solid line is shown to be $-\frac{5\pi}{3}$ and the clockwise angle from the positive x-axis to the dashed line is shown to be $-\frac{2\pi}{3}$.

Ikkinchi koordinatali juftlikda biz nuqtaga borish uchun soat yo'nalishi bo'yicha aylantirdik. Biz soat yo'nalishi bo'yicha aylanishni unutmasligimiz kerak. Ba'zan bu biz qilishimiz kerak bo'lgan narsadir.

Oxirgi ikkita koordinata juftligi shundan foydalanadiki, agar biz nuqtadan qarama-qarshi kvadrantga tushsak, nuqtaga qaytish uchun manfiy \(r\) dan foydalanishimiz mumkin va, albatta, burchakka erishish uchun soat miliga teskari va soat yo'nalishi bo'yicha aylanish mavjud.

Ushbu to'rtta nuqta tizim atrofida bir necha marta aylanmasdan faqat nuqtaning koordinatalarini ifodalaydi. Agar burchakni o'q tizimi atrofida biz xohlagancha to'liq aylanishga imkon bersak, xuddi shu nuqta uchun cheksiz koordinatalar mavjud. Haqiqatdan ham \(\left( {r,\theta } \right)\) nuqtasi quyidagi koordinata juftlarining har biri bilan ifodalanishi mumkin.

\[\left( {r,\theta + 2\pi n} \right)\hspace{0.25in}\hspace{0.25in}\left( { - r,\theta + \left( {2n + 1} \right)\pi } \right),\hspace{0.25in}\,\,\,\,\,{\mbox{where }}n{\mbox{ is any integer}}{\mbox{.}}\]

Keyinchalik, koordinata tizimining kelib chiqishi haqida gapirishimiz kerak. Polar koordinatalarda kelib chiqish ko'pincha deyiladi qutb. Biz aslida asl/qutbdan uzoqlashmaganimiz uchun \(r = 0\) ekanligini bilamiz. Biroq, biz hali ham tizim atrofida istalgan burchak bilan aylana olamiz va shuning uchun boshlang'ich/qutbning koordinatalari \(\left( {0,\theta } \right)\).

Endi biz qutb koordinatalarini tushunganimizdan so'ng, ikkita koordinata tizimi o'rtasida konvertatsiya qilish haqida o'ylashimiz kerak. Ikkala koordinata tizimi qanday ishlashini eslatuvchi quyidagi eskizdan boshlang.

This is the same sketch from the first image except the point is now labeled (r, $\theta$) and there is a line from the origin to the point that is labeled r indicating the distance of the point from the origin. The angle from the positive x-axis and the line for r is labeled $\theta$.

E'tibor bering, bizda yuqorida to'g'ri burchakli uchburchak bor va shu bilan biz qutb koordinatalarini dekart koordinatalariga aylantiradigan quyidagi tenglamalarni olishimiz mumkin.

Qutbdan kartezianga aylantirish formulalari

\[x = r\cos \theta \hspace{1.0in}y = r\sin \theta \]

Dekartdan aylantirish deyarli oson. Keling, avvalo quyidagilarga e'tibor beraylik.

\[\begin{align*}{x^2} + {y^2} & = {\left( {r\cos \theta } \right)^2} + {\left( {r\sin \theta } \right)^2}\\ & = {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta \\ & = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) = {r^2}\end{align*}\]

Bu biz eslashimiz kerak bo'lgan juda foydali formula, ammo biz \(r\) uchun tenglamadan keyin turibmiz, shuning uchun ikkala tomonning kvadrat ildizini olaylik. Bu beradi,

\[r = \sqrt {{x^2} + {y^2}} \]

Texnik jihatdan ildiz oldida ortiqcha yoki minus bo'lishi kerak, chunki biz \(r\) ijobiy yoki salbiy bo'lishi mumkinligini bilamiz. Biz bu erda ijobiy \(r\) konventsiyasi bilan ishlaymiz.

\(\theta \) uchun tenglamani olish deyarli oddiy. Biz boshlaymiz,

\[\frac{y}{x} = \frac{{r\sin \theta }}{{r\cos \theta }} = \tan \theta \]

Ikkala tomonning teskari tangensini olish,

\[\theta = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\]

Biz bunga ehtiyot bo'lishimiz kerak, chunki teskari tangenslar faqat \( - \frac{\pi }{2} < \theta < \frac{\pi }{2}\) diapazonidagi qiymatlarni qaytaradi. Eslatib o'tamiz, ikkinchi mumkin bo'lgan burchak bor va ikkinchi burchak \(\theta + \pi \) tomonidan berilgan.

Xulosa qilish, dekart koordinatalaridan qutb koordinatalariga aylantirish uchun quyidagi formulalarni beradi.

Karteziandan qutbga aylantirish formulalari

\[\begin{align*}{r^2} & = {x^2} + {y^2}\hspace{1.45in} r = \sqrt {{x^2} + {y^2}} \\ \theta_{1} & = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\hspace{0.35in}\mbox{OR}\hspace{0.35in}\theta_{2} = \theta_{1} + \pi\end{align*}\]

Tez misol bilan ishlaylik.

1-misol Quyidagi nuqtalarning har birini berilgan koordinatalar tizimiga aylantiring.
  1. \(\left( { - 4,\frac{{2\pi }}{3}} \right)\) ni dekart koordinatalariga aylantiring.
  2. \(\left( { - 1,-1} \right)\) ni qutb koordinatalariga aylantiring.
a \(\left( { - 4,\frac{{2\pi }}{3}} \right)\) ni kartezian koordinatalariga aylantiring. Yechim

Ushbu konvertatsiya qilish juda oson. Biz qilishimiz kerak bo'lgan yagona narsa nuqtalarni formulalarga ulashdir.

\[\begin{align*}x & = - 4\cos \left( {\frac{{2\pi }}{3}} \right) = - 4\left( { - \frac{1}{2}} \right) = 2\\ y & = - 4\sin \left( {\frac{{2\pi }}{3}} \right) = - 4\left( {\frac{{\sqrt 3 }}{2}} \right) = - 2\sqrt 3 \end{align*}\]

Demak, Dekart koordinatalarida bu nuqta \(\left( {2, - 2\sqrt 3 } \right)\).


b \(\left( { - 1,-1} \right)\) ni qutb koordinatalariga aylantiring. Yechim

Avval \(r\) ni olamiz.

\[r = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} = \sqrt 2 \]

Keling, \(\theta \) ni olaylik.

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

Biroq, bu to'g'ri burchak emas. \(\theta \) ning bu qiymati birinchi kvadrantda va bizga berilgan nuqta uchinchi kvadrantda. Yuqorida aytib o'tilganidek, biz bunga \(\pi\) qo'shish orqali to'g'ri burchakka ega bo'lishimiz mumkin. Shunday qilib, haqiqiy burchak,

\[\theta = \frac{\pi }{4} + \pi = \frac{{5\pi }}{4}\]

Shunday qilib, qutb koordinatalarida nuqta \(\left( {\sqrt 2 ,\frac{{5\pi }}{4}} \right)\). Shuni ham yodda tutingki, biz salbiy \(r\) yordamida olgan birinchi \(\theta \) dan foydalanishimiz mumkin edi. Bu holda nuqta qutb koordinatalarida \(\left( { - \sqrt 2 ,\frac{\pi }{4}} \right)\) sifatida ham yozilishi mumkin.

Yuqoridagi formulalardan tenglamalarni bir koordinata tizimidan ikkinchisiga o'tkazish uchun ham foydalanishimiz mumkin.

2-misol Quyidagilarning har birini berilgan koordinatalar tizimidagi tenglamaga aylantiring.
  1. \(2x - 5{x^3} = 1 + xy\) ni qutb koordinatalariga aylantiring.
  2. \(r = - 8\cos \theta \) ni dekart koordinatalariga aylantiring.
a \(2x - 5{x^3} = 1 + xy\) ni qutb koordinatalariga aylantiring. Yechim

Bunday holda, \(x\) va \(y\) formulalarini kiritishdan boshqa ko'p narsa qilish kerak emas (i.e. Dekart koordinatalari) \(r\) va \(\theta \) (i.e. qutb koordinatalari).

\[\begin{align*}2\left( {r\cos \theta } \right) - 5{\left( {r\cos \theta } \right)^3} & = 1 + \left( {r\cos \theta } \right)\left( {r\sin \theta } \right)\\ 2r\cos \theta - 5{r^3}{\cos ^3}\theta & = 1 + {r^2}\cos \theta \sin \theta \end{align*}\]

b \(r = - 8\cos \theta \) ni kartezian koordinatalariga aylantiring. Yechim

Bu biroz hiyla-nayrang, lekin unchalik emas. Birinchidan, biz \(r\) ni to'g'ridan-to'g'ri almashtirishimiz mumkinligiga e'tibor bering. Biroq, kosinusni to'g'ridan-to'g'ri almashtirish yo'q, bu bizga faqat Dekart koordinatalarini beradi. Agar bizda kosinus bilan birga o'ng tomonda \(r\) bo'lsa, biz to'g'ridan-to'g'ri almashtirishni amalga oshirishimiz mumkin edi. Shunday qilib, agar o'ng tarafdagi \(r\) qulay bo'lsa, uni o'sha erga qo'yaylik, chap tomonga ham qo'yishni unutmang.

\[{r^2} = - 8r\cos \theta \]

Endi biz buni Kartezian koordinatalariga aylantiradigan ba'zi almashtirishlarni amalga oshirishimiz mumkin.

\[{x^2} + {y^2} = - 8x\]

Keyingi mavzuga o'tishdan oldin, oldingi misolning ikkinchi qismi ustida biroz ko'proq ish qilaylik.

Ikkinchi qismda berilgan tenglama aslida juda yaxshi ma'lum bo'lgan grafikdir; u shunchaki ko'pchilik tezda tanib oladigan shaklda emas. Uni aniqlash uchun dekart koordinata tenglamasini olaylik va biroz o'zgartiramiz.

\[{x^2} + 8x + {y^2} = 0\]

Endi tenglamaning \(x\) qismidagi kvadratni to'ldiring.

\[\begin{align*}{x^2} + 8x + 16 + {y^2} & = 16\\ {\left( {x + 4} \right)^2} + {y^2} & = 16\end{align*}\]

Demak, bu radiusi 4 va markaz \(\left( { - 4,0} \right)\) aylana edi.

Bu bizni ushbu bo'limning yakuniy mavzusiga olib boradi.

Umumiy qutbli koordinata grafiklari

Keling, qutbli koordinatalarda keng tarqalgan bir nechta grafiklarni aniqlaymiz. Shuningdek, biz bir nechta maxsus qutbli grafiklarni ko'rib chiqamiz.

Chiziqlar
Ba'zi chiziqlar qutb koordinatalarida juda oddiy tenglamalarga ega.

  1. \(\theta = \beta \).
    Bu chiziq ekanligini Dekart koordinatalariga quyidagi tarzda o'tkazish orqali ko'rishimiz mumkin \[\begin{align*}\theta & = \beta \\ {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) & = \beta \\ \frac{y}{x} & = \tan \beta \\ y & = \left( {\tan \beta } \right)x\end{align*}\]

    Bu koordinata boshidan o'tadigan va musbat \(x\) o'qi bilan \(\beta \) burchak hosil qiladigan chiziq. Yoki boshqacha qilib aytganda, bu \(\tan \beta \) qiyaligi bilan koordinatali chiziq.

  2. \(r\cos \theta = a\)
    Bu Kartezian koordinatalarini \(x = a\) ga aylantirish uchun etarlicha oson. Demak, bu vertikal chiziq.

  3. \(r\sin \theta = b\)
    Xuddi shunday, bu \(y = b\) ga aylanadi va gorizontal chiziq ham xuddi shunday.
3-misol Xuddi shu o'q tizimidagi \(\theta = \frac{{3\pi }}{4}\), \(r\cos \theta = 4\) va \(r\sin \theta = - 3\) grafiklari.
Yechim

Haqiqatan ham buning uchun grafikni bajarishdan boshqa ortiqcha narsa yo'q, shuning uchun bu erda.

This graph contains all the graphs from this example. The graph of $\theta =\frac{3\pi }{4}$ goes through the origin and forms a counter clockwise angle of $\frac{3\pi }{4}$ with the positive x-axis. The graph of $r\cos \theta =4$ is a vertical line going through x=4 on the x-axis. The graph of $r\sin \theta =-3$ is a horizontal line that goes through y=-3.

Davralar
Keling, qutb koordinatalaridagi doiralar tenglamalarini ko'rib chiqaylik.

  1. \(r = a\).
    Bu tenglama shuni anglatadiki, biz qaysi burchakka ega bo'lishimizdan qat'iy nazar, boshlang'ichdan masofa \(a\) bo'lishi kerak. Agar siz bu haqda o'ylab ko'rsangiz, bu aynan boshlang'ich markazida joylashgan \(a\) radiusli doiraning ta'rifi.

    Demak, bu koordinata boshida joylashgan \(a\) radiusi doirasi. Bu, shuningdek, qutbli koordinatalarda ishlashni xohlashimizning sabablaridan biridir. Boshida markazlashgan aylana tenglamasi Dekart koordinatalaridagi mos tenglamadan farqli o'laroq, juda yaxshi tenglamaga ega.

  2. \(r = 2a\cos \theta \).
    Biz tenglamalarni dekart koordinatalariga o'tkazayotganimizda ulardan birining aniq misolini ko'rib chiqdik.

    Bu \(\left| a \right|\) radiusi va \(\left( {a,0} \right)\) markazining aylanasi. E'tibor bering, \(a\) salbiy bo'lishi mumkin (yuqoridagi misolimizda bo'lgani kabi) va shuning uchun radiusda mutlaq qiymat chiziqlari talab qilinadi. Biroq, ular markazda ishlatilmasligi kerak.

  3. \(r = 2b\sin \theta \).
    Bu avvalgisiga o'xshaydi. Bu \(\left| b \right|\) radiusi va \(\left( {0,b} \right)\) markazining aylanasidir.

  4. \(r = 2a\cos \theta + 2b\sin \theta \).
    Bu avvalgi ikkitasining kombinatsiyasi va kvadratni ikki marta to'ldirish orqali bu \(\sqrt {{a^2} + {b^2}} \) radiusi va markazi \(\left( {a,b} \right)\) doirasi ekanligini ko'rsatish mumkin. Boshqacha qilib aytadigan bo'lsak, bu boshning markazida bo'lmagan aylananing umumiy tenglamasidir.
4-misol Xuddi shu eksa tizimidagi \(r = 7\), \(r = 4\cos \theta \) va \(r = - 7\sin \theta \) grafiklari.
Yechim

Birinchisi, radiusi 7 bo'lgan aylana bo'lib, uning markazida joylashgan. Ikkinchisi - \(\left( {2,0} \right)\) markazida joylashgan 2 radiusli doira. Uchinchisi - \(\left( {0, - \frac{7}{2}} \right)\) markazida joylashgan \(\frac{7}{2}\) radiusli doira. Mana uchta tenglamaning grafigi.

This graph contains all the graphs from this example. The graph of r=7 is a circle of radius 7 centered at the origin. The graph of $r=4\cos \theta $ is a circle of radius 2 centered at (2,0). The graph of $r=-7\sin \theta $ is a circle of radius 7/2 centered at (0,-7/2).

E'tibor bering, \(r = a\) ning to'liq grafigi uchun \(0 \le \theta \le 2\pi \) diapazoni kerak bo'ladi va bu erda berilgan boshqa doiralarning grafigini tuzish uchun faqat \(0 \le \theta \le \pi \) diapazoni kerak bo'ladi. Agar xohlasangiz, buni tezkor qiymatlar jadvali bilan tekshirishingiz mumkin.

Kardioidlar va Limakonlar
Bularni quyidagi uchta holatga bo'lish mumkin.

  1. Kardioidlar: \(r = a \pm a\cos \theta \) va \(r = a \pm a\sin \theta \).
    Ularda yurak shakli noaniq bo'lgan va har doim kelib chiqishi mavjud bo'lgan grafik mavjud.

  2. Ichki halqali limakonlar: \(r = a \pm b\cos \theta \) va \(a < b\) bilan \(r = a \pm b\sin \theta \).
    Ular ichki halqaga ega bo'ladi va har doim kelib chiqishni o'z ichiga oladi.

  3. Ichki halqasiz limakonlar: \(r = a \pm b\cos \theta \) va \(a > b\) bilan \(r = a \pm b\sin \theta \).
    Ularning ichki halqasi yo'q va kelib chiqishini o'z ichiga olmaydi.
5-misol \(r = 5 - 5\sin \theta \), \(r = 7 - 6\cos \theta \) va \(r = 2 + 4\cos \theta \) grafiklari.
Yechim

Bularning barchasi \(0 \le \theta \le 2\pi \) diapazonida bir marta chiziladi. Bu erda har biri uchun qiymatlar jadvali va har birining grafiklari keltirilgan.

\(\theta \) \(r = 5 - 5\sin \theta \) \(r = 7 - 6\cos \theta \) \(r = 2 + 4\cos \theta \)
0 5 1 6
\(\displaystyle \frac{\pi }{2}\) 0 7 2
\(\pi \) 5 13 -2
\(\displaystyle \frac{{3\pi }}{2}\) 10 7 2
\(2\pi \) 5 1 6


This is the graph of $r=5-5\sin \theta $ that goes through the points given in the table above. It is a vaguely heart shaped graph that is symmetric about the y-axis. The “crease” in the heart is at the origin and is in the 1st and 2nd quadrants. The vast majority of the graph appears below the x-axis.


This is the graph of $r=7-6\cos \theta $ that goes through the points given in the table above. It is a vaguely heart shaped graph that is symmetric about the x-axis. The “crease” in the heart is at (1,0) and is in the 1st and 4th quadrants. The vast majority of the graph appears to the left of the y-axis.


This is the graph of $r=2+4\cos \theta $ that goes through the points given in the table above. It is a vaguely heart shaped graph (with one major difference to be discussed shortly) that is symmetric about the x-axis. The “crease” in the heart is at the origin and is in the 2nd and 3rd quadrants. The vast majority of the graph appears to the right of the y-axis. The major difference mentioned above is that when the graph goes into the crease/origin in the 2nd quadrant it doesn’t just pop back out into the 3rd quadrant. Instead when the graph goes into the crease/origin it continues through the origin into the 4th quadrant and then goes up through the point (2,0) and then loops around through the 1st quadrant and goes back out through the origin into the 3rd quadrant. This forms what is usually called the inner loop of the graph.

Ushbu bo'limda biz qilishimiz kerak bo'lgan yakuniy narsa bor. Oldingi misoldagi uchinchi grafikda bizda ichki pastadir bor edi. Ba'zan biz \(\theta \) qiymatini bilishimiz kerak bo'ladi, buning uchun grafik boshlang'ichdan o'tadi. Bularni topish uchun tenglamani nolga tenglashtirib, quyidagi tarzda yechishimiz kifoya:

\[0 = 2 + 4\cos \theta \hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,\cos \theta = - \frac{1}{2}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\theta = \frac{{2\pi }}{3},\frac{{4\pi }}{3}\]