2.3-bo'lim: Trig formulalari
Bu trig formulalarining to'liq ro'yxati emas. Bu men Calculus kursida eng foydali deb topgan formulalar ro'yxati. Trig formulalarining to'liq ro'yxati uchun siz mening trigonometrik formulalar jadvaliimni yuklab olishingiz mumkin.
Quyidagi formulalarni to‘ldiring.
- \({\sin ^2}\left( \theta \right) + {\cos ^2}\left( \theta \right) = \)
Yechim
\[{\sin ^2}\left( \theta \right) + {\cos ^2}\left( \theta \right) = 1\]E'tibor bering, bu sinus va kosinusda bir xil bo'lsa, HAR QANDAY argumentga to'g'ri keladi. Shunday qilib, masalan:
\[{\sin ^2}\left( {3{x^4} - 5{x^2} + 87} \right) + {\cos ^2}\left( {3{x^4} - 5{x^2} + 87} \right) = 1\] - \({\tan ^2}\left( \theta \right) + 1 = \)
Yechim
\[{\tan ^2}\left( \theta \right) + 1 = {\sec ^2}\left( \theta \right)\]Agar siz ushbu bo'limdagi 1-masala formulasini bilsangiz, uni bepul olishingiz mumkin.
\[\begin{align*}{\sin ^2}\left( \theta \right) + {\cos ^2}\left( \theta \right) & = 1\\ \frac{{{{\sin }^2}\left( \theta \right)}}{{{{\cos }^2}\left( \theta \right)}} + \frac{{{{\cos }^2}\left( \theta \right)}}{{{{\cos }^2}\left( \theta \right)}} & = \frac{1}{{{{\cos }^2}\left( \theta \right)}}\\ {\tan ^2}\left( \theta \right) + 1 & = {\sec ^2}\left( \theta \right)\end{align*}\]\({\cot ^2}\left( \theta \right)\) va \({\csc ^2}\left( \theta \right)\) bilan bog'liq shunga o'xshash formulani topa olasizmi?
- \(\sin \left( {2t} \right) = \)
Yechim
\[\sin \left( {2t} \right) = 2\sin \left( t \right)\cos \left( t \right)\]Ushbu formula ko'pincha teskari holatda qo'llaniladi, shuning uchun sinus va kosinusning mahsuloti (albatta bir xil argument bilan) bitta sinus sifatida yozilishi mumkin. Masalan,
\[\begin{align*}{\sin ^3}\left( {3{x^2}} \right){\cos ^3}\left( {3{x^2}} \right) & = {\left( {\sin \left( {3{x^2}} \right)\cos \left( {3{x^2}} \right)} \right)^3}\\ & = {\left( {\frac{1}{2}\sin \left( {2\left( {3{x^2}} \right)} \right)} \right)^3}\\ & = \frac{1}{8}{\sin ^3}\left( {6{x^2}} \right)\end{align*}\]Ushbu formuladan teskari foydalanish hisob-kitob sinfida duch keladigan ba'zi muammolarning murakkabligini sezilarli darajada kamaytirishi mumkinligini bilib olasiz.
- \(\cos \left( {2x} \right) = \) (Uchta mumkin boʻlgan formulalar)
Yechim
Yuqorida aytib o'tilganidek, bu erda uchta formuladan foydalanish mumkin.
\[\begin{align*}\cos \left( {2x} \right) & = {\cos ^2}\left( x \right) - {\sin ^2}\left( x \right)\\ \cos \left( {2x} \right) & = 2{\cos ^2}\left( x \right) - 1\\ \cos \left( {2x} \right) & = 1 - 2{\sin ^2}\left( x \right)\end{align*}\]Ikkinchi formulani birinchisiga \({\sin ^2}\left( x \right) = 1 - {\cos ^2}\left( x \right)\) (ushbu bo'limdagi 1-masalaga qarang) ni almashtirish orqali olishingiz mumkin. Xuddi shunday, uchinchi formulani olish uchun \({\cos ^2}\left( x \right) = 1 - {\sin ^2}\left( x \right)\) ni birinchi formulaga almashtirishingiz mumkin.
- \({\cos ^2}(x) = \) (Birinchi darajaga kosinus boʻyicha)
Yechim
\[{\cos ^2}(x) = \frac{1}{2}\left( {1 + \cos \left( {2x} \right)} \right)\]Bu haqiqatan ham ushbu bo'limdagi 4-muammoning ikkinchi formulasi qayta tartibga solingan va kosinuslarning hatto kuchlarini yo'q qilish uchun JUDA foydalidir. Masalan,
\[\begin{align*}5{\cos ^2}\left( {3x} \right) & = 5\left( {\frac{1}{2}\left( {1 + \cos \left( {2\left( {3x} \right)} \right)} \right)} \right)\\ & = \frac{5}{2}\left( {1 + \cos \left( {6x} \right)} \right)\end{align*}\]E'tibor bering, siz ushbu formulani shunday yozilganini ko'rgansiz
\[\cos \left( {\frac{x}{2}} \right) = \pm \sqrt {\frac{1}{2}\left( {1 + \cos \left( x \right)} \right)} \]trig sinfida va yarim burchakli formula deb ataladi.
- \({\sin ^2}(x) = \) (Birinchi darajaga kosinus boʻyicha)
Yechim
\[{\sin ^2}(x) = \frac{1}{2}\left( {1 - \cos \left( {2x} \right)} \right)\]Oldingi muammoda bo'lgani kabi, bu haqiqatan ham ushbu bo'limdagi 4-masalaning uchinchi formulasi qayta tartibga solingan va hatto sinus kuchlarini yo'q qilish uchun juda foydali. Masalan,
\[\begin{align*}4{\sin ^4}\left( {2t} \right) & = 4{\left( {{{\sin }^2}\left( {2t} \right)} \right)^2}\\ & = 4{\left( {\frac{1}{2}\left( {1 - \cos \left( {4t} \right)} \right)} \right)^2}\\ & = 4\left( {\frac{1}{4}} \right)\left( {1 - 2\cos \left( {4t} \right) + {{\cos }^2}\left( {4t} \right)} \right)\\ & = 1 - 2\cos \left( {4t} \right) + \frac{1}{2}\left( {1 + \cos \left( {8t} \right)} \right)\\ & = \frac{3}{2} - 2\cos \left( {4t} \right) + \frac{1}{2}\cos \left( {8t} \right)\end{align*}\]Ushbu misolda ko'rsatilganidek, agar quvvat 2 dan katta bo'lsa va javob ko'pincha turli argumentlarga ega bo'lgan bir nechta kosinuslarga ega bo'lsa, siz ikkala formuladan ham bir necha marta foydalanishingiz mumkin.
Shunga qaramay, trig sinfida bu, ehtimol, yarim burchakli formula deb nomlangan va shunday yozilgan:
\[\sin \left( {\frac{x}{2}} \right) = \pm \sqrt {\frac{1}{2}\left( {1 - \cos \left( x \right)} \right)} \]