Question
\( 11 f(x)=\cos ^{2} 2 x-\sin ^{2} 2 x \) funksiyaning eng kichik musbat davrini toping. \( \begin{array}{llll}\text { A) } \frac{\pi}{4} & \text { B) } \frac{\pi}{2} & \text { C) } \pi & \text { D) } 2 \pi\end{array} \)
Ask by Horton Mcdonald. in Uzbekistan
Jan 21,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
Funksiyaning eng kichik musbat davri **B) \( \frac{\pi}{2} \)**.
Solution
Berilgan funksiya:
\[ 11f(x) = \cos^{2}2x - \sin^{2}2x \]
Avvalo, trigonometrik identifikni qo‘llaymiz:
\[ \cos^{2}\theta - \sin^{2}\theta = \cos2\theta \]
Shunday qilib:
\[ 11f(x) = \cos4x \]
\[ f(x) = \frac{1}{11}\cos4x \]
Kosinus funksiyaning umumiy davri \( \frac{2\pi}{k} \) formulasi bilan hisoblanadi, bu yerda \( k \) — burilish soni. Bizda \( k = 4 \), demak:
\[ T = \frac{2\pi}{4} = \frac{\pi}{2} \]
Shunday qilib, funksiyaning eng kichik musbat davri:
**B) \( \frac{\pi}{2} \)**
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Beyond the Answer
Keling, berilgan funktsiyani ko'rib chiqamiz: \( 11 f(x) = \cos^2 2x - \sin^2 2x \). Bu ifodani simplifikat qilish uchun \( \cos^2 2x - \sin^2 2x = \cos 4x \) formulasi yordamida yozamiz. Shunday qilib, \( f(x) = \frac{1}{11} \cos 4x \). Kosinus funksiyasining eng kichik musbat davri \( 2\pi \) ga teng, lekin bizda \( \cos 4x \) funksiyasi bor, shuning uchun bu davr $ \frac{2\pi}{4} = \frac{\pi}{2} $ ga qisqaradi. Natijada, eng kichik musbat davri \( \frac{\pi}{2} \) bo'ladi. To'g'ri javob: B) \( \frac{\pi}{2} \).
