Toán Giúp em tìm min max bài toán này với ạ Sin(x+pi/7)+sin(x-pi/7) 05/10/2021 By Reagan Giúp em tìm min max bài toán này với ạ Sin(x+pi/7)+sin(x-pi/7)
Đáp án: Giải thích các bước giải: \(y = \sin \left( {x + \dfrac{\pi }{7}} \right) + \sin \left( {x – \dfrac{\pi }{7}} \right)\) Ta có: \(y = \sin \left( {x + \dfrac{\pi }{7}} \right) + \sin \left( {x – \dfrac{\pi }{7}} \right)\)\( = 2\sin x\cos \dfrac{\pi }{7}\) Mà \( – 1 \le \sin x \le 1\) nên \( – 2\cos \dfrac{\pi }{7} \le 2\cos \dfrac{\pi }{7}\sin x \le 2\cos \dfrac{\pi }{7}\) Do đó \( – 2\cos \dfrac{\pi }{7} \le y \le 2\cos \dfrac{\pi }{7}\). Vậy \(\max y = 2\cos \dfrac{\pi }{7}\) khi \(\sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \). \(\min y = – 2\cos \dfrac{\pi }{7}\) khi \(\sin x = – 1 \Leftrightarrow x = – \dfrac{\pi }{2} + k2\pi \). Trả lời
$y’ = cos(x+ \pi/7) + cos(x-\pi/7)$. y’=0 <-> $cos(x+\pi/7) = -cos(x-\pi/7)$ <-> $cos(x+\pi/7) = cos(\pi – x + \pi/7)$ <->$x + \pi/7 = 8\pi/7 -x + 2k pi$ hoac $x + \pi/7 = -8\pi/7 + x + 2k\pi$. <->$x = \pi/2 + k\pi$ Vay y(\pi/2 + k\pi) = sin((9\pi)/14) + sin((5\pi)/14) neu k chan va bang -sin((9\pi)/14) – sin((5\pi)/14) neu k le. Trả lời
Đáp án:
Giải thích các bước giải:
\(y = \sin \left( {x + \dfrac{\pi }{7}} \right) + \sin \left( {x – \dfrac{\pi }{7}} \right)\)
Ta có: \(y = \sin \left( {x + \dfrac{\pi }{7}} \right) + \sin \left( {x – \dfrac{\pi }{7}} \right)\)\( = 2\sin x\cos \dfrac{\pi }{7}\)
Mà \( – 1 \le \sin x \le 1\) nên \( – 2\cos \dfrac{\pi }{7} \le 2\cos \dfrac{\pi }{7}\sin x \le 2\cos \dfrac{\pi }{7}\)
Do đó \( – 2\cos \dfrac{\pi }{7} \le y \le 2\cos \dfrac{\pi }{7}\).
Vậy \(\max y = 2\cos \dfrac{\pi }{7}\) khi \(\sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \).
\(\min y = – 2\cos \dfrac{\pi }{7}\) khi \(\sin x = – 1 \Leftrightarrow x = – \dfrac{\pi }{2} + k2\pi \).
$y’ = cos(x+ \pi/7) + cos(x-\pi/7)$.
y’=0 <-> $cos(x+\pi/7) = -cos(x-\pi/7)$
<-> $cos(x+\pi/7) = cos(\pi – x + \pi/7)$
<->$x + \pi/7 = 8\pi/7 -x + 2k pi$ hoac $x + \pi/7 = -8\pi/7 + x + 2k\pi$.
<->$x = \pi/2 + k\pi$
Vay y(\pi/2 + k\pi) = sin((9\pi)/14) + sin((5\pi)/14) neu k chan va bang -sin((9\pi)/14) – sin((5\pi)/14) neu k le.