$f(x)=A\sin (\omega x+\varphi )\quad (\omega \gt 0,\left | \varphi \right | \lt \cfrac{\pi}{2})的图像如图所示,f(x)$
2025-03-04T02:16:34.png
显然$A=2,f(0)=1$
$f(0)=1\Rightarrow \varphi=\cfrac{\pi}{6} ,故f(x)=2\sin (\omega x+\cfrac{\pi}{6}),令f(x)=0\Rightarrow \omega x+\cfrac{\pi}{6}=0 $
$\Rightarrow x=-\cfrac{\pi}{6\omega } \Rightarrow -\cfrac{\pi}{6\omega } -(-\pi)=\cfrac{T}{4} =\cfrac{2\pi}{4\omega }$
$\Rightarrow \pi=\cfrac{\pi}{2\omega } +\cfrac{\pi}{6\omega }\Rightarrow \omega =\cfrac{2}{3} $
$\therefore \quad f(x)=2\sin ( \cfrac{2}{3} x+\cfrac{\pi}{6})$

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