Toán Xét tính tuần hoan chu kì: y= $\frac{1}{sin2x}$ 07/07/2021 By Eloise Xét tính tuần hoan chu kì: y= $\frac{1}{sin2x}$
$y=\dfrac{1}{\sin2x}$ $=\dfrac{1}{2}.\dfrac{1}{\sin x\cos x}$ $=\dfrac{1}{2}.\dfrac{\sin^2x+\cos^2x}{\sin x\cos x}$ $=\dfrac{1}{2}\Big( \dfrac{\sin^2x}{\sin x\cos x}+\dfrac{\cos^2x}{\sin x\cos x}\Big)$ $=\dfrac{1}{2}(\tan x+\cot x)$ Hàm $y=\tan x$, $y=\cot x$ tuần hoàn chu kì $\pi$ Vậy hàm $y=\dfrac{1}{\sin2x}$ tuần hoàn chu kì $T=BCNN(\pi,\pi)=\pi$ Trả lời
$y=\dfrac{1}{\sin2x}$
$=\dfrac{1}{2}.\dfrac{1}{\sin x\cos x}$
$=\dfrac{1}{2}.\dfrac{\sin^2x+\cos^2x}{\sin x\cos x}$
$=\dfrac{1}{2}\Big( \dfrac{\sin^2x}{\sin x\cos x}+\dfrac{\cos^2x}{\sin x\cos x}\Big)$
$=\dfrac{1}{2}(\tan x+\cot x)$
Hàm $y=\tan x$, $y=\cot x$ tuần hoàn chu kì $\pi$
Vậy hàm $y=\dfrac{1}{\sin2x}$ tuần hoàn chu kì $T=BCNN(\pi,\pi)=\pi$