- TXĐ : D = R .
- Ta có : \(y=\sin^3x.\cos x-\cos^3x.\sin x\)
\(=\dfrac{1}{2}\sin2x.\sin^2x-\dfrac{1}{2}\sin2x.\cos^2x\)
\(=-\dfrac{1}{2}\sin2x\left(\cos^2x-\sin^2x\right)\)
\(=-\dfrac{1}{2}\sin2x.\cos2x=-\dfrac{1}{4}\sin4x\)
Đề bài yêu cầu điều gì nhỉ?
\(y=sinx.cosx\left(sin^2x-cos^2x\right)=-\dfrac{1}{2}.\left(2sinx.cosx\right)\left(cos^2x-sin^2x\right)\)
\(=-\dfrac{1}{2}sin2x.cos2x=-\dfrac{1}{4}sin4x\)
Do \(-1\le sin4x\le1\Rightarrow-\dfrac{1}{4}\le-\dfrac{1}{4}sin4x\le\dfrac{1}{4}\)
\(y_{max}=\dfrac{1}{4}\) khi \(sin4x=-1\)
\(y_{min}=-\dfrac{1}{4}\) khi \(sin4x=1\)
