Question

Biết sin x. cos x=0,48. Tính A= \(sin^3x+cos^3x\)

Answer 1

ta có : \(sin^2x+cos^2x=1\Leftrightarrow\left(sinx+cosx\right)^2-2sinx.cosx=1\)

\(\Leftrightarrow\left(sinx+cosx\right)^2-0,96=1\) \(\Leftrightarrow sinx+cosx=\pm\sqrt{1,96}=\pm1,4\)

ta có : \(sin^3x+cos^3x=\left(sinx+cosx\right)^3-3sinx.cosx\left(sinx+cosx\right)\)

th1: \(sinx+cosx=1,4\Rightarrow sin^3x+cos^3x=0,728\)

th2: \(sinx+cosx=-1,4\Rightarrow sin^3x+cos^3x=-0,728\)

vậy ............................................................................................................