Question

вычислите sin^3α+cos^3α, если sinα+cosα=1/3 ​

Answer 1

Объяснение:

$sin\alpha +cos\alpha =\frac{1}{3}\ \ \ \ sin^3\alpha +cos^3\alpha=?$

$sin^3\alpha +cos^3\alpha =(sin\alpha +cos\alpha )*(sin^2\alpha -sin\alpha *cos\alpha +cos^2\alpha )=\\=\frac{1}{3}*(1-sin\alpha *cos\alpha )=\frac{1-sin\alpha *cos\alpha }{3} .\\(sin\alpha +cos\alpha )^2=(\frac{1}{3} )^2\\sin^2\alpha +2*sin\alpha *cos\alpha +cos^2\alpha =\frac{1}{9} \\1+2*sin\alpha *cos\alpha =\frac{1}{9}\\2*sin\alpha *cos\alpha =-\frac{8}{9}\ |:2\\sin\alpha *cos\alpha =-\frac{4}{9} .\ \ \ \ \Rightarrow$

$sin^3\alpha +cos^3\alpha =\frac{1-(-\frac{4}{9}) }{3}=\frac{1+\frac{4}{9} }{3} =\frac{\frac{13}{9} }{3} =\frac{13}{27} .$