Question

cho \(tan\alpha=4\) tính \(\frac{sin^3\alpha+cos^3\alpha}{sin^3\alpha-cos^3\alpha}\)

Answer 1

ta có tan=4=>\(\frac{sin}{cos}=4\) =>sin=4cos

ta có \(\frac{sin^3+cos^3}{sin^3-cos^3}=\frac{\left(sin+cos\right)^3-3sin.cos\left(sin+cos\right)}{\left(sin-cos\right)^3+3sin.cos\left(sin-cos\right)}\)

\(=\frac{125cos^3-3.4cos.cos.5cos}{27cos^3+3.4cos.cos.3cos}\)\(=\frac{125cos^3-60cos^3}{27cos^3+36cos^3}\)

\(=\frac{65cos^3}{63cos^3}=\frac{65}{63}\)