Question

Если ctga=4, тогда найдите решение sin^3a-2cos^3a/cos^3a+2sin^3a​

Answer 1

$\frac{Sin^{3}\alpha-2Cos^{3}\alpha}{Cos^{3}\alpha+2Sin^{3}\alpha}=\frac{\frac{Sin^{3}\alpha}{Sin^{3}\alpha}-\frac{2Cos^{3}\alpha}{Sin^{3}\alpha}}{\frac{Cos^{3}\alpha}{Sin^{3}\alpha}+\frac{2Sin^{3}\alpha}{Sin^{3}\alpha}}=\frac{1-2Ctg^{3}\alpha}{Ctg^{3}\alpha+2}=\frac{1-2*4^{3}}{4^{3}+2 }=\frac{-127}{66}=-1\frac{5}{66}$