Question

Daca ctg x=2, calculati: $\frac{cos x+3sin x}{cos^{3}x-sin^{3}x}$

Answer 1

Explicație pas cu pas:

Scriem cotangenta ca fiind:

$ctgx=2 => \frac{cosx}{sinx}=2=>cosx=2sinx$

Inlocuim cosinus in expresie:

$E(x)=\frac{2sinx+3sinx}{(2sinx)^3-sin^3x}=\frac{5sinx}{7sin^3x}=\frac{5}{7sin^2x}=\frac{5}{7}*\frac{1}{sin^2x}$

Cautam sa descoperim valoarea pentru $\frac{1}{sin^2x}$ si folosim si formula $sin^2x+cos^2x=1$.

$ctgx=2\\ctg^2x=4\\\frac{cos^2x}{sin^2x}=4\\\frac{1-sin^2x}{sin^2x}=4\\\frac{1}{sin^2x}-\frac{sin^2x}{sin^2x}=4\\\frac{1}{sin^2x}-1=4\\\frac{1}{sin^2x}=5$

Finalizam:

$E(x)=\frac{5}{7}*\frac{1}{sin^2x}=\frac{5}{7}*5=\frac{25}{7}$