oblicz
: 28 wrz 2007, o 12:35
\(\displaystyle{ \frac{sin^6 (x)+cos^6 x}{sin^4 (x)+cos^4 x}}\)
tg(x)=2
tg(x)=2
\(\displaystyle{ \frac{sin^6x+cos^6x}{sin^4x+cos^4x}=
\frac{\cos^6x(\tan^6x+1)}{\cos^4x(\tan^4x+1)}=
\frac{\cos^2x(\tan^6x+1)}{\tan^4x+1}=
\frac{\tan^6x+1}{(\tan^2+1)\cdot(\tan^4x+1)}=\frac{65}{5\cdot17}=\frac{13}{17}}\)