Question

Cho \(\tan=3\)

Chứng minh \(\frac{\sin^3\alpha-\cos^3\alpha}{\sin^2\alpha+\cos^2\alpha}=\frac{13}{14}\)

Answer 1

Lời giải:

Ta có: \(3=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=3\cos a\)

Khi đó:

\(\frac{\sin ^3a-\cos ^3a}{\sin ^3a+\cos ^3a}=\frac{(3\cos a)^3-\cos ^3a}{(3\cos a)^3+\cos ^3a}=\frac{\cos ^3a(3^3-1)}{\cos ^3a(3^3+1)}=\frac{3^3-1}{3^3+1}=\frac{13}{14}\)

Ta có đpcm.