Question

Biết \(\dfrac{Sin^4x-Cos^4x+Cos^2x}{2\left(1-Cosx\right)}=Cos^k\dfrac{mx}{n}\).Trong đó \(\dfrac{m}{n}\) tối giản. Tìm m,k,n.

Answer 1

\(\dfrac{sin^4x-cos^4x+cos^2x}{2\left(1-cosx\right)}=\dfrac{\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)+cos^2x}{2\left(1-cosx\right)}\)=\(\dfrac{sin^2x}{2\left(1-cosx\right)}=\dfrac{1-cos^2x}{2\left(1-cosx\right)}=\dfrac{\left(1-cosx\right)\left(1+cosx\right)}{2\left(1-cosx\right)}=\dfrac{1}{2}\left(1+cosx\right)\)=\(cos^2\left(\dfrac{x}{2}\right)\).

Suy ra k=2, m=1 và n=2.