Question

cho tam giác ABC  có C =2B=4A.CMR : \(\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\)

Answer 1

ta có : A+C+B=\(_{\pi}\) \(\Rightarrow\) A+2A+4A=\(\pi\) \(\Rightarrow\) A=\(\frac{\pi}{7}\) ,B=\(\frac{2\pi}{7}\),C=\(\frac{4\pi}{7}\) 

Do đó : \(\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\Rightarrow\) BC=AB+AC

\(\Leftrightarrow\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}=\sin\frac{\pi}{7}\sin\frac{2\pi}{7}+\sin\frac{\pi}{7}\sin\frac{4\pi}{7}\) 

\(\Leftrightarrow\cos\frac{2\pi}{7}-\cos\frac{6\pi}{7}=\cos\frac{\pi}{7}-\cos\frac{3\pi}{7}+\cos\frac{3\pi}{7}-\cos\frac{5\pi}{7}\)

 \(\Leftrightarrow\cos\frac{2\pi}{7}-\cos\frac{6\pi}{7}=\cos\frac{\pi}{7}-\cos\frac{5\pi}{7}\) (điều hiển nhiên)