【用数学归纳法证明:cos(x/2)×cos(x/2^2)×...×cos(x/2^n)=sinx/[2^n×sin(x/2^n)]当n=k+1时,怎么证的?】

查开德回答：

　　设当n=k时成立,则有cos(x/2)×cos(x/2^2)×...×cos(x/2^k)=sinx/[2^k×sin(x/2^k)]

　　则当n=k+1时,cos(x/2)×cos(x/2^2)×...×cos(x/2^k)×cos[x/2^(k+1)]=sinx×cos[x/2^(k+1)]/[2^k×sin(x/2^k)]

　　将sin(x/2^k)＝2×cos[x/2^(k+1)]×sin[x/2^(k+1)]代入式子消掉cos[x/2^(k+1)]得

　　cos(x/2)×cos(x/2^2)×...×cos(x/2^k)×cos[x/2^(k+1)]＝sinx/[2^(k+1)×sin(x/2^(k+1))]

　　即当n=k+1时等式也成立