前面验证初值,假设就省略了.
当n=k+1时,
左边=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)]
由于,sin(x/2^k)=2×sin[x/2^(k+1)]×cos[x/2^(k+1)]
所以左边=sinx×cos[x/2^(k+1)]/{2^k×2×sin[x/2^(k+1)]×cos[x/2^(k+1)]}=sinx/{2^(k+1)×sin[x/2^(k+1)]}=右边
所以等式对于n=k+1时也成立.