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