如果|sinkx|<=k|sinx|,那么(sinkx)^2<=k^2(sinx)^2,依歸納法需證
|sin(k+1)x|<=(k+1)|sinx|,也就是(sin(k+1)x)^2<=(k+1)^2(sinx)^2
sin(k+1)x=sinkxcosx+coskxsinx
(sin(k+1)x)^2=(sinkx)^2(cosx)^2+(sinx)^2(coskx)^2+2sinkxsinxcosxcoskx
(cosx)^2<=1,(coskx)^2<=1,|cosx|<=1,|coskx|<=1,
(sinkx)^2(cosx)^2<=(sinkx)^2
(sinx)^2(coskx)^2<=(sinx)^2
sinkxsinxcosxcoskx<=|sinkx||sinx||cosx||coxkx|<=|sinkx||sinx|
(sin(k+1)x)^2<=(sinkx)^2+(sinx)^2+2|sinkx||sinx|
由假設(shè)|sinkx|<=k|sinx|和(sinkx)^2<=k^2(sinx)^2
(sin(k+1)x)^2<=k^2(sinx)^2+(sinx)^2+2k(sinx)^2
(sin(k+1)x)^2<=(k^2+2k+1)(sinx)^2=(k+1)^2(sinx)^2
得證|sin(k+1)x|<(k+1)|sinx|