三角函数归纳法证明题 数学三角函数题（数学归纳法）
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cos2x=2(cosx)^2-1\uff0ccos3x=cos(x+2x)=cosxcos2s-sinxsin2x=cosx[2(cosx)^2-1]-sinx(2sinxcosx)=cosx[2(cosx)^2-1]-2cosx(1-(cosx)^2)=4(cosx)^3-3cosx\uff0ccos4x=2(cos2x)^2-1=8(cosx)^4-8(cosx)^2+1\uff0c\u6545n=1,2,3,4\u65f6\u6c42\u8bc1\u6210\u7acb
\u8bbecos(n-1),cosnx\u90fd\u53ef\u4ee5\u8868\u793a\u4e3a\u4ec5\u542bcosx\u7684\u6574\u6b21\u5e42\u7684\u591a\u9879\u5f0f
\u5219cos(n+1)x=cosxcosnx-sinxsinnx=cosxcosnx-sinx[sin(n-1)xcosx+cos(n-1)sinx]=cosxcosnx-sinxsin(n-1)xcosx-cos(n-1)(1-(cosx)^2)=cosxcosnx-sinxsin(n-1)xcosx-cos(n-1)+cos(n-1)(cosx)^2=cosxcosnx-cos(n-1)-[sinxsin(n-1)x-cos(n-1)(cosx)]cosx=2cosxcosnx-cos(n-1)
\u56e0\u4e3acos(n-1),cosnx\u90fd\u53ef\u4ee5\u8868\u793a\u4e3a\u4ec5\u542bcosx\u7684\u6574\u6b21\u5e42\u7684\u591a\u9879\u5f0f\uff0ccos(n+1)x\u53ef\u4ee5\u8868\u793a\u4e3a\u4ec5\u542bcosx\u7684\u6574\u6b21\u5e42\u7684\u591a\u9879\u5f0f\uff0c\u5f97\u8bc1\u3002
∵左边=cosx，
右边=[sin((1+1/2)x)-sin(x/2)]/[2sin(x/2)]=2cosx*sin(x/2)/[2sin(x/2)] (和差化积)
=cosx，
∴左边=右边，命题成立。
（2）假设当n=k时，命题成立。
即cosx+cos(2x)+cos(3x)+......+cos(kx)=[sin((k+1/2)x)-sin(x/2)]/[2sin(x/2)]成立。
那么，当n=k+1时，
∵左边=cosx+cos(2x)+cos(3x)+......+cos(kx)+cos((k+1)x)
=[sin((k+1/2)x)-sin(x/2)]/[2sin(x/2)]+cos((k+1)x)
=[sin((k+1/2)x)-sin(x/2)+2sin(x/2)*cos((k+1)x)]/[2sin(x/2)]
=[sin((k+1/2)x)-sin(x/2)+sin((k+1+1/2)x-sin((k+1/2)x)]/[2sin(x/2)] (积化和差)
=[sin((k+1+1/2)x-sin(x/2))]/[2sin(x/2)]
右边=[sin((k+1+1/2)x-sin(x/2))]/[2sin(x/2)]
∴左边=右边，命题成立。
故由数学归纳法，得
cosx+cos(2x)+cos(3x)+......+cos(nx)=[sin((n+1/2)x)-sin(x/2)]/[2sin(x/2)]成立。
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