f《1－cosx》＝sinx的平方。。求f《x》
\u5df2\u77e5 f\uff081-cosx\uff09=sinx\u7684\u5e73\u65b9\uff0c\u5219f\uff08x\uff09=\uff1f\u5177\u4f53\u7684\u5206\u6790\u2026\u2026f\uff081-cosx\uff09=sin²x
f\uff081-cosx\uff09=1-cos²x
f\uff081-cosx\uff09=\uff081+cosx\uff09\uff081-cosx\uff09
f\uff081-cosx\uff09=-\uff08-1-cosx\uff09\uff081-cosx\uff09
f\uff081-cosx\uff09=-\uff081-cosx-2\uff09\uff081-cosx\uff09
\u81ea\u53d8\u91cf\u53d8\u6210\u4e861-cosx\uff0c\u4e8e\u662f\u6211\u4eec\u6362\u5143 \u4ee4t=1-cosx
f\uff08t\uff09=-\uff08t-2\uff09*t=-t²+2t
\u5373\u6709\uff1a
f\uff08x\uff09=-x²+2x
\u4ee41-cosx=t \u5219cosx=t-1 \u56e0\u4e3asinx=1-cosx \u6240\u4ee5sinx=1-(1-t)=-t+2t \u6240\u4ee5\u5c31\u662ff(t)=-t+2t \u5373f(x)=-x+2x \u53e6\u59161-cosx\u2208[0,2] f(x)=-x+2x,x\u2208[0,2] \u8fd8\u6709\u95ee\u9898\u5417?
f《1－cosx》＝sinx的平方=1-cos²x=(1+cosx)(1-cosx)令y=1-cosx,cosx=1-y
f(y)=f《1－cosx》
=(1+cosx)(1-cosx)
=(1+1-y)(1-1+y)
=y(2-y)
=2y-y²
求f《x》=2x-x²
f《1－cosx》＝sinx的平方
设t=1-cosx,则cosx=1-t
sin^2x=1-cos^2x=1-(1-t)^2=2t-t^2
即f(t)=2t-t^2
那么f(x)=2x-x^2
f(1-cosx)=(sinx)^2=1-(cosx)^2
令1-cosx=t得cosx=1-t
∴f(t)=1-(1-t)^2
=-t^2+2t
即f(x)=-x^2+2x
解：令t=1-cosx
则cosx=1-t
sin²x=1-cos²x=1-(1-t)²=2t-t²
所以f(x)=2x-x²
用替换法就可以求了
高一得题吧
f(x)=2x-x^2
1-cosx=t cosx=1-t x=arcos(1-t) f(t)=sin[arcos(1-t)] 把t换成x
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