已知fn=cos(nπ/5),(n属于N正),求f1+f2+……f2010
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已知fn=cos(nπ/5),(n属于N正),求f1+f2+……f2010
已知fn=cos(nπ/5),(n属于N正),求f1+f2+……f2010
已知fn=cos(nπ/5),(n属于N正),求f1+f2+……f2010
f的周期是10
因为 cosx=-cos(x+π)
所以 f1+f6=0
f2+f7=0
f3+f8=0
f4+f9=0
f5+f10=0
因此一个周期内 T=0
当n=2010时 fn共有201个周期
所以 f1+f2+……f2010=0
附诱导公式(口诀:奇变偶不变,符号看象限.)
sin(-α)=-sinα
cos(-α)=cosα
tan(-α)=-tanα
cot(-α)=-cotα
sin(π/2-α)=cosα
cos(π/2-α)=sinα
tan(π/2-α)=cotα
cot(π/2-α)=tanα
sin(π/2+α)=cosα
cos(π/2+α)=-sinα
tan(π/2+α)=-cotα
cot(π/2+α)=-tanα
sin(π-α)=sinα
cos(π-α)=-cosα
tan(π-α)=-tanα
cot(π-α)=-cotα
sin(π+α)=-sinα
cos(π+α)=-cosα
tan(π+α)=tanα
cot(π+α)=cotα
sin(3π/2-α)=-cosα
cos(3π/2-α)=-sinα
tan(3π/2-α)=cotα
cot(3π/2-α)=tanα
sin(3π/2+α)=-cosα
cos(3π/2+α)=sinα
tan(3π/2+α)=-cotα
cot(3π/2+α)=-tanα
sin(2π-α)=-sinα
cos(2π-α)=cosα
tan(2π-α)=-tanα
cot(2π-α)=-cotα
sin(2kπ+α)=sinα
cos(2kπ+α)=cosα
tan(2kπ+α)=tanα
cot(2kπ+α)=cotα
(其中k∈Z)