cos(π+π/2)等于多少

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cos(π+π/2)等于多少cos(π+π/2)等于多少cos(π+π/2)等于多少=cos(3π/2)=0cos(3∏/2)=0因为cosx的周期为2π所以cos(π+π/2)=cos(2π-π/2

cos(π+π/2)等于多少
cos(π+π/2)等于多少

cos(π+π/2)等于多少
=cos(3π/2)=0

cos(3∏/2)=0

因为cosx的周期为2π
所以cos(π+π/2)=cos(2π-π/2)=cos(-π/2)=0

=cos(3π/2)=0, =cos(3π/2)=0 ,=cos(3π/2)=0, =cos(3π/2)=0 ,