已知sin(π-α)=-1/2,计算: (1)cos(3π+α) (2)tan(5π-α)

sin(π-α)=sinα=-1/2,
cos(3π+α)=cos(π+α)=-cosα=根号3/2或-根号3/2
tan(5π-α)=tan(π-α)=-tanα=根号3/3或-根号3/3

(1)∵sin(π-α)=-1/2,∴sinα=-1/2∴α=2kπ-π/6或2kπ-5π/6 ∴cos(3π+α)=cos(π+α)=cos(2kπ+5π/6)或cos(2kπ+π/6) ∴原试=√￣3/2或-√￣3/2
(2)tan(5π-α)=tan(π-α)=-tanα ∵α=2kπ-π/6或2kπ-5π/6 ∴-tanα=1/√￣3 或-1/√￣3

(1)正负二分之（根3）
（2）正负三分之（根3）