高中数学,第1小题,谢谢
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1)
= sin(π+π/3) + cos(4π-π/6) - (sin(2π-π/6) - cos(3π+π/4)^2)/tan(2π-π/3)
= -sin(π/3) + cos(π/6) - (-sin(π/6) - cos(π/4)^2)/-tanπ/3
= -sin(π/3) + sin(π/2-π/6) - (1/2+1/2)/sqrt(3)
= - sqrt(3)/3
= sin(π+π/3) + cos(4π-π/6) - (sin(2π-π/6) - cos(3π+π/4)^2)/tan(2π-π/3)
= -sin(π/3) + cos(π/6) - (-sin(π/6) - cos(π/4)^2)/-tanπ/3
= -sin(π/3) + sin(π/2-π/6) - (1/2+1/2)/sqrt(3)
= - sqrt(3)/3
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