搜搜考试网已知cos(π/3-α)=根号3/3,求cos(2π/3+α)+cos2(7π/6+α)
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已知cos(π/3-α)=根号3/3,求cos(2π/3+α)+cos2(7π/6+α)
cos(π/3-α)=√3/3
cos(2π/3+α)+cos2(7π/6+α)
=-cos(π-(2π/3+α)) + cos(7π/3 +2α)
=-cos(π/3-α) + cos(π/3 +2α)
=-√3/3 - cos(2π/3-2α)
=-√3/3 - cos2(π/3-α)
=-√3/3 - 2[cos(π/3-α)]^2 +1
=-√3/3 - 2(1/3) + 1
=-√3/3 + 1/3
再问： 2是平方
再答： cos(π/3-α) =√3/3cos(2π/3+α)+[cos(7π/6+α)]^2 =-cos(π-(2π/3+α)) + (1+ cos2(7π/6 +α)) /2 =-cos(π/3-α) +1/2 + cos(7π/3 +2α) /2 =-√3/3 +1/2 + cos(π/3+2α)/2 =-√3/3 +1/2 - cos(2π/3-2α)/2 =-√3/3 +1/2 - cos2(π/3-α)/2 =-√3/3 +1/2 - [cos(π/3-α)]^2 +1/2 =-√3/3+1-(1/3) =-√3/3 +2/3
cos(2π/3+α)+cos2(7π/6+α)
=-cos(π-(2π/3+α)) + cos(7π/3 +2α)
=-cos(π/3-α) + cos(π/3 +2α)
=-√3/3 - cos(2π/3-2α)
=-√3/3 - cos2(π/3-α)
=-√3/3 - 2[cos(π/3-α)]^2 +1
=-√3/3 - 2(1/3) + 1
=-√3/3 + 1/3
再问： 2是平方
再答： cos(π/3-α) =√3/3cos(2π/3+α)+[cos(7π/6+α)]^2 =-cos(π-(2π/3+α)) + (1+ cos2(7π/6 +α)) /2 =-cos(π/3-α) +1/2 + cos(7π/3 +2α) /2 =-√3/3 +1/2 + cos(π/3+2α)/2 =-√3/3 +1/2 - cos(2π/3-2α)/2 =-√3/3 +1/2 - cos2(π/3-α)/2 =-√3/3 +1/2 - [cos(π/3-α)]^2 +1/2 =-√3/3+1-(1/3) =-√3/3 +2/3
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