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已知cos(π/3-α)=根号3/3,求cos(2π/3+α)+cos2(7π/6+α)
人气:478 ℃ 时间:2020-06-24 02:01:27
解答
cos(π/3-α)=√3/3cos(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是平方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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