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求cos87°cos432°-sin93°cos198°的值
人气:298 ℃ 时间:2020-09-24 05:51:09
解答
cos87°cos432°-sin93°cos198°=cos87°cos(360°+72°)-sin(180°-87°)cos(180°+18°)=cos87°cos72°+sin87°cos18°=cos87°cos72°+sin87°sin72°=cos(87°-72°)=cos15°=(√6+√2)/4
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