cos²5π/12+cos²π/12+5π/12+cos5π/12×cosπ/12

zimeia2022-10-04 11:39:541条回答

cos²5π/12+cos²π/12+5π/12+cos5π/12×cosπ/12

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gxliuliu 共回答了16个问题 | 采纳率93.8%: cos²5π/12+cos²(π/12+5π/12)+cos5π/12×cosπ/12
=cos²(3π/12+2π/12)+cos²π/2+cos(3π/12+2π/12)×cos(3π/12-2π/12)
=cos²(π/4+π/6)+cos(π/4+π/6)×cos(π/4-π/6)
=cos(π/4+π/6)[cos(π/4+π/6)+cos(π/4-π/6)]
=(√6-√2)/4[(√6-√2)/4+(√6+√2)/4]
=(√6-√2)/4x(√6/2)
=(6-2√3)/8
=(3-√3)/4; 1年前

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cos²5π/12+cos²π/12+cosπ/12*cos5π/12
由于cos5π/12=sin（π/2-5π/12）=sinπ/12
原式=sin²π/12+cos²π/12+cosπ/12sinπ/12
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