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求值cosπ/9cos2π/9cos4π/9还有一题：2·（根号下1+sin4）-（根号下2-2cos4）
题目详情
求值 cosπ/9cos2π/9cos4π/9
还有一题：
2·（根号下1+sin4）-（根号下2-2cos4）
还有一题：
2·（根号下1+sin4）-（根号下2-2cos4）
▼优质解答
答案和解析
1.原式=sin(π/9)*[cos(π/9)cos(2π/9)cos(4π/9)]/sin(π/9)
=1/2*[sin(2π/9)cos(2π/9)cos(4π/9)]/sin(π/9)
=1/8cos(8π/9)/sin(π/9)
=1/8
2.原式=2*根号下(1+2*sin2cos2)-根号下[2-2*(1-2(sin2)^2)]
=2*根号下(sin2+cos2)^2-根号下[4(sin2)^2)]
=2*(sin2+cos2)-2sin2
=2cos2
=1/2*[sin(2π/9)cos(2π/9)cos(4π/9)]/sin(π/9)
=1/8cos(8π/9)/sin(π/9)
=1/8
2.原式=2*根号下(1+2*sin2cos2)-根号下[2-2*(1-2(sin2)^2)]
=2*根号下(sin2+cos2)^2-根号下[4(sin2)^2)]
=2*(sin2+cos2)-2sin2
=2cos2
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