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已知:α-β=π/3,证明 cos^2α+cos^2β+sinαsinβ为定值,谢谢
人气:210 ℃ 时间:2020-01-25 05:10:26
解答
cos²α+cos²β+sinαsinβ
=[1+cos(2α)]/2+[1+cos(2β)]/2+sinαsinβ
=1+[cos(2α)+cos(2β)]/2+sinαsinβ
=1+2cos(α+β)cos(α-β)/2 +sinαsinβ/运用了和差化积公式
=1+cos(α+β)cos(π/3)+sinαsinβ
=1+(1/2)cos(α+β)+sinαsinβ
=1+(1/2)(cosαcosβ-sinαsinβ)+sinαsinβ
=1+(1/2)(cosαcosβ+sinαsinβ)
=1+(1/2)cos(α-β)
=1+(1/2)cos(π/3)
=1+(1/2)(1/2)
=1+1/4
=5/4
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