若sinθ+cosθ=2分之1,求sin3θ-cos3θ之值為 數學作業！！謝謝?
回答 (2)
2016-09-16 2:31 am
Sol
a^3+b^3
=(a+b)^3－3ab(a+b)
=(a+b)^3－3(a+b)[(a+b)^2－(a^2+b^2)]/2
Sinθ+Cosθ=1/2
Sin^3 θ+Cos^3 θ
=(1/2)^3－3*(1/2)*[(1/2)^2-1]/2
=1/8-(3/2)*(－3/8)
=2/16+9/16
=11/16
Sin3θ=3Sinθ－4Sin^3 θ
cos3θ=4Cos^3 θ－3Cosθ
Sin3θ－Cos3θ
=3Sinθ－4Sin^3 θ－4Cos^3 θ+3Cosθ
=3(Sinθ+Cosθ)－4(Sin^3 θ+Cos^3 θ)
=3*(1/2)－4*(11/16)
=3/2－11/4
=－5/4
a^3+b^3
=(a+b)^3－3ab(a+b)
=(a+b)^3－3(a+b)[(a+b)^2－(a^2+b^2)]/2
Sinθ+Cosθ=1/2
Sin^3 θ+Cos^3 θ
=(1/2)^3－3*(1/2)*[(1/2)^2-1]/2
=1/8-(3/2)*(－3/8)
=2/16+9/16
=11/16
Sin3θ=3Sinθ－4Sin^3 θ
cos3θ=4Cos^3 θ－3Cosθ
Sin3θ－Cos3θ
=3Sinθ－4Sin^3 θ－4Cos^3 θ+3Cosθ
=3(Sinθ+Cosθ)－4(Sin^3 θ+Cos^3 θ)
=3*(1/2)－4*(11/16)
=3/2－11/4
=－5/4
2016-09-16 12:08 am
答案是負4分之5，求過程
收錄日期: 2021-04-30 21:51:40
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https://hk.answers.yahoo.com/question/index?qid=20160915160013AAE3d7b