作业帮数学证明恒等式sin(a+b)cos(a-b)=sinacosa+sinbcosb
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数学证明恒等式
sin(a+b)cos(a-b)=sinacosa+sinbcosb
sin(a+b)cos(a-b)=sinacosa+sinbcosb
sin(a+b)cos(a-b)
=(sinacosb+cosasinb)(cosacosb+sinasinb)
=sinacosa(cosb)^2+(sina)^2sinbcosb+(cosa)^2sinbcosb+(sinb)^2sinacosa
=sinacosa[(cosb)^2+(sinb)^2]+sinbcosb[(sina)^2+(cosa)^2]
=sinacosa*1+sinbcosb*1
=sinacosa+sinbcos
=(sinacosb+cosasinb)(cosacosb+sinasinb)
=sinacosa(cosb)^2+(sina)^2sinbcosb+(cosa)^2sinbcosb+(sinb)^2sinacosa
=sinacosa[(cosb)^2+(sinb)^2]+sinbcosb[(sina)^2+(cosa)^2]
=sinacosa*1+sinbcosb*1
=sinacosa+sinbcos
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