已知sin(α+β)sin(α-β)=m,求cos^2α-cos^2β
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已知sin(α+β)sin(α-β)=m,求cos^2α-cos^2β
解答:sin(α+β)sin(α-β)=m=sin^2α-sin^2β
cos^2α-cos^2β=(1-sin^2α)-(1-sin^2β)=sin^2α-sin^2β=m
感谢网友批评指正 弄错了一个符号嘿嘿 回答了几十个问题 眼花了~~
修正解答:sin(α+β)sin(α-β)=m=sin^2α-sin^2β
cos^2α-cos^2β=(1-sin^2α)-(1-sin^2β)=-sin^2α+sin^2β=-m
补充:sin(α+β)sin(α-β)=(sinacosb+cosasinb)(sinacosb-cosasinb)
=(sinacosb)^2-(cosasinb)^2=sin^2α-sin^2β
解答:sin(α+β)sin(α-β)=m=sin^2α-sin^2β
cos^2α-cos^2β=(1-sin^2α)-(1-sin^2β)=sin^2α-sin^2β=m
感谢网友批评指正 弄错了一个符号嘿嘿 回答了几十个问题 眼花了~~
修正解答:sin(α+β)sin(α-β)=m=sin^2α-sin^2β
cos^2α-cos^2β=(1-sin^2α)-(1-sin^2β)=-sin^2α+sin^2β=-m
补充:sin(α+β)sin(α-β)=(sinacosb+cosasinb)(sinacosb-cosasinb)
=(sinacosb)^2-(cosasinb)^2=sin^2α-sin^2β
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