求证恒等式sin(A+B)sin(A-B)=sin²A-sin²B

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求证恒等式sin(A+B)sin(A-B)=sin²A-sin²B
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求证恒等式sin(A+B)sin(A-B)=sin²A-sin²B
求证恒等式sin(A+B)sin(A-B)=sin²A-sin²B

求证恒等式sin(A+B)sin(A-B)=sin²A-sin²B
sinαsinβ=-[cos(α+β)-cos(α-β)]/2  积化和差后简单

(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)=
sin²Acos²B-cos²Asin²B=
sin²A(1-sin²B)-(1-sin²A)sin²B=
sin²A-sin²B

a/sinA=b/sinB=c/sinC=2R=2√2 =>a=2RsinA,b=2RsinB,c=2RsinC 2√2(sin 2;A-sin 2;C)=(a-b)sinB =>4R 2;(sin 2;A-sin 2;C