在△ABC中,若sin²A + sin²B = sin C,且A,B都是锐角,求A+B=
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在△ABC中,若sin²A + sin²B = sin C,且A,B都是锐角,求A+B=
在△ABC中,若sin²A + sin²B = sin C,且A,B都是锐角,求A+B=
在△ABC中,若sin²A + sin²B = sin C,且A,B都是锐角,求A+B=
sin²A + sin²B = sin C
即sin²A + sin²B =sin(a+b)
由倍角公式,得(1-cos2a)/2+(1-cos2b)/2=sin(a+b)
即1-(cos2a+cos2b)/2=sin(a+b)
由和差化积,得1-cos(a+b)cos(a-b)=sin(a+b)
即cos(a+b)cos(a-b)+sin(a+b)=1
因为sin(a+b)=0,而cos(a-b)>0,故cos(a+b)>=0,
所以0<a+b<=90
sin(a+b)>=[sin(a+b)]^2,1=cos(a+b)cos(a-b)+sin(a+b)>=cos(a+b)cos(a-b)+[sin(a+b)]^2-1+1=cos(a+b)cos(a-b)-[cos(a+b)]^2+1,
即cos(a+b)cos(a-b)
应该是sin²A + sin²B = sin ²C
a/sinA=b/sinB=c/sinC
所以就是a²+b²=c²
所以A+B=90度
sin²A + sin²B = sin C
即sin²A + sin²B =sin(a+b)
由倍角公式,得(1-cos2a)/2+(1-cos2b)/2=sin(a+b)
即1-(cos2a+cos2b)/2=sin(a+b)
由和差化积,得1-cos(a+b)cos(a-b)=sin(a+b)
即cos(a+...
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sin²A + sin²B = sin C
即sin²A + sin²B =sin(a+b)
由倍角公式,得(1-cos2a)/2+(1-cos2b)/2=sin(a+b)
即1-(cos2a+cos2b)/2=sin(a+b)
由和差化积,得1-cos(a+b)cos(a-b)=sin(a+b)
即cos(a+b)cos(a-b)+sin(a+b)=1
因为sin(a+b)<=1,故cos(a+b)cos(a-b)>=0,而cos(a-b)>0,故cos(a+b)>=0,
所以0<a+b<=90
sin(a+b)>=[sin(a+b)]^2,1=cos(a+b)cos(a-b)+sin(a+b)>=cos(a+b)cos(a-b)+[sin(a+b)]^2-1+1=cos(a+b)cos(a-b)-[cos(a+b)]^2+1,
即cos(a+b)cos(a-b)<=[cos(a+b)]^2,cos(a+b)[cos(a-b)-cos(a+b)]<=0,
显然0<|a-b|cos(a+b),所以cos(a+b)<=0。
因cos(a+b)>=0,故cos(a+b)=0
∴a+b=90
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