三角函数
三角形ABC中,A=C+90,sinA+sinC=√2sinB,求角C.
因为A+B+C=180°,已知A=C+90° 所以:B=180°-(A+C)=180°-(2C+90°)=90°-2C 已知sinA+sinC=√2sinB ===> sin(C+90°)+sinC=√2sin(90°-2C) ===> cosC+sinC=√2cos2C ===> cosC+sinC=√2(cos^2 C-sin^2 C) ===> cosC+sinC=√2(cosC+sinC)*(cosC-sinC) ===> (cosC+sinC)*[√2(cosC-sinC)-1]=0 因为C是锐角,所以:sinC、cosC>0 所以,cosC+sinC>0 所以,cosC-sinC=1/√2=√2/2 ===> (cosC-sinC)^2=1/2 ===> cos^2 C+sin^2 C-2sinC*cosC=1/2 ===> 1-sin2C=1/2 ===> sin2C=1/2 ===> 2C=30° ===> C=15°。
A=C+90 B=180-A-C=180-(C+90)-C=90-2C sinA+sinC=√2sinB cosC+sinC=√2cos2C=√2[(cosC)^2-(sinC)^2] (1)cosC+sinC=0, C=135,不合题意,舍去。 (2)1=√2(cosC-sinC),2sin(C-45)=-1 sin(C-45)=-1/2 -45
答:tanAtanB =3 >0 ==> A,B sinAsinB =3/4, cosAcosB =1/4 cosC =cos(180-A-B) = sinAsi...详情>>
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