优质解答
1、A,B,C为三角形内角,已知1+cos2A-cos2B-cos2C=2sinBsinC,求角A
1+cos2A-cos2B-cos2C=2sinBsinC
2cos²A-1-2cos²B+1+2sin²C=2sinBsinC
cos²A-cos²B+sin² (A+B)=sinBsinC
cos²A-cos²B+sin²Acos²B+2sinAcosAsinBcosB+cos²Asin²B=sinBsinC
cos²A-cos²Acos²B+2sinAcosAsinBcosB+cos²Asin²B=sinBsinC
2cos²AsinB+2sinAcosAcosB=sin(180-A-B)
2cosA(cosAsinB+sinAcosB)-sin(A+B)=0
Sin(A+B)(2cosA-1)=0
cosA=1/2
A=60
在△ABC中,sinB*sinC=cos²(A/2),则△ABC的形状是?
sinBsin(180-A-B)=(1+cosA)/2
2sinBsin(A+B)=1+cosA
2sinB(sinAcosB+cosAsinB)=1+cosA
sin2BsinA+2cosAsin²B-cosA-1=0
sin2BsinA+cosA(2sin²B-1)=1
sin2BsinA-cosAcos2B=1
cos2BcosA-sin2BsinA=-1
cos(2B+A)=-1
因为A,B是三角形内角
2B+A=180
因为A+B+C=180
所以B=C
三角形ABC是等腰三角形
计算cos20°-cos40°+cos60°+cos100°
=cos20°-cos40°-cos80°+1/2
=cos20°-cos40°-cos(20°+60°)+1/2
=cos20°-cos40°+sin20°sin60°-cos20°cos60°+1/2
=cos20°-cos40°-1/2cos20°+√3/2sin20°+1/2
=1/2cos20°+√3/2sin20°-cos40°+1/2
=cos(60°-20°)-cos40°+1/2
=cos40°-cos40°+1/2
=1/2
认为可以的话,发给你!
1、A,B,C为三角形内角,已知1+cos2A-cos2B-cos2C=2sinBsinC,求角A
1+cos2A-cos2B-cos2C=2sinBsinC
2cos²A-1-2cos²B+1+2sin²C=2sinBsinC
cos²A-cos²B+sin² (A+B)=sinBsinC
cos²A-cos²B+sin²Acos²B+2sinAcosAsinBcosB+cos²Asin²B=sinBsinC
cos²A-cos²Acos²B+2sinAcosAsinBcosB+cos²Asin²B=sinBsinC
2cos²AsinB+2sinAcosAcosB=sin(180-A-B)
2cosA(cosAsinB+sinAcosB)-sin(A+B)=0
Sin(A+B)(2cosA-1)=0
cosA=1/2
A=60
在△ABC中,sinB*sinC=cos²(A/2),则△ABC的形状是?
sinBsin(180-A-B)=(1+cosA)/2
2sinBsin(A+B)=1+cosA
2sinB(sinAcosB+cosAsinB)=1+cosA
sin2BsinA+2cosAsin²B-cosA-1=0
sin2BsinA+cosA(2sin²B-1)=1
sin2BsinA-cosAcos2B=1
cos2BcosA-sin2BsinA=-1
cos(2B+A)=-1
因为A,B是三角形内角
2B+A=180
因为A+B+C=180
所以B=C
三角形ABC是等腰三角形
计算cos20°-cos40°+cos60°+cos100°
=cos20°-cos40°-cos80°+1/2
=cos20°-cos40°-cos(20°+60°)+1/2
=cos20°-cos40°+sin20°sin60°-cos20°cos60°+1/2
=cos20°-cos40°-1/2cos20°+√3/2sin20°+1/2
=1/2cos20°+√3/2sin20°-cos40°+1/2
=cos(60°-20°)-cos40°+1/2
=cos40°-cos40°+1/2
=1/2
认为可以的话,发给你!