大师作文网求解数学三角函数证明题
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求解数学三角函数证明题
tg(a-c)/tga=[sin(a-c)cosa]/[cos(a-c)sina]
所以[sin(a-c)cosa]/[cos(a-c)sina]+[sinbsinb]/[sinasina]=1
两边各乘以cos(a-c)sinasina,得到sin(a-c)sinacosa+sinbsinbcos(a-c)=cos(a-c)sinasina
sinacosa=(1/2)sin2a,sinasina=(1/2)-cos2a/2
所以sinbsinbcos(a-c)=(1/2)cos(a-c)-(1/2)cos(a-c)cos2a-(1/2)sin(a-c)sin2a
=(1/2)cos(a-c)-(1/2)[cos(a-c)cos2a+sin(a-c)sin2a]
=(1/2)cos(a-c)-(1/2)cos[2a-(a-c)]=(1/2)[cos(a-c)-cos(a+c)]=sinasinc
tgbtgb=sinbsinb/(1-sinbsinb)=sinasinc/(cosacosc)=tgatgc
所以[sin(a-c)cosa]/[cos(a-c)sina]+[sinbsinb]/[sinasina]=1
两边各乘以cos(a-c)sinasina,得到sin(a-c)sinacosa+sinbsinbcos(a-c)=cos(a-c)sinasina
sinacosa=(1/2)sin2a,sinasina=(1/2)-cos2a/2
所以sinbsinbcos(a-c)=(1/2)cos(a-c)-(1/2)cos(a-c)cos2a-(1/2)sin(a-c)sin2a
=(1/2)cos(a-c)-(1/2)[cos(a-c)cos2a+sin(a-c)sin2a]
=(1/2)cos(a-c)-(1/2)cos[2a-(a-c)]=(1/2)[cos(a-c)-cos(a+c)]=sinasinc
tgbtgb=sinbsinb/(1-sinbsinb)=sinasinc/(cosacosc)=tgatgc