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討論串[中學] 三角函數求解
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Of course you should let A=r*sin(t), B=r*cos(t). for some r and t (it is easy to find such r and t),. then. C = A*sin(d) + B*cos(d). = r*(sin(t)*sin(d
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B^2cos^2(d)=A^2sin^2(d)-2CAsin(d)+C^2. (A^2+B^2)sin^2(d)-2CAsin(d)-(B^2-C^2)=0. sin^2(d)-[2CA/(A^2+B^2)]sin(d)-[(B^2-C^2)/(A^2+B^2)]=0. {sin(d)-[CA/(A
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