求 (sinA+sinB)/(cosA+cosB)=tan(A/2+B/2) 详细化简过程!
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求 (sinA+sinB)/(cosA+cosB)=tan(A/2+B/2) 详细化简过程!
A B 均为锐角
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A B 均为锐角
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sinA+sinB
=sin((A+B)/2+(A-B)/2)+sin((A+B)/2-(A-B)/2)
=sin(A+B)/2 *cos(A-B)/2
cosA+cosB
=cos((A+B)/2+(A-B)/2)+cos((A+B)/2-(A-B)/2)
=cos(A+B)/2 *cos(A-B)/2
(sinA+sinB)/(cosA+cosB)
=[sin(A+B)/2 *cos(A-B)/2 ]/[cos(A+B)/2 *cos(A-B)/2]
=[sin(A+B)/2]/[cos(A+B)/2]
=tan(A/2+B/2)
=sin((A+B)/2+(A-B)/2)+sin((A+B)/2-(A-B)/2)
=sin(A+B)/2 *cos(A-B)/2
cosA+cosB
=cos((A+B)/2+(A-B)/2)+cos((A+B)/2-(A-B)/2)
=cos(A+B)/2 *cos(A-B)/2
(sinA+sinB)/(cosA+cosB)
=[sin(A+B)/2 *cos(A-B)/2 ]/[cos(A+B)/2 *cos(A-B)/2]
=[sin(A+B)/2]/[cos(A+B)/2]
=tan(A/2+B/2)
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