tan[(a+b)/2]等于什么
1个回答
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答案1:按公式tan(A+B) = (tanA+tanB)/(1-tanAtanB)展开
tan[(a+b)/2]=tan(a/2+b/2)=(tana/2+tanb/2)/(1-tana/2tanb/2)
答案2:tan(A/2+B/2)=(sinA+sinB)/(cosA+cosB)
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)
tan[(a+b)/2]=tan(a/2+b/2)=(tana/2+tanb/2)/(1-tana/2tanb/2)
答案2:tan(A/2+B/2)=(sinA+sinB)/(cosA+cosB)
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)
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