三角形ABC所对的三边长为abc cos=1/4求sin方[(B+C)/2]+cos2A求a=4 ,b+c=6 且b
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三角形ABC所对的三边长为abc cos=1/4求sin方[(B+C)/2]+cos2A求a=4 ,b+c=6 且b
三角形ABC所对的三边长为abc cos=1/4
求sin方[(B+C)/2]+cos2A
求a=4 ,b+c=6 且b
三角形ABC所对的三边长为abc cos=1/4求sin方[(B+C)/2]+cos2A求a=4 ,b+c=6 且b
三角形ABC所对的三边长为abc cos=1/4
求sin方[(B+C)/2]+cos2A
求a=4 ,b+c=6 且b
cosA=1/4
A+B+C=180
(B+C)/2=180-A/2
1.
sin^2[(B+C)/2]+cos2A=cos^2A/2+cos2A
=(cosA+1)/2+2cos^2A-1
=-1/4
2.
余弦定理
a^2=b^2+c^2-2bccosA
(b+c)^2=b^2+c^2+2bc
a^2=36-2bc-bc/2
bc=8
(1)cosA=(b^2+c^2-a^2)/2bc=1/4
sin方[(B+C)/2]+cos2A=sin^2[(180-A)/2]+cos2A=sin^2(90-A/2)+cos2A
=(1-cos[2(90-1/2A)])/2+cos2A
cos2A=2cos^2A-1=-7/8
代入即可
(2)cosA=[(b+c)^2-2bc-a^2]/2bc ...
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(1)cosA=(b^2+c^2-a^2)/2bc=1/4
sin方[(B+C)/2]+cos2A=sin^2[(180-A)/2]+cos2A=sin^2(90-A/2)+cos2A
=(1-cos[2(90-1/2A)])/2+cos2A
cos2A=2cos^2A-1=-7/8
代入即可
(2)cosA=[(b+c)^2-2bc-a^2]/2bc 将条件带入就可以了
做数学题目讲究的是思路,有思路题目就做出来了
顺便再告诉你,一般的每一句话都是条件,把条件转化为式子就可以了
收起
cos
=>
cosA=1/4
A+B+C=180
(B+C)/2=180-A/2
1.
sin^2[(B+C)/2]+cos2A=cos^2A/2+cos2A
=(cosA+1)/2+2cos^2A-1
=-1/4
2.
余弦定理
a^2=b^2+c^2-2bccosA
(b+c)^2=b^2+c^2+2bc
a^2=36-2bc-bc/2
bc=8
1.sin^2【(B+C)/2】+cos2A=(1-cos(π-A))/2+2cos^2A-1,下边自己算
2.a^2=b^2+c^2-2bccosA=〖(b+c)〗^2-2bc(1-cosA),下边自己算