急急急！高一数学题 求详细过程 可用正弦余弦定理

\u6025\u6025\u6025\uff01\u9ad8\u4e00\u6570\u5b66\u9898 \u6c42\u8be6\u7ec6\u8fc7\u7a0b \u53ef\u7528\u6b63\u5f26\u4f59\u5f26\u5b9a\u7406

1\u3001\u2460|m|=1\uff0c|n|=1\uff0c\u5219m*n=cos²(C/2)\uff0dsin²(C/2)=cosC=|m||n|cos60\u00b0=1/2\uff0c\u6240\u4ee5C=60\u00b0\uff1b\u2461S=(1/2)absinC=(3\u221a3)/2\uff0c\u89e3\u51faab=6\uff0c\u518d\u9488\u5bf9\u89d2C\u7528\u4f59\u5f26\u5b9a\u7406\uff0c\u6709C²=a²\uff0bb²\uff0d2abcosC=a²\uff0bb²\uff0dab=(a\uff0bb)²\uff0d3ab\uff0c\u6240\u4ee5(7/2)²=(a\uff0bb)²\uff0d18\uff0c\u89e3\u5f97a\uff0bb=11/2\u3002
2\u3001\u2460sinC\uff1asinA=c\uff1aa\uff0c\u6240\u4ee5c\uff1aa=2\uff1a1\uff0c\u6240\u4ee5AB=c=2a=2\u221a5\u3002\u2461\u5229\u7528\u4f59\u5f26\u5b9a\u7406\uff0c\u6709cosA=2\u221a5/5\uff0c\u6240\u4ee5sinA=\u221a5/5\uff0c\u6240\u4ee5cos2A=2cos²A\uff0d1=3/5\uff0csin2A=2sinAcosA=4/5\uff0csin(2A\uff0d\u03c0/4)=(\u221a2/2)(4/5\uff0d3/5)=\u221a2/10\u3002

\u2460\u89e3\uff1a
\u7531A\uff0bB\uff0bC\uff1d180\u5ea6\uff0cA+C=2B\uff0c\u53ef\u5f97
3B\uff1d180\u5ea6
B\uff1d60\u5ea6
\u6839\u636e\u6b63\u5f26\u5b9a\u7406\uff0c\u53ef\u5f97
a/sinA=b/sinB
1/sinA=\u221a3/sin60
sinA=\u221a3/2/\u221a3=1/2

\u2461\u89e3\uff1a
cos(A-C)+cosB
=cos(A-C)-cos(A+C)
=cosAcosC+sinAsinC-cosAcosC+sinAsinC
=2sinAsinC=3/2
sinAsinC=3/4
\u6839\u636e\u6b63\u5f26\u5b9a\u7406\uff0ca/sinA=b/sinB=c/sinC=2R
b^2=4R^2sin^2B
a=2RsinA
c=2RsinC
\u56e0\u4e3ab^2=ac
\u6240\u4ee5\uff0csin^2B=sinAsinC=3/4
\u56e0\u4e3aB<180\u00b0
\u6240\u4ee5\uff0csinB=\u221a3/2
B=60\u00b0\u6216120\u00b0
\u4f46\u662f\u5f53B=120 \u00b0\u65f6 cosB=-1/2
cos(A-C)-1/2=3/2
cos(A-C)=2(\u4e0d\u6210\u7acb)
\u6240\u4ee5\uff0cB=60\u00b0

解：一，①cosA=(b^2+c^2-a^2)/2bc
3b^2+3c^2-3a^2=（4根号2）bc
(b^2+c^2-a^2)/2bc=（2根号2）/3
cosA=（2根号2）/3
sinA=1/3
②2sin(A+π/4)sin(B+C+π/4)/（1-cos2A)
=2sin(A+π/4)sin(π-A+π/4)/（2sin^2A)
=sin(A+π/4)sin(π-A+π/4)/（sin^2A)
=sin(A+π/4)sin(A-π/4)/（sin^2A)
=1/2(cos^2A-sin^2A)/sin^2A
=1/2(8/9-1/9)/1/9
=7/2
二，（1）cos2A=3/5，cos2A=1-2（sinA）^2=3/5，sinA=（根号5）/5，cosA=（2根号5）/5，sinB=（根号10）/10，cosB=（3根号10）/10，sin(A+B)=sinAcosB+cosAsinB=(根号2)/2，因为2A为锐角，所以0<∠A<45°，∠B为锐角，所以A+B不会为135°，则A+B=45°。
（2）a/sinA=b/sinB，又有a-b=（根号2）-1，所以a=根号2，b=1，又有a/sinA=c/sinC，A+B=45°，则C=135°，代入算得c=根号5。

直接在卷子上面写百度一下 你就知道 绝对满分

①cosA=(b^2+c^2-a^2)/2bc
3b^2+3c^2-3a^2=四倍根号二倍的bc
(b^2+c^2-a^2)/2bc=2根号2/3
cosA=2根号2/3
sinA=1/3
②2sin(A+π/4)sin(B+C+π/4)除以（1-cos2A)
=2sin(A+π/4)sin(π-A+π/4)除以（2sin^2A)
=sin(A+π/4)sin(π-A+π/4)除以（sin^2A)
=sin(A+π/4)sin(A-π/4)除以（sin^2A)
=1/2(cos^2A-sin^2A)/sin^2A
=1/2(8/9-1/9)/1/9
=7/2

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