高一数学。三角函数
\u9ad8\u4e00\u6570\u5b66 \u4e09\u89d2\u51fd\u6570tanA+tanB=2(log8 72 + log9 72)
=2(log8 9+log8 8 + log9 8+log9 9)
=2[(2/3)log2 3 +(3/2)log3 2 +2]
\u6240\u4ee5(sinAcosB+cosAsinB)/cosAcosB=sin(A+B)/cosAcosB
=2[(2/3)log2 3 +(3/2)log3 2 +2]------------\u2460
tanAtanB=-log8 72\u00b7log9 72
=-[(2/3)log2 3+1][(3/2)log3 2+1]
=-[(2/3)log2 3 +(3/2)log3 2 +2]
\u6240\u4ee5sinAsinB/cosAcosB
=-[(2/3)log2 3 +(3/2)log3 2 +2]---------\u2461
\u7531\u2460\u5f0f\u548c\u2461\u5f0f\u5f97(sinAcosB+cosAsinB)/cosAcosB=-2sinAsinB/cosAcosB
\u6240\u4ee5sinAcosB+cosAsinB+2sinAsinB=0
sin50\u00b0*(1+\u6839\u53f73\u00d7tan10\u00b0)
=sin50\u00b0*(1+\u6839\u53f73*sin10\u00b0/cos10\u00b0)
=sin50\u00b0/cos10\u00b0*(cos10\u00b0+\u6839\u53f73*sin10\u00b0)
=2*sin50\u00b0/cos10\u00b0*(1/2*cos10\u00b0+\u6839\u53f7 3/2*sin10\u00b0)
=2*sin50\u00b0/cos10\u00b0*(sin30\u00b0*cos10\u00b0
+cos30\u00b0*sin10\u00b0)
=(2*sin50\u00b0/cos10\u00b0)* sin40\u00b0
=(2*cos40\u00b0/cos10\u00b0)* sin40\u00b0
=(2* sin40\u00b0cos40\u00b0)/ cos10\u00b0
=sin80\u00b0/cos10\u00b0
=cos10\u00b0/cos10\u00b0
=1
=sin50[1+(sin60/cos60)*(sin10/cos10)]
=sin50[1+(sin60*sin10)/(cos60*cos10)]
=sin50[(sin10*sin60+cos10*cos60)/(cos10*cos60)]
=sin50[cos(60-10)/(cos10*cos60)]
=sin50[cos50/(cos10*cos60)]
=(sin50*cos50)/(cos10*cos60)
=sin100/(2*cos10*cos60)
=sin80/(2*cos10*cos60)
=cos10/(2*cos10*cos60)
=1/(2cos60)
=1
所以sinAcosC+sinCcosA=sinAcosC
所以sinCcosA=0 （sinC=0或cosA=0)
因为在三角形ABC中，C不可能等于0或180，则sinC不等于0
所以cosA=0 即A＝90 A为最大角
A所对的边为最长边12，设B或C中任意一个为最小角，此处设B为最小
在直角三角形ABC中，sinB=1/3＝b/12 所以b=4
用勾股定理或解直角三角形都可以算出c＝8根号2
所以这个三角形是直角三角形
面积为 1/2*b*c=16根号2
sin(A+C)-sinC
=sin[180-(A+C)]-sinC
=sinB-sinC [证毕]
sin(A+C)-sinC
=2cos[(A+C+C)/2]sin[(A+C-C)/2]
=2cos[A/2 + C)]sin(A/2)
=2[cos(A/2)cosC-sin(A/2)sinC]sin(A/2)
=2sin(A/2)cos(A/2)cosC
-2sin(A/2)sin(A/2)sinC
=sinAcosC-2{[sin(A/2)]^2}sinC
=sinAcosC-[1-cosA]sinC
请楼主核实一下，题目是否有错？
sin(A+C)-sinC=sinB-sinC是恒等式因为sin(A+C)=sin(180-B)=sinB
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