求教两道高中解三角形的题目
\u9ad8\u4e2d\u89e3\u4e09\u89d2\u5f62\u9898\u76ee \u5728\u7ebf\u7b49 \u6c42\u89e3\u7b54\u6839\u636e\u6b63\u5f26\u5b9a\u7406
sinA/a=sinC/c
A=2C
\u6240\u4ee5sin2C/a=sinC/c
2cosCsinC/a=sinC/c
cosC=2a/c
\u6839\u636e\u4f59\u5f26\u5b9a\u7406
c^2=a^2+b^2-2abcosC
\u4ee3\u5165a^2-c^2=64/5 b=4
\u8ba1\u7b97\u5373\u53ef\u5f97\u7b54\u6848
A=60\u5ea6\uff0c\u5219B+C=120\u5ea6\uff0c
(cosB)^2+(cosC)^2
=(1+cos2B)/2+(1+cos2C)/2
=(cos2B+cos2C)/2+1
=1/2*[cos2B+cos(240-2B)]+1
=1/2*[cos2B-1/2*cos2B-\u221a3/2*sin2B]+1
=1/2*[1/2*cos2B-\u221a3/2*sin2B]+1
=1/2*cos(2B+60)+1
\u7531 0<B<120 \u5f97 0<2B<240 \uff0c60<2B+60<300 \uff0c
\u56e0\u6b64 -1<=cos(2B+60)<1/2 \uff0c
\u6240\u4ee5 \u6240\u6c42\u8303\u56f4\u662f [1/2\uff0c5/4) \u3002
设角A对应的边长为a（由已知，a=2√3），角B对应的边长为b，角C对应的边长为c，根据正弦定理，
sinA/a=sinB/b=sinC/c
所以，b=a* sinB/sinA=2√3 * sinx/sin(π/3)=4sinx
c=a* sinC/sinA=2√3 * sin[2π/3 - x]/sin(π/3)=4sin[2π/3 - x]=4sin(2π/3) cosx-4cos(2π/3) sinx=2√3 cosx+2sinx
故y=f(x)=a+b+c=2√3 + 4sinx +2√3 cosx+2sinx
=6sinx +2√3 cosx +2√3 =4√3(√3/2 sinx + 1/2 cosx)+2√3=4√3sin(x+π/6)+2√3
由于角ABC均需满足大于0且小于π，所以定义域为(0,2π/3)
（2）x+π/6∈(π/6,5π/6)
所以sin(x+π/6)的最大值为1，此时x+π/6=π/2，即x=π/3
ymax=f(π/3)=4√3+2√3=6√3
========================================================
第二题，先从证明结论入手反推，
acos(θ-B)+bcos(θ+A)=ccosθ，利用和差化积公式，
acosθcosB+asinθsinB+bcosθcosA-bsinθsinA=ccosθ，移项，整理得
(acosB+bcosA-c)cosθ + (asinB-bsinA)sinθ=0
那么欲证acos(θ-B)+bcos(θ+A)=ccosθ成立，只需证明(acosB+bcosA-c)cosθ + (asinB-bsinA)sinθ=0成立。
根据正弦定理可设：a/sinA=b/sinB=c/sinC=k，k≠0
那么，a=ksinA，b=ksinB，c=ksinC
代入上式，有(ksinAcosB+ksinBcosA-ksinC)cosθ + (ksinAsinB-ksinBsinA)sinθ=0
整理可得
k[sin(A+B)-sinC]cosθ=0
由于A+B+C=π，所以C=π-（A+B）所以sin（A+B）=sinC
所以k[sin(A+B)-sinC]cosθ=0成立，进而(acosB+bcosA-c)cosθ + (asinB-bsinA)sinθ=0成立，
故而acos(θ-B)+bcos(θ+A)=ccosθ成立
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