高一数学正弦余弦题
\u3010\u9ad8\u4e00\u6570\u5b66\u3011\u4e00\u9053\u6b63\u5f26\u4f59\u5f26\u4e09\u89d2\u5f62\u7684\u9898\u76ee\u300b\u300b\u300b\u6839\u636e\u4f59\u5f26\u5b9a\u7406\uff0ca\u5e73\u65b9=b\u5e73\u65b9+c\u5e73\u65b9-2bc*A\u7684\u4f59\u5f26\uff0c\u5373a^2=b^2+c^2-2bc*(1/2),\u4ee3\u51653c=4b,\u6c42\u5f97a\u4e0ec\u7684\u5173\u7cfb\u4e3a\uff1aa/c= (\u221a13)/4......I\u5f0f\uff0c\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u6839\u636e\u6b63\u5f26\u5b9a\u7406\uff1aa/sinA=c/sinC,\u6240\u4ee5a/c=sinA/sinC=(\u221a3)/2sinC......J\u5f0f\uff0c\u7efc\u5408I\u5f0f\u3001J\u5f0f\u53ef\u5f97\uff1asinC=(2\u221a39)/13
\u5229\u7528\u6b63\u5f26\u5b9a\u7406
sinB^2sinC^2+sinC^2sinB^2=2sinBsinCcosBcosC
2sinB^2sinC^2=2sinBsinCcosBcosC
sinBsinC-cosBcosC=0
cos(B+C)=0
B+C=90\u5ea6
\u76f4\u89d2\u4e09\u89d2\u5f62
两个式子分别平方然后相加就得出结果
解：两边平方
(3sinA＋4cosB)^2＝36
得9sin^2A +16cos^2B +24sinAcosB=36 ①
(4sinB＋3cosA)^2＝1
得16sin^2B +9cos^2A +24sinBcosA=1 ②
①+ ②
得：（9sin^2A +9cos^2A） +（16cos^2B+ 16sin^2B） +24sinAcosB+24sinBcosA=37
即 9+16+24sin(A+B)=37
所以sin(A+B)=1/2,
所以A+B=5π／6 或者π/6
若A+B=π／6，则cosA>√3/2
3cosA>3√3/2>1 ，则4sinB＋3cosA>1 舍去
所以A+B=5π／6
因为A+B+C=180
所以 C=π/6
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