高中三角证明题

求复合三角函数y=(sinA+s

我的方法比较粗俗，证得不太漂亮，还望见谅 首先sinA+sinB+sinC=2sin(A/2+B/2)cos(A/2-B/2)+sin(π-A-B) =2sin(A/2+B/2)cos(A/2-B/2)+sin(A+B) =2sin(A/2+B/2)cos(A/2-B/2)+2sin(A/2+B/2)cos(A/2+B/2) =2sin(A/2+B/2)[cos(A/2-B/2)+cos(A/2+B/2)] =2sin(π/2-C/2)*2cos(A/2)cos(B/2) =4cos(A/2)cos(B/2)cos(C/2) 其次cosA+cosB+cosC=2cos(A/2+B/2)co...全部

  我的方法比较粗俗，证得不太漂亮，还望见谅 首先sinA+sinB+sinC=2sin(A/2+B/2)cos(A/2-B/2)+sin(π-A-B) =2sin(A/2+B/2)cos(A/2-B/2)+sin(A+B) =2sin(A/2+B/2)cos(A/2-B/2)+2sin(A/2+B/2)cos(A/2+B/2) =2sin(A/2+B/2)[cos(A/2-B/2)+cos(A/2+B/2)] =2sin(π/2-C/2)*2cos(A/2)cos(B/2) =4cos(A/2)cos(B/2)cos(C/2) 其次cosA+cosB+cosC=2cos(A/2+B/2)cos(A/2-B/2)+cos(π-A-B) =2cos(A/2+B/2)cos(A/2-B/2)-cos(A+B) =2cos(A/2+B/2)cos(A/2-B/2)-2cos²(A/2+B/2)+1 =2cos(A/2+B/2)[cos(A/2-B/2)-cos(A/2+B/2)]+1 =2cos(π/2-C/2)*2sin(A/2)sin(B/2)+1 =4sin(A/2)sin(B/2)sin(C/2)+1 于是y=(sinA+sinB+sinC)/(cosA+cosB+cosC)=4cos(A/2)cos(B/2)cos(C/2)/[4sin(A/2)sin(B/2)sin(C/2)+1] 考虑到00,cos(B/2)>0,cos(C/2)>0 sin(A/2)>0,sin(B/2)>0,sin(C/2)>0 则y=4cos(A/2)cos(B/2)cos(C/2)/[4sin(A/2)sin(B/2)sin(C/2)+1]>0，于是函数y的下界就确定了 由于cos(A/2)>0,cos(B/2)>0,cos(C/2)>0，我们把cos(A/2)cos(B/2)cos(C/2)除到分母上去就得到： y=4/[4tan(A/2)tan(B/2)tan(C/2)+sec(A/2)sec(B/2)sec(C/2)] 我要利用三角恒等式： tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1 下面做出证明： 左=tan(A/2)tan(B/2)+tan(B/2)tan(π/2-A/2-B/2)+tan(A/2)tan(π/2-A/2-B/2)=tan(A/2)tan(B/2)+tan(B/2)cot(A/2+B/2)+ tan(A/2)cot(A/2+B/2)=tan(A/2)tan(B/2)+[tan(A/2)+tan(B/2)]/tan(A/2+B/2)=tan(A/2)tan(B/2)+1-tan(A/2)tan(B/2)=1=右 现在我们令u=tan(A/2),v=tan(B/2),w=tan(C/2) 则uv+vw+wu=1，u>0,v>0,w>0 又由于sec²x=1+tan²x,特别地，当00不等号严格取到 故分母>2,此时y<4/2=2,2就是所要求的函数y的上界 综上所述：y的取值范围是(0,2)。
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