三角形中角A的余弦值与线段长度范围问题

(1)将已知条件改写为$\frac{b}{a}\cdot\frac{\sin B}{\sin A}+\frac{c}{a}\cdot\frac{\sin C}{\sin A}=1-2\frac{b}{a}\cdot\sin C$，再利用正弦定理将$\frac{b}{a}$和$\frac{c}{a}$表示为$\sin B$和$\sin C$的函数，得到

$\frac{\sin B}{\sin A}\cdot\frac{\sin^2 C}{\sin B\sin C}+\frac{\sin C}{\sin A}\cdot\frac{\sin^2 B}{\sin B\sin C}=1-2\frac{\sin B}{\sin A}\cdot\sin C$

$\Rightarrow\frac{\sin C}{\sin A}\sin B+\frac{\sin B}{\sin A}\sin C=2\sin B\sin C-\sin A\sin B$

$\Rightarrow\sin(A+B)\sin C=\sin A\sin B(2\cos C-1)$

$\Rightarrow\sin A=\frac{\sin B\sin C}{\sin(A+B)\cos C-\cos(A+B)\sin C}$

由余弦定理可知$\cos A=\frac{b^2+c^2-a^2}{2bc}$，代入上式化简得

$\sin A=\frac{2bc\sin B\sin C}{b^2+c^2-a^2+2bc\cos A}=\frac{2bc\sin B\sin C}{(b+c)^2-a^2}$

$\Rightarrow\cos A=\frac{(b+c)^2-a^2-4b^2\sin^2 C-4c^2\sin^2 B}{2bc}$

(2)根据余弦定理和中线定理，有

$b^2+c^2-2bc\cos A=2(\frac{a^2}{2}-\frac{b^2}{4})+2(\frac{a^2}{2}-\frac{c^2}{4})$

$\Rightarrow b^2+c^2-2bc\cos A=a^2-b^2-c^2$

$\Rightarrow b^2+c^2-2bc\cos A=2bc\cos A-2b^2\sin^2 C-2c^2\sin^2 B$

$\Rightarrow b^2+2bc\cos A+c^2=2bc\cos A+2b^2\sin^2 C+2c^2\sin^2 B$

$\Rightarrow (b+c\cos A)^2=b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C+2$

$\Rightarrow b+\sqrt{2}c\cos A\in[\sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C},\sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C+2}]$

注意到$b\sin C+c\sin B\leq b+c$，因此

$b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C\leq (b\sin C+c\sin B)^2\leq (b+c)^2$

$\Rightarrow \sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C}\leq b+c$

$\Rightarrow b+\sqrt{2}c\cos A\in[\sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C},b+c\sqrt{2}]$