2.
阅读下面材料:
根据两角和与差的正弦公式,有
\(\sin (α+β)=\sin α\cos β+\cos α\sin β\),\(①\)
\(\sin (α-β)=\sin α\cos β+\cos α\sin β\),\(②\)
由\(①+②\)得\(\sin (α+β)+\sin (α-β)=2\sin α\cos β\),\(③\)
令\(α+β=A\),\(α-β=B\),有\(\alpha =\dfrac{A+B}{2}\),\(\beta =\dfrac{A-B}{2}\),
代入\(③\)得\(\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}\).
\((1)\)类比上述推理方法,根据两角和与差的余弦公式,证明:\(\cos A-\cos B=-2\sin \dfrac{A+B}{2}\sin \dfrac{A-B}{2}\);
\((2)\)若\(\triangle ABC\)的三个内角\(A\),\(B\),\(C\)满足\(\cos 2A-\cos 2B=1-\cos 2C\),试判断\(\triangle ABC\)的形状.