Daily questions _191117

Radius \ (1 \) has three fixed points on the circle \ (A, B, C \) , then \ (\ overrightarrow {AB} \ cdot \ overrightarrow {AC} \) the minimum value of \ ((\ qquad ) \)
\ (\ mathrm {A} .-. 1 \) \ (\ qquad \ mathrm {B} .- \. 3 dfrac {} {}. 4 \) \ (\ qquad \ mathrm {C} .- \ {dfrac {2}}. 1 \) \ (\ qquad \ mathrm {D} .- \ dfrac. 1 {{}}. 4 \)
Analysis:
wish to set \ [A (0,1), B (\ cos \ alpha, \ sin \ alpha), C (\ cos \ beta, \ sin \ beta), \ alpha, \ beta \ in \ left [0,2 \ pi \ right). \] Thus\[ \begin{split} \overrightarrow{AB}\cdot\overrightarrow{AC}&=\left( \cos\alpha,\sin\alpha-1\right)\cdot \left(\cos\beta,\sin\beta-1\right)\\ &=\cos\alpha\cos\beta+\sin\alpha\sin\beta-\left(\sin\alpha+\sin\beta\right)+1\\ &=\cos\left(\alpha-\beta\right)-2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}+1\\ &=2\cos\dfrac{\alpha-\beta}{2}\cdot \left(\cos\dfrac{\alpha-\beta}{2}-\sin\dfrac{\alpha+\beta}{2}\right)\\ &\geqslant -2\cdot \left(\dfrac{1}{2}\sin\dfrac{\alpha+\beta}{2}\right)^2\\ &\geqslant -\dfrac{1}{2}. \end{split} \]
因此当\(\cos\dfrac{\alpha-\beta}{2}=\dfrac{1}{2}\sin\dfrac{\alpha+\beta}{2}\)且\(\sin\dfrac{\alpha+\beta}{2}=1\)时,上述不等式取等.取\[(\alpha,\beta)= \left(\dfrac{5\pi}{6},\dfrac{\pi}{6}\right).\]此时\ (\ overrightarrow {AB} \ cdot \ overrightarrow {AC} \) to obtain the minimum value \ (- \ dfrac. 1} {2} {\) .