\[\eqalign{& {\cos ^2}[x - a] + {\sin ^2}[x - b] \cr&= {{1 + \cos 2\left[ {x - a} \right]} \over 2} + {{1 - \cos 2\left[ {x - b} \right]} \over 2} \cr& = 1 + {1 \over 2}\left[ {\cos 2\left[ {x - a} \right] - \cos 2\left[ {x - b} \right]} \right] \cr&= 1 + \frac{1}{2}.\left[ { - 2} \right]\sin \left[ {2x - a - b} \right]\sin \left[ {b - a} \right] \cr&= 1 - \sin \left[ {2x - a - b} \right]\sin \left[ {b - a} \right]\cr&= 1 + \sin \left[ {2x - a - b} \right]\sin \left[ {a - b} \right] \cr} \]
Đề bài
Chứng minh rằng
\[{\cos ^2}[x - a] + {\sin ^2}[x - b] \]\[- 2\cos [x - a]\sin [x - b]\sin [a - b] \]\[= {\cos ^2}[a - b]\]
Lời giải chi tiết
Ta có:
\[\eqalign{
& {\cos ^2}[x - a] + {\sin ^2}[x - b] \cr&= {{1 + \cos 2\left[ {x - a} \right]} \over 2} + {{1 - \cos 2\left[ {x - b} \right]} \over 2} \cr
& = 1 + {1 \over 2}\left[ {\cos 2\left[ {x - a} \right] - \cos 2\left[ {x - b} \right]} \right] \cr&= 1 + \frac{1}{2}.\left[ { - 2} \right]\sin \left[ {2x - a - b} \right]\sin \left[ {b - a} \right] \cr&= 1 - \sin \left[ {2x - a - b} \right]\sin \left[ {b - a} \right]\cr&= 1 + \sin \left[ {2x - a - b} \right]\sin \left[ {a - b} \right] \cr} \]
Do đó