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Câu hỏi: Chứng minh rằng
\({\cos ^2}(x - a) + {\sin ^2}(x - b) \)\(- 2\cos (x - a)\sin (x - b)\sin (a - b) \)\(= {\cos ^2}(a - b)\)
\({\cos ^2}(x - a) + {\sin ^2}(x - b) \)\(- 2\cos (x - a)\sin (x - b)\sin (a - b) \)\(= {\cos ^2}(a - b)\)
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
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