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Câu hỏi: Cho \(\sin \alpha = \dfrac{8}{{17}},\sin \beta = \dfrac{{15}}{{17}}\)với \(0 < \alpha < \dfrac{\pi }{3}, 0 < \beta < \dfrac{\pi }{2}\). Chứng minh rằng \(\alpha + \beta = \dfrac{\pi }{2}\)
Lời giải chi tiết
Từ giải thiết suy ra \(\cos \alpha >0, \cos \beta >0\). Ta có:
\(\cos \alpha = \sqrt {1 - \dfrac{{64}}{{289}}} = \sqrt {\dfrac{{225}}{{289}}} = \dfrac{{15}}{{17}};\) \(\cos \beta = \sqrt {1 - \dfrac{{225}}{{289}}} = \sqrt {\dfrac{{64}}{{289}}} = \dfrac{8}{{17}}\).
Do đó
\(\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
=\(\dfrac{8}{{17}}.\dfrac{8}{{17}} + \dfrac{{15}}{{17}}.\dfrac{{15}}{{17}} = \dfrac{{289}}{{289}} = 1\).
Vì \(0 < \alpha < \dfrac{\pi }{2}, 0 < \beta < \dfrac{\pi }{2}\)nên từ đó suy ra \(\alpha + \beta = \dfrac{\pi }{2}\).
Từ giải thiết suy ra \(\cos \alpha >0, \cos \beta >0\). Ta có:
\(\cos \alpha = \sqrt {1 - \dfrac{{64}}{{289}}} = \sqrt {\dfrac{{225}}{{289}}} = \dfrac{{15}}{{17}};\) \(\cos \beta = \sqrt {1 - \dfrac{{225}}{{289}}} = \sqrt {\dfrac{{64}}{{289}}} = \dfrac{8}{{17}}\).
Do đó
\(\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
=\(\dfrac{8}{{17}}.\dfrac{8}{{17}} + \dfrac{{15}}{{17}}.\dfrac{{15}}{{17}} = \dfrac{{289}}{{289}} = 1\).
Vì \(0 < \alpha < \dfrac{\pi }{2}, 0 < \beta < \dfrac{\pi }{2}\)nên từ đó suy ra \(\alpha + \beta = \dfrac{\pi }{2}\).