Chứng minh rằng: - bài 46 trang 215 sgk đại số 10 nâng cao
\(\eqalign{& \sin \alpha \sin ({\pi \over 3} - \alpha )\sin ({\pi \over 3} + \alpha ) \cr&= \sin \alpha .\frac{1}{2}\left[ {\cos \left( {\frac{\pi }{3} - \alpha - \frac{\pi }{3} - \alpha } \right) - \cos \left( {\frac{\pi }{3} - \alpha + \frac{\pi }{3} + \alpha } \right)} \right]\cr &= sin\alpha .{1 \over 2}(cos2\alpha - \cos {{2\pi } \over 3}) \cr& = {1 \over 2}\sin \alpha (1 - 2{\sin ^2}\alpha + {1 \over 2}) \cr &= {1 \over 4}\sin \alpha (3 - 4{\sin ^2}\alpha ) \cr& = {1 \over 4}\sin 3\alpha \cr& \cos \alpha \cos ({\pi \over 3} - \alpha )cos({\pi \over 3} + \alpha ) \cr&= \cos \alpha .\frac{1}{2}\left[ {\cos \left( {\frac{\pi }{3} - \alpha + \frac{\pi }{3} + \alpha } \right) + \cos \left( {\frac{\pi }{3} - \alpha - \frac{\pi }{3} - \alpha } \right)} \right]\cr &= \cos \alpha .{1 \over 2}(\cos 2\alpha + \cos {{2\pi } \over 3}) \cr& = {1 \over 2}\cos \alpha (2{\cos ^2}\alpha - 1 - {1 \over 2}) \cr&= {1 \over 4}\cos \alpha (4{\cos ^2}\alpha - 3) \cr &= {1 \over 4}\cos 3\alpha \cr} \)
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Chứng minh rằng: LG a \(sin3α = 3sinα 4si{n^3}\alpha \); \( cos3α =4co{s^3}\alpha 3cosα\) Lời giải chi tiết: Ta có: \(sin3α = sin (2α + α) \) \(= sin 2α cosα + sinα cos 2α\) \( = {\rm{ }}2{\rm{ }}sin\alpha {\rm{ }}co{s^2}\alpha {\rm{ }} + {\rm{ }}sin\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}2si{n^2}\alpha )\) \(= {\rm{ }}2sin\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}si{n^2}\alpha ){\rm{ }} + {\rm{ }}sin\alpha (1{\rm{ }}-{\rm{ }}si{n^2}\alpha ){\rm{ }}\) \(= {\rm{ }}3sin\alpha {\rm{ }}-{\rm{ }}4si{n^3}\alpha \) \(cos3α = cos (2α + α) \) \(= cos 2α cosα - sin2α sinα\) \(= {\rm{ }}(2co{s^2}\alpha {\rm{ }}-{\rm{ }}1)cos\alpha {\rm{ }}-{\rm{ }}2si{n^2}\alpha {\rm{ }}cos\alpha \) \( = {\rm{ }}2co{s^3}\alpha {\rm{ }}-{\rm{ }}cos\alpha {\rm{ }}-{\rm{ }}2cos\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}co{s^2}\alpha ){\rm{ }} \) \(= {\rm{ }}4co{s^3}\alpha {\rm{ }}-{\rm{ }}3cos\alpha \) LG b \(\eqalign{ Ứng dụng: Tính: sin 200sin 400sin 800và tan 200tan 400tan 800 Lời giải chi tiết: Ta có: \(\eqalign{ Ứng dụng: \(\eqalign{ Vậy : \(\tan {20^0}\tan {40^0}\tan {80^0} = \sqrt 3 \)
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