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Câu hỏi: Chứng minh rằng:
Lời giải chi tiết:
Ta có:
\(sin3α = sin (2α + α) \) \(= sin 2α cosα + sinα cos 2α\)
\(= {\rm{ }}2{\rm{ }}sin\alpha {\rm{ }}co{s^2}\alpha {\rm{ }} + {\rm{ }}sin\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}2si{n^2}\alpha)\)
\(= {\rm{ }}2sin\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}si{n^2}\alpha){\rm{ }} + {\rm{ }}sin\alpha (1{\rm{ }}-{\rm{ }}si{n^2}\alpha){\rm{ }}\)
\(= {\rm{ }}3sin\alpha {\rm{ }}-{\rm{ }}4si{n^3}\alpha \)
\(cos3α = cos (2α + α) \) \(= cos 2α cosα - sin2α sinα\)
\(= {\rm{ }}(2co{s^2}\alpha {\rm{ }}-{\rm{ }}1)cos\alpha {\rm{ }}-{\rm{ }}2si{n^2}\alpha {\rm{ }}cos\alpha \)
\(= {\rm{ }}2co{s^3}\alpha {\rm{ }}-{\rm{ }}cos\alpha {\rm{ }}-{\rm{ }}2cos\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}co{s^2}\alpha){\rm{ }} \)
\(= {\rm{ }}4co{s^3}\alpha {\rm{ }}-{\rm{ }}3cos\alpha \)
& \sin \alpha \sin ({\pi \over 3} - \alpha)\sin ({\pi \over 3} + \alpha)\cr & = {1 \over 4}\sin 3\alpha \cr
& \cos \alpha \cos ({\pi \over 3} - \alpha)cos({\pi \over 3} + \alpha) \cr &= {1 \over 4}\cos 3\alpha \cr} \)
Ứng dụng: Tính: sin 200 sin 400 sin 800 và tan 200 tan 400 tan 800
Lời giải chi tiết:
Ta có:
\(\eqalign{
& \sin \alpha \sin ({\pi \over 3} - \alpha)\sin ({\pi \over 3} + \alpha) \cr& = \sin \alpha .\frac{1}{2}\left[ {\cos \left({\frac{\pi }{3} - \alpha - \frac{\pi }{3} - \alpha } \right) - \cos \left({\frac{\pi }{3} - \alpha + \frac{\pi }{3} + \alpha } \right)} \right]\cr &= sin\alpha .{1 \over 2}(cos2\alpha - \cos {{2\pi } \over 3}) \cr
& = {1 \over 2}\sin \alpha (1 - 2{\sin ^2}\alpha + {1 \over 2}) \cr &= {1 \over 4}\sin \alpha (3 - 4{\sin ^2}\alpha) \cr
& = {1 \over 4}\sin 3\alpha \cr
& \cos \alpha \cos ({\pi \over 3} - \alpha)cos({\pi \over 3} + \alpha) \cr& = \cos \alpha .\frac{1}{2}\left[ {\cos \left({\frac{\pi }{3} - \alpha + \frac{\pi }{3} + \alpha } \right) + \cos \left({\frac{\pi }{3} - \alpha - \frac{\pi }{3} - \alpha } \right)} \right]\cr &= \cos \alpha .{1 \over 2}(\cos 2\alpha + \cos {{2\pi } \over 3}) \cr
& = {1 \over 2}\cos \alpha (2{\cos ^2}\alpha - 1 - {1 \over 2}) \cr&= {1 \over 4}\cos \alpha (4{\cos ^2}\alpha - 3) \cr &= {1 \over 4}\cos 3\alpha \cr} \)
Ứng dụng:
\(\eqalign{
& \sin {20^0}\sin {40^0}\sin {80^0} \cr&= \sin {20^0}\sin ({60^0} - {20^0})\sin ({60^0} + {20^0}) \cr
& = {1 \over 4}\sin ({3.20^0}) = {1 \over 4}\sin {60^0} = {{\sqrt 3 } \over 8} \cr
& \cos {20^0}\cos {40^0}\cos {80^0} = {1 \over 4}\cos ({3.20^0}) = {1 \over 8} \cr} \)
Vậy : \(\tan {20^0}\tan {40^0}\tan {80^0} = \sqrt 3 \)
Câu a
\(sin3α = 3sinα – 4si{n^3}\alpha \) ; \(cos3α =4co{s^3}\alpha – 3cosα\)Lời giải chi tiết:
Ta có:
\(sin3α = sin (2α + α) \) \(= sin 2α cosα + sinα cos 2α\)
\(= {\rm{ }}2{\rm{ }}sin\alpha {\rm{ }}co{s^2}\alpha {\rm{ }} + {\rm{ }}sin\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}2si{n^2}\alpha)\)
\(= {\rm{ }}2sin\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}si{n^2}\alpha){\rm{ }} + {\rm{ }}sin\alpha (1{\rm{ }}-{\rm{ }}si{n^2}\alpha){\rm{ }}\)
\(= {\rm{ }}3sin\alpha {\rm{ }}-{\rm{ }}4si{n^3}\alpha \)
\(cos3α = cos (2α + α) \) \(= cos 2α cosα - sin2α sinα\)
\(= {\rm{ }}(2co{s^2}\alpha {\rm{ }}-{\rm{ }}1)cos\alpha {\rm{ }}-{\rm{ }}2si{n^2}\alpha {\rm{ }}cos\alpha \)
\(= {\rm{ }}2co{s^3}\alpha {\rm{ }}-{\rm{ }}cos\alpha {\rm{ }}-{\rm{ }}2cos\alpha {\rm{ }}(1{\rm{ }}-{\rm{ }}co{s^2}\alpha){\rm{ }} \)
\(= {\rm{ }}4co{s^3}\alpha {\rm{ }}-{\rm{ }}3cos\alpha \)
Câu b
\(\eqalign{& \sin \alpha \sin ({\pi \over 3} - \alpha)\sin ({\pi \over 3} + \alpha)\cr & = {1 \over 4}\sin 3\alpha \cr
& \cos \alpha \cos ({\pi \over 3} - \alpha)cos({\pi \over 3} + \alpha) \cr &= {1 \over 4}\cos 3\alpha \cr} \)
Ứng dụng: Tính: sin 200 sin 400 sin 800 và tan 200 tan 400 tan 800
Lời giải chi tiết:
Ta có:
\(\eqalign{
& \sin \alpha \sin ({\pi \over 3} - \alpha)\sin ({\pi \over 3} + \alpha) \cr& = \sin \alpha .\frac{1}{2}\left[ {\cos \left({\frac{\pi }{3} - \alpha - \frac{\pi }{3} - \alpha } \right) - \cos \left({\frac{\pi }{3} - \alpha + \frac{\pi }{3} + \alpha } \right)} \right]\cr &= sin\alpha .{1 \over 2}(cos2\alpha - \cos {{2\pi } \over 3}) \cr
& = {1 \over 2}\sin \alpha (1 - 2{\sin ^2}\alpha + {1 \over 2}) \cr &= {1 \over 4}\sin \alpha (3 - 4{\sin ^2}\alpha) \cr
& = {1 \over 4}\sin 3\alpha \cr
& \cos \alpha \cos ({\pi \over 3} - \alpha)cos({\pi \over 3} + \alpha) \cr& = \cos \alpha .\frac{1}{2}\left[ {\cos \left({\frac{\pi }{3} - \alpha + \frac{\pi }{3} + \alpha } \right) + \cos \left({\frac{\pi }{3} - \alpha - \frac{\pi }{3} - \alpha } \right)} \right]\cr &= \cos \alpha .{1 \over 2}(\cos 2\alpha + \cos {{2\pi } \over 3}) \cr
& = {1 \over 2}\cos \alpha (2{\cos ^2}\alpha - 1 - {1 \over 2}) \cr&= {1 \over 4}\cos \alpha (4{\cos ^2}\alpha - 3) \cr &= {1 \over 4}\cos 3\alpha \cr} \)
Ứng dụng:
\(\eqalign{
& \sin {20^0}\sin {40^0}\sin {80^0} \cr&= \sin {20^0}\sin ({60^0} - {20^0})\sin ({60^0} + {20^0}) \cr
& = {1 \over 4}\sin ({3.20^0}) = {1 \over 4}\sin {60^0} = {{\sqrt 3 } \over 8} \cr
& \cos {20^0}\cos {40^0}\cos {80^0} = {1 \over 4}\cos ({3.20^0}) = {1 \over 8} \cr} \)
Vậy : \(\tan {20^0}\tan {40^0}\tan {80^0} = \sqrt 3 \)
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