- LG a
- LG b
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
\[\eqalign{
& \sin \alpha \sin [{\pi \over 3} - \alpha ]\sin [{\pi \over 3} + \alpha ]\cr & = {1 \over 4}\sin 3\alpha \cr
& \cos \alpha \cos [{\pi \over 3} - \alpha ]cos[{\pi \over 3} + \alpha ] \cr &= {1 \over 4}\cos 3\alpha \cr} \]
Ứng dụng: Tính: sin 200sin 400sin 800và tan 200tan 400tan 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 : \[\tan {20^0}\tan {40^0}\tan {80^0} = \sqrt 3 \]