- LG a
- LG b
- LG c
- LG d
Chứng minh rằng nếu \[ + β + γ = π\] thì
LG a
\[\sin \alpha + \sin \beta + \sin \gamma = 4\cos {\alpha \over 2}\cos {\beta \over 2}\cos {\gamma \over 2}\]
Lời giải chi tiết:
Ta có:
\[\eqalign{
& \sin \alpha + \sin \beta + \sin \gamma\cr& = \sin \alpha + 2\sin {{\beta + \gamma } \over 2}\cos {{\beta - \gamma } \over 2} \cr
& = \sin \alpha + 2\sin {{\pi - \alpha } \over 2}\cos {{\beta - \gamma } \over 2} \cr&= 2\sin {\alpha \over 2}\cos {\alpha \over 2} + 2\cos {\alpha \over 2} \cos {{\beta - \gamma } \over 2} \cr
& = 2\cos {\alpha \over 2}[\sin {\alpha \over 2} + \cos {{\beta - \gamma } \over 2}]\cr& = 2\cos {\alpha \over 2}{\rm{[sin}}{{\pi - [\beta + \gamma ]} \over 2} + \cos{{\beta - \gamma } \over 2}{\rm{]}} \cr
& = 2\cos {\alpha \over 2}[cos{{\beta + \gamma } \over 2} + \cos {{\beta - \gamma } \over 2}] \cr
& =4\cos {\alpha \over 2}\cos {\beta \over 2}\cos {\gamma \over 2} \cr} \]
LG b
\[\cos \alpha + \cos \beta + \cos \gamma = 1 + 4\sin {\alpha \over 2}\sin {\beta \over 2}\sin {\gamma \over 2}\]
Lời giải chi tiết:
Ta có:
\[\eqalign{
& \cos \alpha + \cos \beta + \cos \gamma \cr&= 2\cos {{\alpha + \beta } \over 2}\cos {{\alpha - \beta } \over 2} + 1 - 2\sin ^2{{\gamma } \over 2} \cr
& = 2\cos [{\pi \over 2} - {\gamma \over 2}]cos{{\alpha - \beta } \over 2} + 1 - 2{\sin ^2}{\gamma \over 2} \cr&= 2\sin \frac{\gamma }{2}\cos \frac{{\alpha - \beta }}{2} + 1 - 2{\sin ^2}\frac{\gamma }{2}\cr &= 1 + 2\sin {\gamma \over 2}[cos{{\alpha - \beta } \over 2} - \sin {\gamma \over 2}] \cr
& = 1 + 2\sin \frac{\gamma }{2}\left[ {\cos \frac{{\alpha - \beta }}{2} - \sin \left[ {\frac{\pi }{2} - \frac{{\alpha + \beta }}{2}} \right]} \right]\cr &= 1 + 2\sin {\gamma \over 2}[cos{{\alpha - \beta } \over 2} - cos{{\alpha + \beta } \over 2}] \cr
& = 1 + 4\sin {\alpha \over 2}\sin {\beta \over 2}\sin {\gamma \over 2} \cr} \]
LG c
\[sin2 + sin2β + sin2γ = 4sin sinβ sin γ\]
Lời giải chi tiết:
\[sin2 + sin2β + sin2γ\]
\[= 2sin [ + β]cos[ - β ] + 2sinγcosγ\]
\[ = 2\sin \left[ {{{180}^0} - \gamma } \right]\cos \left[ {\alpha - \beta } \right] \] \[+ 2\sin \gamma \cos \left[ {{{180}^0} - \left[ {\alpha + \beta } \right]} \right] \]
\[= 2\sin \gamma \cos \left[ {\alpha - \beta } \right] - 2\sin \gamma \cos \left[ {\alpha + \beta } \right]\]
\[= 2sinγ [cos[ - β ] - cos[ + β]] \]
\[= 4sin sinβ sin γ\]
LG d
\[co{s^2} \propto + {\rm{ }}co{s^2}\beta + co{s^2}\gamma {\rm{ }}= 1 2cos cosβ cosγ\]
Lời giải chi tiết:
Ta có:
\[\eqalign{
& co{s^2} \propto + {\rm{ }}co{s^2}\beta + co{s^2}\gamma {\rm{ }} \cr
& {\rm{ = }}{{1 + \cos 2\alpha } \over 2} + {{1+\cos 2\beta } \over 2} + {\cos ^2}\gamma \cr
& = 1 + {1 \over 2}[cos2\alpha + \cos 2\beta ] + {\cos ^2}\gamma \cr
& = 1 + \cos [\alpha + \beta ]cos[\alpha - \beta ] + {\cos ^2}\gamma \cr
&= 1 + \cos \left[ {\pi - \gamma } \right]\cos \left[ {\alpha - \beta } \right] + {\cos ^2}\gamma \cr &= 1 - \cos \gamma \cos \left[ {\alpha - \beta } \right] + {\cos ^2}\gamma \cr &= 1 - \cos \gamma [ \cos [\alpha - \beta ]-\cos \gamma ] \cr&= 1 - \cos \gamma {\rm{[cos[}}\alpha {\rm{ - }}\beta {\rm{] + cos[}}\alpha {\rm{ + }}\beta ]{\rm{]}} \cr
& = {\rm{ }}1{\rm{ }}-{\rm{ }}2cos \propto {\rm{ }}cos\beta {\rm{ }}cos\gamma \cr} \]