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