# Bài 51 trang 216 SGK Đại số 10 Nâng cao

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## Chứng minh rằng nếu ∝ + β + γ = π thì

Chứng minh rằng nếu $$∝ + β + γ = π$$ thì

a) $$\sin \alpha + \sin \beta + \sin \gamma = 4\cos {\alpha \over 2}\cos {\beta \over 2}\cos {\gamma \over 2}$$

b) $$\cos \alpha + \cos \beta + \cos \gamma = 1 + 4\sin {\alpha \over 2}\sin {\beta \over 2}\sin {\gamma \over 2}$$

c) $$sin2∝ + sin2β + sin2γ = 4sin∝ sinβ sin γ$$

d) $$co{s^2} \propto + {\rm{ }}co{s^2}\beta + co{s^2}\gamma {\rm{ }}= 1 – 2cos∝ cosβ cosγ$$

Đáp án

a) 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}

b) 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&= 1 + 2\sin {\gamma \over 2}(cos{{\alpha - \beta } \over 2} - \sin {\gamma \over 2}) \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}

c) $$sin2∝ + sin2β + sin2γ$$

$$= 2sin (∝ + β)cos(∝ - β ) + 2sinγcosγ$$

$$= 2sinγ (cos(∝ - β ) - cos(∝ + β))$$

$$= 4sin∝ sinβ sin γ$$

d) 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 \gamma (\cos \gamma - \cos (\alpha - \beta )) \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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