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Câu hỏi: Chứng minh rằng với mọi $\alpha, \beta, \gamma$ ta có:
$\cos (\alpha+\beta) \cdot \sin (\alpha-\beta)+\cos (\beta+\gamma) \cdot \sin (\beta-\gamma)+\cos (\gamma+\alpha) \cdot \operatorname{sin}(\gamma-\alpha)=0$
$\cos (\alpha+\beta) \cdot \sin (\alpha-\beta)+\cos (\beta+\gamma) \cdot \sin (\beta-\gamma)+\cos (\gamma+\alpha) \cdot \operatorname{sin}(\gamma-\alpha)=0$
Lời giải chi tiết
Ta có:
$\begin{array}{l}
\cos (\alpha + \beta ) \cdot \sin (\alpha - \beta ) + \cos (\beta + \gamma ) \cdot \sin (\beta - \gamma ) + \cos (\gamma + \alpha ) \cdot \sin (\gamma - \alpha )\\
= 1/2[\sin (\alpha - \beta - \alpha - \beta ) + \sin (\alpha - \beta + \alpha + \beta ) + \sin (\beta - \gamma - \beta - \gamma ) + \sin (\beta - \gamma + \beta + \gamma ) + \sin (\gamma - \alpha - \gamma - \alpha ) + \sin (\gamma - \alpha + \gamma + \alpha )]\\
= 1/2[ - \sin 2\beta + \sin 2\alpha - \sin 2\gamma + \sin 2\beta - \sin 2\alpha + \sin 2\gamma ] = 0(\;dpcm)
\end{array}$
Ta có:
$\begin{array}{l}
\cos (\alpha + \beta ) \cdot \sin (\alpha - \beta ) + \cos (\beta + \gamma ) \cdot \sin (\beta - \gamma ) + \cos (\gamma + \alpha ) \cdot \sin (\gamma - \alpha )\\
= 1/2[\sin (\alpha - \beta - \alpha - \beta ) + \sin (\alpha - \beta + \alpha + \beta ) + \sin (\beta - \gamma - \beta - \gamma ) + \sin (\beta - \gamma + \beta + \gamma ) + \sin (\gamma - \alpha - \gamma - \alpha ) + \sin (\gamma - \alpha + \gamma + \alpha )]\\
= 1/2[ - \sin 2\beta + \sin 2\alpha - \sin 2\gamma + \sin 2\beta - \sin 2\alpha + \sin 2\gamma ] = 0(\;dpcm)
\end{array}$