#### Question

If \alpha, \beta , \gamma are acute angles, determine the value of \alpha -\beta such that \sin \alpha +\sin \gamma = \sin \beta , \cos \alpha -\cos \gamma = \cos \beta

Collected in the board: Trigonometry

Steven Zheng posted 2 months ago

Express \gamma in terms of \alpha and \beta

\sin \gamma = \sin \beta - \sin \alpha

\cos \gamma =\cos \alpha - \cos \beta

Apply Pythagorean Identity for \gamma

\sin^2\gamma+\cos^2\gamma = ( \sin \beta - \sin \alpha)^2+(\cos \alpha - \cos \beta)^2 = 1

2- 2( \cos \alpha\cos \beta +\sin \alpha \sin \beta) = 1

\cos(\alpha -\beta ) = \dfrac{1}{2}

Since \alpha, \beta , \gammaare acute angles,

\sin \beta - \sin \alpha = \sin \gamma > 0

\alpha - \beta < 0

\therefore \alpha -\beta = - \dfrac{\pi}{3}

Steven Zheng posted 2 months ago

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