Question

If \cos \alpha = \dfrac{1}{7}, \cos(\alpha +\beta )=-\dfrac{11}{14} and 0< \alpha< 90\degree, 0< \beta < 90\degree, find the value of \sin \beta

Collected in the board: Trigonometry

Steven Zheng posted 2 months ago


Answer

\because 0< \alpha< 90\degree, 0< \beta < 90\degree


\therefore 0 < \alpha+\beta<180\degree

\therefore \sin (\alpha+\beta) > 0

\sin (\alpha+\beta) = \sqrt{1-\cos^2 (\alpha+\beta) }

=\sqrt{1-(-\dfrac{11}{14})^2}

= \dfrac{5\sqrt{3} }{14}


\because \cos \alpha = \dfrac{1}{7}

\therefore \sin \alpha = \sqrt{1-(\dfrac{1}{7} )^2} = \dfrac{4\sqrt{3} }{7}


\sin \beta = \sin (\alpha +\beta - \alpha)

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


= \dfrac{5\sqrt{3} }{14}\cdotp \dfrac{1}{7} +(-\dfrac{11}{14})\cdotp \dfrac{4\sqrt{3} }{7}

=\dfrac{\sqrt{3} }{2}

Steven Zheng posted 2 months ago

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