If $$\cos \alpha = \dfrac{1}{7}$$, $$ \cos(\alpha +\beta )=-\dfrac{11}{14} $$ and $$ 0

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 $$\sin \beta      $$

Uniteasy Admin · Accepted Answer

$$\because  0< \alpha< 90\degree$$, $$ 0< \beta < 90\degree$$
$$ 	herefore  0 < \alpha+\beta<180\degree $$
$$ 	herefore \sin (\alpha+\beta) > 0$$
$$\sin (\alpha+\beta) = \sqrt{1-\cos^2 (\alpha+\beta) } $$
$$=\sqrt{1-(-\dfrac{11}{14})^2} $$
$$= \dfrac{5\sqrt{3} }{14} $$
$$\because \cos \alpha = \dfrac{1}{7}$$
$$	herefore \sin \alpha = \sqrt{1-(\dfrac{1}{7} )^2} = \dfrac{4\sqrt{3} }{7}   $$
$$\sin \beta = \sin (\alpha +\beta - \alpha)$$ 
 
$$ = \sin (\alpha+\beta)\cos \alpha -\cos (\alpha+\beta) \sin \alpha$$
$$ = \dfrac{5\sqrt{3} }{14}\cdotp   \dfrac{1}{7} +(-\dfrac{11}{14})\cdotp  \dfrac{4\sqrt{3} }{7}  $$
$$=\dfrac{\sqrt{3} }{2} $$

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