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Find sin(arcsin(3/5) + arccos(15/17)).

Dec 10, 2019

#1
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Find

$$\sin\left( \arcsin\left(\dfrac{3}{5}\right) + \arccos\left(\dfrac{15}{17}\right) \right)$$.

$$\text{Let x=\dfrac{3}{5},\ y=\dfrac{15}{17} }$$

$$\begin{array}{|rcll|} \hline && \mathbf{\sin\left( \arcsin\left(\dfrac{3}{5}\right) + \arccos\left(\dfrac{15}{17}\right) \right)} \\\\ &=& \sin\Big( \underbrace{\arcsin\left(x\right)}_{\alpha} + \underbrace{\arccos\left(y\right)}_{\beta} \Big) \\ && \boxed{\text{Formula: } \sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+ \cos(\alpha)\sin(\beta)} \\\\ &=& \sin\Big(\arcsin\left(x\right)\Big)\cos\Big(\arccos\left(y\right)\Big) + \cos\Big(\arcsin\left(x\right)\Big)\sin\Big(\arccos\left(y\right)\Big) \\ &=& xy + \mathbf{\cos\Big(\arcsin\left(x\right)\Big)}\sin\Big(\arccos\left(y\right)\Big) \\ && \boxed{ \arcsin\left(x\right) = \alpha\\ x=\sin(\alpha)\\ x^2=\sin^2(\alpha)\\ x^2=1-\cos^2(\alpha)\\ \cos^2(\alpha)=1-x^2 \\ \cos(\alpha) =\sqrt{1-x^2} \\ \mathbf{\cos\Big(\arcsin\left(x\right)\Big) =\sqrt{1-x^2} } } \\\\ &=& xy + \sqrt{1-x^2}\sin\Big(\arccos\left(y\right)\Big) \\ && \boxed{ \arccos\left(y\right) = \beta\\ y=\cos(\beta)\\ y^2=\cos^2(\beta)\\ y^2=1-\sin^2(\beta)\\ \sin^2(\beta)=1-y^2 \\ \sin(\beta) =\sqrt{1-y^2} \\ \mathbf{\sin\Big(\arccos\left(y\right)\Big) =\sqrt{1-y^2} } } \\\\ &=& xy + \sqrt{1-x^2}\sqrt{1-y^2} \quad | \quad x=\dfrac{3}{5},\ y=\dfrac{15}{17} \\ &=& \dfrac{3}{5}*\dfrac{15}{17} + \sqrt{1-\dfrac{3^2}{5^2}}\sqrt{1-\dfrac{15^2}{17^2}} \\ &=& \dfrac{45}{85} + \dfrac{4}{5} * \dfrac{8}{17} \\ &=& \dfrac{45}{85} + \dfrac{32}{85} \\ &=& \mathbf{ \dfrac{77}{85} } \\ \hline \end{array}$$

Dec 11, 2019