Find the value of $\sin \left(\cos^{-1} \left(\dfrac {1}{2}\right) + \sin^{-1} \left(\dfrac {3}{5}\right)\right)$
My Attempt:\begin{align*} \sin \left(\cos^{-1} \left(\dfrac {1}{2}\right) + \sin^{-1} \left(\dfrac {3}{5}\right)\right)&=\sin \left(\cos^{-1} \left(\dfrac {1}{2}\right)+\cos^{-1} \left(\dfrac {4}{5}\right)\right)\\ &=\sin\left(\cos^{-1} \left(\dfrac {2}{5} - \dfrac {3\sqrt {3}}{10}\right)\right)\\ &=\sin \left(\cos^{-1} \left(\dfrac {4-3\sqrt {3}}{10}\right)\right) \end{align*}
$\endgroup$ 42 Answers
$\begingroup$$$\sin \left(\cos^{-1} \left(\dfrac {1}{2}\right) + \sin^{-1} \left(\dfrac {3}{5}\right)\right)=\sin \left(\cos^{-1} \left(\dfrac {1}{2}\right)+\cos^{-1} \left(\dfrac {4}{5}\right)\right)$$ Now let $\cos^{-1} \left(\dfrac {1}{2}\right)=\alpha$ and $\sin^{-1} \left(\dfrac {3}{5}\right)=\beta$
$$\cos\alpha=\frac12,\sin\beta=\frac35$$ $$\sin\alpha=\frac{\sqrt{3}}{2},\cos\beta=\frac{4}{5}$$ Now we get, $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$ $$=\frac{\sqrt{3}}{2}\times\frac45+\frac12\times\frac35=\frac{4\sqrt{3}}{10}$$
$\endgroup$ $\begingroup$Hints:
if $\theta=\arccos \frac12\;$ and $\;\varphi=\arcsin \frac35$, then
- $\enspace\theta\in[0,\pi]\,$ and $\;\varphi\in\Bigl[-\dfrac\pi2,\dfrac\pi2\Bigr]$.
- $\;$Pythagoras' identity.
- $\enspace\sin(\theta+\varphi)=\sin\theta\,\cos\varphi+\sin\varphi\,\cos\theta$.
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