Esercizio 14 
. Applicando le formule di prostaferesi dimostrare che
![\[\sin\left(\dfrac{\pi}{3}-\alpha\right)+\sin\left(\dfrac{4\pi}{3}+\alpha\right)=-\sin(\alpha).\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-1116713908f0798fbe49a6e8d49b501f_l3.png "Rendered by QuickLaTeX.com")
Svolgimento. Applicando le formule di prostaferesi si ha:
![\[\begin{aligned} &\sin\left(\dfrac{\pi}{3}-\alpha\right)+\sin\left(\dfrac{4\pi}{3}+\alpha\right)=\\ &=2\sin\left(\dfrac{\left(\dfrac{\pi}{3}-\alpha\right)+\left(\dfrac{4\pi}{3}+\alpha\right)}{2}\right)\cos\left(\dfrac{\left(\dfrac{\pi}{3}-\alpha\right)-\left(\dfrac{4\pi}{3}+\alpha\right)}{2}\right) \\ &=2\sin\left(\dfrac{\dfrac{5\pi}{3}}{2}\right)\cos\left(\dfrac{-\dfrac{3\pi}{3}-2\alpha}{2}\right)=2\sin\left(\dfrac{5\pi}{6}\right)\cos\left(-\dfrac{\pi}{2}-\alpha\right), \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c0b002d00b485875350cab7b744b005a_l3.png "Rendered by QuickLaTeX.com")
da cui applicando gli angoli associati si ottiene
![\[\begin{aligned} &\sin\left(\dfrac{\pi}{3}-\alpha\right)+\sin\left(\dfrac{4\pi}{3}+\alpha\right)=\\ &=2\sin\left(\dfrac{5\pi}{6}\right)\cos\left(-\left(\dfrac{\pi}{2}+\alpha\right)\right)= 2\sin\left(\dfrac{5\pi}{6}\right)\cos\left(\dfrac{\pi}{2}+\alpha\right)=\\ &=2\cdot\dfrac{1}{2}\cdot\cos\left(\dfrac{\pi}{2}+\alpha\right)=-\sin(\alpha), \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-3607a0db2a9a0ed788662c9cc3e8ead5_l3.png "Rendered by QuickLaTeX.com")
cioè la tesi.
Fonte: Corso base blu di matematica di Massimo Bergami-Anna Trifone-Graziella Barozzi