Esercizio 14 ripasso goniometria e trigonometria

Ripasso goniometria e trigonometria

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Esercizio 14   (\bigstar\largewhitestar\largewhitestar\largewhitestar\largewhitestar). Applicando le formule di prostaferesi dimostrare che

    \[\sin\left(\dfrac{\pi}{3}-\alpha\right)+\sin\left(\dfrac{4\pi}{3}+\alpha\right)=-\sin(\alpha).\]

 

Svolgimento. Applicando le formule di prostaferesi si ha:

    \[\begin{aligned} &\sin\left(\dfrac{\pi}{3}-\alpha\right)+\sin\left(\dfrac{4\pi}{3}+\alpha\right)=\\ &=2\sin\left(\dfrac{\left(\dfrac{\pi}{3}-\alpha\right)+\left(\dfrac{4\pi}{3}+\alpha\right)}{2}\right)\cos\left(\dfrac{\left(\dfrac{\pi}{3}-\alpha\right)-\left(\dfrac{4\pi}{3}+\alpha\right)}{2}\right) \\ &=2\sin\left(\dfrac{\dfrac{5\pi}{3}}{2}\right)\cos\left(\dfrac{-\dfrac{3\pi}{3}-2\alpha}{2}\right)=2\sin\left(\dfrac{5\pi}{6}\right)\cos\left(-\dfrac{\pi}{2}-\alpha\right), \end{aligned}\]

da cui applicando gli angoli associati si ottiene

    \[\begin{aligned} &\sin\left(\dfrac{\pi}{3}-\alpha\right)+\sin\left(\dfrac{4\pi}{3}+\alpha\right)=\\ &=2\sin\left(\dfrac{5\pi}{6}\right)\cos\left(-\left(\dfrac{\pi}{2}+\alpha\right)\right)= 2\sin\left(\dfrac{5\pi}{6}\right)\cos\left(\dfrac{\pi}{2}+\alpha\right)=\\ &=2\cdot\dfrac{1}{2}\cdot\cos\left(\dfrac{\pi}{2}+\alpha\right)=-\sin(\alpha), \end{aligned}\]

cioè la tesi.

 

Fonte: Corso base blu di matematica di Massimo Bergami-Anna Trifone-Graziella Barozzi