Esercizio 11 
Verifica la seguente identità
![\[\sin(\alpha)\cos(2\alpha)-\cos(\alpha)\sin(2\alpha)=\cos\left(\dfrac{\pi}{2}+\alpha\right).\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-f6759446d9c98e3bf2c916c3fd66d75a_l3.png "Rendered by QuickLaTeX.com")
Svolgimento. Si osserva che
;
sostituendo nell’equazione iniziale otteniamo
![\[\begin{aligned} &\sin(\alpha)\cos(2\alpha)-\cos(\alpha)\sin(2\alpha)=\cos\left(\dfrac{\pi}{2}+\alpha\right)\quad \Leftrightarrow \quad \\ &\quad \Leftrightarrow \quad \sin(\alpha)\cos(2\alpha)-\cos(\alpha)2\sin(\alpha)\cos(\alpha)=-\sin(\alpha)\quad \Leftrightarrow \quad \\ &\quad \Leftrightarrow \quad \sin(\alpha)\cos(2\alpha)-2\sin(\alpha)\cos^2(\alpha)=-\sin(\alpha). \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-65d6be99b51e1355546516f3a189b8c6_l3.png "Rendered by QuickLaTeX.com")
Sostituiamo 
e otteniamo:
![\[\begin{aligned} &\sin(\alpha)\cos(2\alpha)-2\sin(\alpha)\cos^2(\alpha)=-\sin(\alpha)\quad \Leftrightarrow \quad\\ &\quad \Leftrightarrow \quad\sin(\alpha)(2\cos^2(\alpha)-1)-2\sin(\alpha)\cos^2(\alpha)=-\sin(\alpha)\quad \Leftrightarrow \quad\\ &\quad \Leftrightarrow \quad2\sin(\alpha)\cos^2(\alpha)-\sin(\alpha)-2\sin(\alpha)\cos^2(\alpha)=-\sin(\alpha)\quad \Leftrightarrow \quad\\ &-\sin \left(\alpha\right)=-\sin \alpha, \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-65823514246d5e21b533ec1468430451_l3.png "Rendered by QuickLaTeX.com")
cioè la tesi.
Fonte: Corso base blu di matematica di Massimo Bergami-Anna Trifone-Graziella Barozzi