Esercizio 13 
. Dimostrare che
![\[\sin(75^\circ)+\cos(165^\circ)=0.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-f397bc092a82f73992684420993f3dc7_l3.png "Rendered by QuickLaTeX.com")
Svolgimento. Osserviamo quanto segue
![\[\cos\left(165^\circ\right)=\cos\left(90^\circ+75^\circ\right)=-\sin\left(75^\circ\right),\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-49cf3cdcbf3c7788389646e7f9f00ace_l3.png "Rendered by QuickLaTeX.com")
da cui
![\[\sin(75^\circ)+\cos(165^\circ)=\sin(75^\circ)-\sin\left(75^\circ\right)=0,\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-0ce700a5f1b2d5f7a79e45c019dd0d9d_l3.png "Rendered by QuickLaTeX.com")
cioè la tesi.
Altrimenti si possono applicare le formule di bisezione nel seguente modo:
![\[\begin{aligned} \sin(75)&=\sin\left(\dfrac{250}{2}\right)=\sqrt{\dfrac{1-\cos(150)}{2}}=\sqrt{\dfrac{1-\left(-\dfrac{\sqrt{3}}{2}\right)}{2}}=\\ &=\sqrt{\dfrac{1+\dfrac{\sqrt{3}}{2}}{2}}= \sqrt{\dfrac{2+\sqrt{3}}{4}}= \dfrac{\sqrt{2+\sqrt{3}}}{2} \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-934ede5f113706da243f524616ffeaf4_l3.png "Rendered by QuickLaTeX.com")
e
![\[\begin{aligned} \cos(165)&=\cos\left(\dfrac{330}{2}\right)=-\sqrt{\dfrac{1+\cos(330)}{2}}=-\sqrt{\dfrac{1+\dfrac{\sqrt{3}}{2}}{2}}=\\ &=-\sqrt{\dfrac{2+\sqrt{3}}{4}}=-\dfrac{\sqrt{2+\sqrt{3}}}{2}, \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d87b660d3c2e9c63dbe43897aba6058b_l3.png "Rendered by QuickLaTeX.com")
da cui
![\[\sin(75)+\cos(105)=\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{\sqrt{2+\sqrt{3}}}{2}=0,\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-13b21e1f7527860b4b9bfdee9490cfac_l3.png "Rendered by QuickLaTeX.com")
cioè la tesi.
Fonte: Corso base blu di matematica di Massimo Bergami-Anna Trifone-Graziella Barozzi