Esercizio 12 ripasso goniometria e trigonometria

Ripasso goniometria e trigonometria

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Esercizio 12   (\bigstar\largewhitestar\largewhitestar\largewhitestar\largewhitestar). Semplificare la seguente espressione:

    \[\left(\sin\left(\dfrac{\alpha}{2}\right)+\cos\left(\dfrac{\alpha}{2}\right)\right)^2-\sin(\alpha)+1.\]

 

Svolgimento. Si ha:

    \begin{align*} \left(\sin\left(\dfrac{\alpha}{2}\right)+\cos\left(\dfrac{\alpha}{2}\right)\right)^2=\sin^2\left(\dfrac{\alpha}{2}\right)+\cos^2\left(\dfrac{\alpha}{2}\right)+2\sin\left(\dfrac{\alpha}{2}\right)\cos\left(\dfrac{\alpha}{2}\right), \end{align*}

e per la regola fondamentale si ottiene

    \[\sin^2\left(\dfrac{\alpha}{2}\right)+\cos^2\left(\dfrac{\alpha}{2}\right)=1,\]

mentre per duplicazione si ha

    \[2\sin\left(\dfrac{\alpha}{2}\right)\cos\left(\dfrac{\alpha}{2}\right)=\sin\left(2\cdot\dfrac{\alpha}{2}\right)=\sin(\alpha),\]

quindi

    \[\left(\sin\left(\dfrac{\alpha}{2}\right)+\cos\left(\dfrac{\alpha}{2}\right)\right)^2=1+\sin(\alpha),\]

da cui

    \[\left(\sin\left(\dfrac{\alpha}{2}\right)+\cos\left(\dfrac{\alpha}{2}\right)\right)^2-\sin(\alpha)+1= 1+\sin(\alpha)-\sin(\alpha)+1=2.\]

 

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