Esercizio 6 ripasso goniometria e trigonometria

Ripasso goniometria e trigonometria

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Esercizio 6.   (\bigstar\bigstar\largewhitestar\largewhitestar\largewhitestar) Semplificare la seguente espressione:

    \[\dfrac{\sin\left(\alpha-\dfrac{\pi}{2}\right)\sin(-\alpha)+\cos\left(\dfrac{3\pi}{2}-\alpha\right)\sin\left(\dfrac{11\pi}{2}+\alpha\right)+\cos(3\pi+\alpha)} {-\tan\left(\alpha+\dfrac{\pi}{2}\right)\cot\left(\alpha-\dfrac{3\pi}{2}\right)-\sin(\alpha+\pi)+\sin(7\pi-\alpha)}.\]

 

Svolgimento. Applicando gli archi associati a tutti termini del numeratore si ha

 

  • \sin\left(\alpha-\dfrac{\pi}{2}\right)= \sin\left(-\left(\dfrac{\pi}{2}-\alpha\right)\right)= -\sin\left(\dfrac{\pi}{2}-\alpha\right)=-\cos(\alpha);
  • \sin(-\alpha)=-\sin(\alpha);
  • \cos\left(\dfrac{3\pi}{2}-\alpha\right)=-\sin(\alpha);
  • \sin\left(\dfrac{11\pi}{2}+\alpha\right)=\sin\left(4\pi+\dfrac{3\pi}{2}+\alpha\right)=\sin\left(\dfrac{3\pi}{2}+\alpha\right)=-\cos(\alpha);
  • \cos(3\pi+\alpha)=\cos(2\pi+\pi+\alpha)= \cos(\pi+\alpha)=-\cos(\alpha),

e il denominatore

  • \tan\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{\sin\left(\alpha+\dfrac{\pi}{2}\right)}{\cos\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{\cos(\alpha)}{-\sin(\alpha)}=-\dfrac{\cos(\alpha)}{\sin(\alpha)};
  • \cot\left(\alpha-\dfrac{3\pi}{2}\right)=\dfrac{\cos\left(\alpha-\dfrac{3\pi}{2}\right)}{\sin\left(\alpha-\dfrac{3\pi}{2}\right)}=\dfrac{-\sin(\alpha)}{-(-\cos(\alpha)}=-\dfrac{\sin(\alpha)}{\cos(\alpha)};
  • \sin(\alpha+\pi)=-\sin(\alpha);
  • \sin(7\pi-\alpha)=\sin(6\pi+\pi-\alpha)= \sin(\pi-\alpha)=\sin(\alpha).

Adesso possiamo sostituire tutti i termini ottenuti nell’espressione iniziale ottenendo:

    \[\dfrac{\sin\left(\alpha-\dfrac{\pi}{2}\right)\sin(-\alpha)+\cos\left(\dfrac{3\pi}{2}-\alpha\right)\sin\left(\dfrac{11\pi}{2}+\alpha\right)+\cos(3\pi+\alpha)} {-\tan\left(\alpha+\dfrac{\pi}{2}\right)\cot\left(\alpha-\dfrac{3\pi}{2}\right)-\sin(\alpha+\pi)+\sin(7\pi-\alpha)}=\]

    \[=\dfrac{-\cos(\alpha)(-\sin(\alpha))-\sin(\alpha)\cdot(-\cos(\alpha))-\cos(\alpha)} {-\left(-\dfrac{\cos(\alpha)}{\sin(\alpha)}\right)\cdot\left(-\dfrac{\sin(\alpha)}{\cos(\alpha)}\right)-(-\sin(\alpha))+\sin(\alpha)}=\]

    \[=\dfrac{\cos(\alpha)\sin(\alpha)+\sin(\alpha)\cos(\alpha)-\cos(\alpha)} {-1+\sin(\alpha)+\sin(\alpha)}=\]

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

 

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