Exprima em função de sen b e cos b
Trigonometria: Infinito 11 A - Parte 1 Pág. 96 Ex. 52
Enunciado
Recorrendo ao círculo trigonométrico, exprima, em função de sen b e cos b, as expressões:
- $sen\,(b+\pi )+sen\,(b+2\pi )+sen\,(b-\pi )$
- $\cos (b+\pi )+sen\,(b+\frac{\pi }{2})+\cos (b-\pi )+sen\,(b+\frac{3\pi }{2})$
- $sen\,(-b-\pi )-2\cos (-\frac{\pi }{2}-b)+sen\,(-\frac{3\pi }{2}+b)$
Resolução
- Ora,
\[\begin{array}{*{35}{l}}
sen\,(b+\pi )+sen\,(b+2\pi )+sen\,(b-\pi ) & = & -sen\,b+sen\,b-sen\,(\pi -b) \\
{} & = & 0-sen\,b \\
{} & = & -sen\,b \\
\end{array}\] - Ora,
\[\begin{array}{*{35}{l}}
\cos (b+\pi )+sen\,(b+\frac{\pi }{2})+\cos (b-\pi )+sen\,(b+\frac{3\pi }{2}) & = & -\cos b+sen\,(\frac{\pi }{2}-(-b))+\cos (\pi -b)+sen\,(2\pi -\frac{\pi }{2}+b) \\
{} & = & -\cos b+\cos \,(-b)-\cos b+sen\,(-\frac{\pi }{2}+b) \\
{} & = & -\cos b+\cos \,b-\cos b-sen\,(\frac{\pi }{2}-b) \\
{} & = & -\cos b-\cos b \\
{} & = & -2\cos b \\
\end{array}\] - Ora,
\[\begin{array}{*{35}{l}}
sen\,(-b-\pi )-2\cos (-\frac{\pi }{2}-b)+sen\,(-\frac{3\pi }{2}+b) & = & -sen\,(\pi +b)-2\cos (\frac{\pi }{2}-(-b))+sen\,(2\pi -\frac{3\pi }{2}+b) \\
{} & = & sen\,b-2\,sen\,(-b)+sen\,(\frac{\pi }{2}-(-b)) \\
{} & = & sen\,b+2\,sen\,b+\cos (-b) \\
{} & = & 3\,sen\,b+\cos b \\
\end{array}\]
