Exprima A(x) em função de senx e cos x
Trigonometria: Infinito 11 A - Parte 1 Pág. 96 Ex. 53
Enunciado
Exprima A(x) em função de sen x e cos x.
- $A(x)=sen\,(-x)-sen\,(\pi -x)$
- $A(x)=\cos (-x)+\cos (\pi +x)$
- $A(x)=sen\,(\frac{\pi }{2}-x)+\cos (\frac{5\pi }{2}-x)$
- $A(x)=\cos (\frac{3\pi }{2}+x)+sen\,(x-\frac{5\pi }{2})$
Resolução
- Ora,
\[\begin{array}{*{35}{l}}
A(x) & = & sen\,(-x)-sen\,(\pi -x) \\
{} & = & -sen\,x-sen\,x \\
{} & = & -2sen\,x \\
\end{array}\] - Ora,
\[\begin{array}{*{35}{l}}
A(x) & = & \cos (-x)+\cos (\pi +x) \\
{} & = & \cos x-\cos x \\
{} & = & 0 \\
\end{array}\] - Ora,
\[\begin{array}{*{35}{l}}
A(x) & = & sen\,(\frac{\pi }{2}-x)+\cos (\frac{5\pi }{2}-x) \\
{} & = & \cos x+\cos (\frac{\pi }{2}-x) \\
{} & = & \cos x+sen\,x \\
\end{array}\] - Ora,
\[\begin{array}{*{35}{l}}
A(x) & = & \cos (\frac{3\pi }{2}+x)+sen\,(x-\frac{5\pi }{2}) \\
{} & = & \cos (2\pi -\frac{\pi }{2}+x)+sen\,(x-\frac{\pi }{2}) \\
{} & = & \cos (\frac{\pi }{2}-x)-sen\,(\frac{\pi }{2}-x) \\
{} & = & sen\,x-\cos x \\
\end{array}\]
