Ars Mathematica: Solución del problema 44

s e n x = A, s e n y = B, c o s x = \sqrt{1 - A^{2}}, c o s y = \sqrt{1 - B^{2}}

s e n (x + y) = s e n x

\begin{matrix} A \sqrt{1 - B^{2}} + \sqrt{1 - A^{2}} B & = A + B \end{matrix}

\begin{matrix} A^{2} - A^{2} B^{2} & = & A^{2} + B^{2} + B^{2} - A^{2} B^{2} + 2 A B - 2 A B \sqrt{1 - A^{2}} - 2 B^{2} \sqrt{1 - A^{2}} \\ A^{2} - A^{2} B^{2} & = & A^{2} + B^{2} + B^{2} - A^{2} B^{2} + 2 A B - \sqrt{1 - A^{2}} (2 B^{2} + 2 A B) \\ 0 & = & 2 B^{2} + 2 A B - \sqrt{1 - A^{2}} (2 B^{2} + 2 A B) \\ 0 & = & 2 B (B + A - \sqrt{1 - A^{2}} (B + A)) \end{matrix}

{\begin{matrix} B = 0 \\ B + A - \sqrt{1 - A^{2}} (B + A) = 0 \Rightarrow (B + A) (1 - \sqrt{1 - A^{2}}) = 0 \end{matrix} .

\Rightarrow {\begin{matrix} B = - A \\ \sqrt{1 - A^{2}} = 1 \end{matrix} .

{\begin{matrix} B = 0 \Rightarrow s e n y = 0 \Rightarrow y = {\begin{matrix} y = 2 k π \\ y = π + 2 k π \end{matrix} . \\ {\begin{matrix} B = - A \Rightarrow s e n y = - s e n x \Rightarrow {\begin{matrix} y = - x + 2 k π \\ y = π + x + 2 k π \end{matrix} . \\ \sqrt{1 - A^{2}} = 1 \Rightarrow A = 0 \Rightarrow s e n x = 0 \Rightarrow {\begin{matrix} x = 2 k π \\ x = π + 2 k π \end{matrix} . \end{matrix} . \end{matrix} .

Ars Mathematica