Ensino Médio ⇒ Trigonometria: Adição e Subtração de Arcos

[naty_naty_n]

Calcular [tex3]\text{sen}\, x,[/tex3] sabendo-se que [tex3]x+y=\frac{\pi}{4}[/tex3] e [tex3]\text{sen}\, y = \frac{3}{5}[/tex3] [tex3](0\lt x \lt \frac{\pi}{2}).[/tex3]

[Thales Gheós]

[tex3]\{\text{sen}\,(y)=\frac{3}{5}\\ cos (y)=\sqrt{1-\frac{9}{25}}\\ cos y=\frac{4}{5}[/tex3]

[tex3]x+y=\frac{\pi}{4}\\x=\frac{\pi}{4}-y[/tex3]

[tex3]\text{sen}\,(x)=\text{sen}\,(\frac{\pi}{4}-y)=\text{sen}\,(\frac{\pi}{4}).cos (y)+\text{sen}\,(y). cos (\frac{\pi}{4})\\\text{sen}\,(x)=\frac{\sqrt{2}}{2}\cdot\frac{4}{5}+\frac{\sqrt{2}}{2}\cdot\frac{3}{5}[/tex3]

agora é só operar...

[naty_naty_n]

ao invés de...
[tex3]\text{sen}\,(\frac{\pi}{4}-y) = \text{sen}\, (\frac{\pi}{4}) cos (y) + \text{sen}\,(y). \cos(\frac{\pi}{4})[/tex3]

não seria
[tex3]\text{sen}\,(\frac{\pi}{4}-y) = \text{sen}\, (\frac{\pi}{4}) cos (y) - \text{sen}\,(y). cos(\frac{\pi}{4})?[/tex3]