Prove que…
Trigonometria: Infinito 11 A - Parte 1 Pág. 94 Ex. 41
Enunciado
Prove que, para todo o $\theta $, se tem:
- ${{(sen\,\theta +\cos \theta )}^{2}}=1+2\,sen\,\theta \times \cos \theta $
- $\cos \theta -se{{n}^{2}}\,\theta \times \cos \theta ={{\cos }^{2}}\theta $
- $se{{n}^{4}}\,\theta +{{\cos }^{4}}\theta +2\times se{{n}^{2}}\,\theta \times {{\cos }^{2}}\theta =1$
- ${{(\cos \theta -sen\,\theta )}^{2}}+{{(\cos \theta +sen\,\theta )}^{2}}=2$
- $(\cos \theta -sen\,\theta )+(\cos \theta +sen\,\theta )-1=-2\times se{{n}^{2}}\,\theta $
Resolução
- Ora,
\[\begin{array}{*{35}{l}}
{{(sen\,\theta +\cos \theta )}^{2}} & = & se{{n}^{2}}\,\theta +2\,sen\,\theta \times \cos \theta +{{\cos }^{2}}\theta \\
{} & = & (se{{n}^{2}}\,\theta +{{\cos }^{2}}\theta )+2\,sen\,\theta \times \cos \theta \\
{} & = & 1+2\,sen\,\theta \times \cos \theta \\
\end{array}\] - Ora,
\[\begin{array}{*{35}{l}}
\cos \theta -se{{n}^{2}}\,\theta \times \cos \theta & = & \cos \theta \times (1-se{{n}^{2}}\,\theta ) \\
{} & = & \cos \theta \times ({{\cos }^{2}}\theta )\,\,\,\text{, pela F}\text{.F}\text{.T}\text{.} \\
{} & = & {{\cos }^{3}}\theta \\
\end{array}\] - Ora,
\[\begin{array}{*{35}{l}}
se{{n}^{4}}\,\theta +{{\cos }^{4}}\theta +2\times se{{n}^{2}}\,\theta \times {{\cos }^{2}}\theta & = & {{(se{{n}^{2}}\,\theta +{{\cos }^{2}}\theta )}^{2}} \\
{} & = & {{1}^{2}} \\
{} & = & 1 \\
\end{array}\] - Ora,
\[\begin{array}{*{35}{l}}
{{(\cos \theta -sen\,\theta )}^{2}}+{{(\cos \theta +sen\,\theta )}^{2}} & = & ({{\cos }^{2}}\theta -2\times sen\,\theta \times \cos \theta +se{{n}^{2}}\,\theta )+({{\cos }^{2}}\theta +2\times sen\,\theta \times \cos \theta +se{{n}^{2}}\,\theta ) \\
{} & = & ({{\cos }^{2}}\theta +se{{n}^{2}}\,\theta )+({{\cos }^{2}}\theta +se{{n}^{2}}\,\theta ) \\
{} & = & 1+1 \\
{} & = & 2 \\
\end{array}\] - Ora,
\[\begin{array}{*{35}{l}}
(\cos \theta -sen\,\theta )+(\cos \theta +sen\,\theta )-1 & = & ({{\cos }^{2}}\theta -se{{n}^{2}}\,\theta )-1 \\
{} & = & -(1-{{\cos }^{2}}\theta )-se{{n}^{2}}\,\theta \\
{} & = & -se{{n}^{2}}\,\theta -se{{n}^{2}}\,\theta \\
{} & = & -2\times se{{n}^{2}}\,\theta \\
\end{array}\]
