1) Teniendo en cuenta que $$ \forall \alpha \in \mathbf{R}: \operatorname{sen}^{2} \alpha+\cos ^{2} \alpha=1 $$ y utilizando la máquina de calcular verifique para el caso $$ \begin{array}{c}\alpha=\frac{\pi}{2}, \alpha-\frac{7}{6} \pi, \alpha=\frac{\pi}{3}+2 k \pi \ \alpha=-\frac{\pi}{2}, \alpha=-\frac{\pi}{6}\end{array} $$

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Para verificar la identidad $$ \sin^2 \alpha + \cos^2 \alpha = 1 $$ para los valores de $$ \alpha $$ dados, calcularemos $$ \sin^2 \alpha $$ y $$ \cos^2 \alpha $$ para cada caso y sumaremos los resultados.

1. **Para $$ \alpha = \frac{\pi}{2} $$**:
   $$
   \sin\left(\frac{\pi}{2}\right) = 1 \quad \text{y} \quad \cos\left(\frac{\pi}{2}\right) = 0
   $$
   $$
   \sin^2\left(\frac{\pi}{2}\right) + \cos^2\left(\frac{\pi}{2}\right) = 1^2 + 0^2 = 1
   $$

2. **Para $$ \alpha = -\frac{7}{6} \pi $$**:
   $$
   \sin\left(-\frac{7}{6} \pi\right) = -\frac{1}{2} \quad \text{y} \quad \cos\left(-\frac{7}{6} \pi\right) = -\frac{\sqrt{3}}{2}
   $$
   $$
   \sin^2\left(-\frac{7}{6} \pi\right) + \cos^2\left(-\frac{7}{6} \pi\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1
   $$

3. **Para $$ \alpha = \frac{\pi}{3} + 2k\pi $$** (donde $$ k $$ es un entero):
   $$
   \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \quad \text{y} \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}
   $$
   $$
   \sin^2\left(\frac{\pi}{3}\right) + \cos^2\left(\frac{\pi}{3}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1
   $$

4. **Para $$ \alpha = -\frac{\pi}{2} $$**:
   $$
   \sin\left(-\frac{\pi}{2}\right) = -1 \quad \text{y} \quad \cos\left(-\frac{\pi}{2}\right) = 0
   $$
   $$
   \sin^2\left(-\frac{\pi}{2}\right) + \cos^2\left(-\frac{\pi}{2}\right) = (-1)^2 + 0^2 = 1
   $$

5. **Para $$ \alpha = -\frac{\pi}{6} $$**:
   $$
   \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \quad \text{y} \quad \cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}
   $$
   $$
   \sin^2\left(-\frac{\pi}{6}\right) + \cos^2\left(-\frac{\pi}{6}\right) = \left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1
   $$

En todos los casos, hemos verificado que $$ \sin^2 \alpha + \cos^2 \alpha = 1 $$.
En todos los casos, se verifica que $$ \sin^2 \alpha + \cos^2 \alpha = 1 $$.

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