Question

Calcular sec (15°)

Me ajudem pfvr

Answer 1

Resposta:

$\boxed{\mathtt{\sqrt{6} - \sqrt{2}}}$

Explicação passo-a-passo:

$\\ \displaystyle \mathtt{\sec 15^o = \frac{1}{\cos 15^o}} \\\\\\ \mathtt{\sec 15^o = \frac{1}{\cos (45^o - 30^o)}} \\\\\\ \mathtt{\sec 15^o = \frac{1}{\cos 45^o \cdot \cos 30^o + \sin 45^o \cdot \sin 30^o}} \\\\\\ \mathtt{\sec 15^o = \frac{1}{\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}}}$

$\\ \displaystyle \mathtt{\sec 15^o = \frac{1}{\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}}} \\\\\\ \mathtt{\sec 15^o = \frac{1}{\frac{\left ( \sqrt{6} + \sqrt{2} \right )}{4}}} \\\\\\ \mathtt{\sec 15^o = 1 \div \frac{\left ( \sqrt{6} + \sqrt{2} \right )}{4}} \\\\\\ \mathtt{\sec 15^o = 1 \times \frac{4}{\left ( \sqrt{6} + \sqrt{2} \right )}}$

$\\ \displaystyle \mathtt{\sec 15^o = \frac{4}{\left ( \sqrt{6} + \sqrt{2} \right )} \times \frac{\left ( \sqrt{6} - \sqrt{2} \right )}{\left ( \sqrt{6} - \sqrt{2} \right )}} \\\\\\ \mathtt{\sec 15^o = \frac{4 \cdot \left ( \sqrt{6} - \sqrt{2} \right )}{\left (6 - 2 \right )}} \\\\\\ \mathtt{\sec 15^o = \frac{4 \cdot \left ( \sqrt{6} - \sqrt{2} \right )}{4}} \\\\\\ \boxed{\boxed{\mathsf{\sec 15^o = \sqrt{6} - \sqrt{2}}}}$

A saber,

$\\ \displaystyle \mathtt{\bullet \qquad \sec x = \frac{1}{\cos x}} \\\\\\ \mathtt{\bullet \qquad \cos(x - y) = \cos x \cdot \cos y - \sin x \cdot \sin y}$