Question

Transforme em produto 1 + sen16°

Answer 1

Por Prostaférese temos:

$sen(90) + sen(16) = 2*sen( \frac{90+16}{2}) *cos \frac{90-16}{2} )$

$sen(90) + sen(16)= 2*sen(53)*cos(37)$

Answer 2

$\textrm{Por prostaf\'erese, sabemos que:}\\\\ \mathrm{\sin{a}+\sin{b}=2.sin{\bigg(\dfrac{a+b}{2}\bigg).\cos{\bigg(\dfrac{a-b}{2}\bigg)}}}\\\\ \textrm{Portanto:}\\\\ \mathrm{1+\sin{16\º}=\sin{90\º}+\sin{16\º}=}\\\\ \mathrm{=2.\sin{\bigg(\dfrac{90\º+16\º}{2}\bigg)}.\cos{\bigg(\dfrac{90\º-16\º}{2}\bigg)}=}\\\\ \mathrm{=2.\sin{\bigg(\dfrac{106\º}{2}\bigg)}.\cos{\bigg(\dfrac{74\º}{2}\bigg)}=\mathbf{2.\sin{53\º}.\cos{37\º}}}$