Question

Show that $\sin {28}^o\cos {62}^o+\cos {28}^o\sin {62}^o=1$

Answer 1

$sin{28}^{0}cos{62}^0+cos{28}^{0}sin{62}^0=sin{28}^{0}cos({90}^0-{28}^0)+cos{28}^{0}sin({90}^0-{28}^0)$
$=sin{28}^{0}sin{28}^0+cos{28}^{0}cos{28}^0$
$=si{n}^2{28}^0+co{s}^2{28}^0$
$=1$
Hence proved.