Question

cos(8π +a)*cos(4π-a)+ sin(a +6π)*sin(a +4π)

Answer 1

$cos(8 \pi + \alpha )*cos(4 \pi - \alpha ) + sin( \alpha +6 \pi )*sin( \alpha +4 \pi )$$=cos(8 \pi + \alpha )*cos(4 \pi - \alpha ) + sin(6 \pi + \alpha )*sin(4 \pi + \alpha )= cos \alpha *cos \alpha$$+sin \alpha *sin \alpha = cos^{2} \alpha + sin^{2} \alpha =1$