How do you calculate (3(cos14^circ+isin14^circ))/(2(cos121^circ+isin121^circ)?

1 Answer
Mar 22, 2017

(3(cos14^@+isin14^@))/(2(cos121^@+isin121^@))=3/2(cos107^@-isin107^@)

Explanation:

(3(cos14^@+isin14^@))/(2(cos121^@+isin121^@)

= 3/2xx((cos14^@+isin14^@))/((cos121^@+isin121^@))xx((cos121^@-isin121^@))/((cos121^@-isin121^@)

= 3/2(cos14^@cos121^@-i^2sin14^@sin121^@-icos14^@sin121^@+isin14^@cos121^@)/(cos^2 121^@-i^2sin121^@)

= 3/2(cos14^@cos121^@+sin14^@sin121^@+i(sin14^@cos121^@-cos14^@sin121^@))/(cos^2 121^@+sin121^@)

= 3/2(cos(14^@-121^@)+isin(14^@-121^@))/1

= 3/2(cos(-107^@)+isin(-107^@))

= 3/2(cos107^@-isin107^@)