Question

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

Answer 1

`(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^@)`