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

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Answer 1 · 2017-03-22T15:04:32

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

#(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^@)#