How do you perform the operation in trigonometric form #(12(cos52+isin52))/(3(cos110+isin110))#?

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Answer 1 · 2017-06-01T14:45:02

#(12(cos52+isin52))/(3(cos110+isin110))=4(cos58-isin58)#

#(12(cos52+isin52))/(3(cos110+isin110))#

= #(12(cos52+isin52)(cos110-isin110))/(3(cos110+isin110)(cos110-isin110))#

= #(12(cos52cos110+isin52cos110-icos52sin110-i^2sin52sin110))/(3(cos^2 110-i^2sin^2 110))#

= #(12(cos52cos110+sin52sin110+i(sin52cos110-cos52sin110)))/(3(cos^2 110+sin^2 110))#

= #(12(cos(52-110)+isin(52-110)))/(3(cos^2 110+sin^2 110))#

= #(12(cos(-58)+isin(-58)))/3#

= #4(cos58-isin58)#

Answer 2 · 2017-06-01T18:33:15

#(12(cos52+isin52))/(3(cos110+isin110))=4(cos58-isin58)#

#(12(cos52+isin52))/(3(cos110+isin110))=4xx(cos52+isin52)/(cos110+isin110)#

#cos52+isin52=e^(52i)#

#cos110+isin110=e^(110i)#

#therefore4xx(cos52+isin52)/(cos110+isin110)=4xxe^(52i)/e^(110i)=4e^(-58i)#

#4e^(-58i)=4(cos-58+isin-58)=4(cos58-isin58)#