How do you perform the operation in trigonometric form #(5/3(cos(140)+isin(140)))(2/3(cos(60)+isin(60)))#?

1 Answer
Shwetank Mauria
Dec 20, 2016

#(5/3(cos140^@+isin140^@))(2/3(cos60^@+isin60^@))=10/9(cos200^@+isin200^@)#

Explanation:

#(5/3(cos140^@+isin140^@))(2/3(cos60^@+isin60^@))#

= #10/9[cos140^@cos60^@+icos140^@sin60^@+isin140^@cos60^@+i^2sin140^@sin60^@]#

= #10/9(cos140^@cos60^@-sin140^@sin60^@+icos140^@sin60^@+isin140^@cos60^@]#

= #10/9(cos(140^@+60^@)+isin(140^@+60^@))#

= #10/9(cos200^@+isin200^@)#

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