How do you simplify #2(cos 240+isin240)*3(cos60+isin60)# and express the result in rectangular form?

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Answer 1 · 2016-12-07T08:18:59

The answer is #=3-i3sqrt3#

#cos240=-1/2#

#sin240=-sqrt3/2#

#cos60=1/2#

#sin60=sqrt3/2#

and #i^2=-1#

So,

#2(cos240+isin204)*3(cos60+isin60)#

#=6(-1/2-isqrt3/2)(1/2+isqrt3/2)#

#=6(-1/4-isqrt3/4-isqrt3/4-i^2 3/4)#

#=6(1/2-isqrt3/2)#

#=3-i3sqrt3#