How do you simplify #2(cos 240+isin240)*3(cos60+isin60)# and express the result in rectangular form? Precalculus Complex Numbers in Trigonometric Form Trigonometric Form of Complex Numbers 1 Answer Narad T. Dec 7, 2016 The answer is #=3-i3sqrt3# Explanation: #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# Answer link Related questions How do I find the trigonometric form of the complex number #-1-isqrt3#? How do I find the trigonometric form of the complex number #3i#? How do I find the trigonometric form of the complex number #3-3sqrt3 i#? How do I find the trigonometric form of the complex number #sqrt3 -i#? How do I find the trigonometric form of the complex number #3-4i#? How do I convert the polar coordinates #3(cos 210^circ +i\ sin 210^circ)# into rectangular form? What is the modulus of the complex number #z=3+3i#? What is DeMoivre's theorem? How do you find a trigonometric form of a complex number? Why do you need to find the trigonometric form of a complex number? See all questions in Trigonometric Form of Complex Numbers Impact of this question 3958 views around the world You can reuse this answer Creative Commons License