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

1 Answer

Apr 6, 2017

Multiplication this type is defined as:

#r_1(cos(theta_1)+isin(theta_1)*2(cos(theta_2)+isin(theta_2)=r_1r_2(cos(theta_1+theta_2)+isin(theta_1+theta_2))#

Explanation:

In this case, #r_1 = 3, r_2 = 2, theta_1 = 14^@", and "theta_2 =121^@#:

#3(cos14^circ+isin14^circ)*2(cos121^circ+isin121^circ) = 6(cos(135^@)+isin(135^@))#

This simplifies to:

#-3sqrt2+3sqrt2i#

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