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

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Answer 1 · 2017-04-06T16:23:08

Multiplication this type is defined as:

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

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:

$-3sqrt2+3sqrt2i$

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