Question

what is $3 cos (60^{0} )+ i sin 60^{0} cdot \frac{1}{2} cos (15^{0}) + i sin (15^{0})$?
$\frac{3}{2}cos(75^{\circ}) + i sin (75^{\circ})$
$3\frac{1}{2}cos(75^{\circ}) + i sin (75^{\circ})$
$3\frac{1}{2}cos(45^{\circ}) + i sin (45^{\circ})$
$\frac{3}{2}cos(45^{\circ}) + i sin (45^{\circ})$

Explanation:

Step1: Multiply the moduli

$3 \times \frac{1}{2} = \frac{3}{2}$

Step2: Add the arguments

$60^\circ + 15^\circ = 75^\circ$

Step3: Apply cis multiplication rule

For complex numbers $r_1(\cos\theta_1 + i\sin\theta_1)$ and $r_2(\cos\theta_2 + i\sin\theta_2)$, their product is $r_1r_2(\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2))$. Substitute the values from Step1 and Step2.
$\frac{3}{2}[\cos(75^\circ) + i\sin(75^\circ)]$

Answer:

$\frac{3}{2}[\cos(75^\circ) + i\sin(75^\circ)]$ (first option)