Question

What is the product of 4 left-bracket cosine (StartFraction 2 pi Over 3 EndFraction) + I sine (StartFraction 2 pi Over 3 EndFraction) right-bracket and 2 left-bracket cosine (StartFraction pi Over 3 EndFraction) + I sine (StartFraction pi Over 3 EndFraction right-bracket?
Answer its D im just adding this for others to find

Answer 1

The required product of the function is $8cos\frac{2\pi}{3}cos\frac{\pi}{3}+ 4i cos\frac{2\pi}{3}sin\frac{\pi}{3} + 2i sin\frac{2\pi}{3}cos\frac{\pi}{3} - 4i sin\frac{2\pi}{3}sin\frac{\pi}{3}$

We are to find the product of the expression

$a =4cos(\frac{2\pi}{3} )+isin(\frac{2\pi}{3})\\b=2cos(\frac{\pi}{3} )+isin(\frac{\pi}{3})$

The product of the functions is expressed as;

$ab = 4cos(\frac{2\pi}{3} )+isin(\frac{2\pi}{3})[2cos(\frac{\pi}{3} )+isin(\frac{\pi}{3}]\\ab=8cos\frac{2\pi}{3}cos\frac{\pi}{3}+ 4i cos\frac{2\pi}{3}sin\frac{\pi}{3} + 2i sin\frac{2\pi}{3}cos\frac{\pi}{3} + i^24i sin\frac{2\pi}{3}sin\frac{\pi}{3} \\ab=8cos\frac{2\pi}{3}cos\frac{\pi}{3}+ 4i cos\frac{2\pi}{3}sin\frac{\pi}{3} + 2i sin\frac{2\pi}{3}cos\frac{\pi}{3} - 4i sin\frac{2\pi}{3}sin\frac{\pi}{3}$

Note that i² = -1

Hence the required product of the function is $8cos\frac{2\pi}{3}cos\frac{\pi}{3}+ 4i cos\frac{2\pi}{3}sin\frac{\pi}{3} + 2i sin\frac{2\pi}{3}cos\frac{\pi}{3} - 4i sin\frac{2\pi}{3}sin\frac{\pi}{3}$

Learn more here: brainly.com/question/9258225