find 2(cos 240+isin 240) ^4 (answer choices below)

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ParkDweller ParkDweller · Answer 1 · 2020-04-03T09:34:36+02:00

1. C. -512√3+512i

2. B. 16(cos240°+i sin240°)

3. D. 3√2+3√6i, -3√2-3√6i

4. A. cos60°+i sin60°, cos180°+i sin180°, cos300°+i sin300°

5. D. 2√3(cos π/6+i sin π/6), 2√3(cos 7π/6+i sin 7π/6)

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facundo3141592 facundo3141592 · Answer 2 · 2022-04-07T11:11:41+02:00

We will see that the equivalent expression is:

[tex]8*(cos(240\°) + i*sin(240\°))[/tex]

So the correct option is the first one.

We have the expression:

[2*(cos(240°) + i*sin(240°))]^4

Remember that Euler's formula says that:

[tex]e^{ix} = cos(x) + i*sin(x)[/tex]

Then we can rewrite our expression as:

[tex][2*(cos(240\°) + i*sin(240\°)]^4 = [2*e^{i*240\°}]^4[/tex]

Now we distribute the exponent:

[tex]2^4*e^{4*i*240\°} = 8*e^{i*960\°}[/tex]

Now, we need to find an angle equivalent to 960°.

Remember that the period of the trigonometric functions is 360°, then we can rewrite:

960° - 2*360° = 240°

This means that 960° is equivalent to 240°. Then we can write:

[tex]8*e^{i*960\°} = 8*e^{i*240\°} = 8*(cos(240\°) + i*sin(240\°))[/tex]

So the correct option is the first one.

If you want to learn more about complex numbers, you can read:

https://brainly.com/question/10662770