\"*\"
where k = 0, 1, 2, 3... (n = root; equal to n - 1; dependent on the quantity of roots needed - 0 is included).

You are watching: Find the fourth roots of 16(cos 200° + i sin 200°).

In this case, n = 4.

Therefore, we readjust the polar equation we are given and also modify it to be fixed for the roots.

Part 2: resolving for source #1

To settle for root #1, do k = 0 and substitute every values right into the equation. ~ above the 2nd step, transform the measure in levels to the measure up in radians by multiplying the degrees measurement by

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and also simplify.

\"*\"

\"*\"

\"*\"

Root #1:

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Part 3: fixing for source #2

To deal with for root #2, follow the very same simplifying steps over but change k to k = 1.

\"*\"

\"*\"

\"*\"

Root #2:

\"*\"

Part 4: solving for source #3

To fix for root #3, monitor the exact same simplifying steps above but readjust k to k = 2.

\"*\"

\"*\"

\"*\"

Root #3:

\"*\"

Part 4: fixing for source #4

To deal with for root #4, monitor the exact same simplifying steps above but readjust k come k = 3.

\"*\"

\"*\"

\"*\"

Root #4:

\"*\"

The fourth roots of 16(cos 200° + i(sin 200°) are noted above.



Answer from: AdiGirlBrainly

The 4th roots the 16(cos 200° + ns sin 200°) space .

Step-by-step explanation:

The given expression is

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Using deMoivre\"s Theorem

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^\\frac14\" />

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The four roots space in the form of

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For n=1,

\"*\"

For n=2,

\"*\"

For n=3,

\"*\"

Therefore the 4th roots that 16(cos 200° + ns sin 200°) are .



Answer from: JaredO
y=2(cos(50=90n)=isin(50+90n))take n=0 ,y=2(cos(50)+isin(50)) take n=1,y=2(cos(140)+ isin (140))take n=2,y=2(cos(230)+ isin (230)) take n=3,y=2(cos(320)+isin(320)) 
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