Trigonometry help please?

Find the cube roots of 125(cos 288° + i sin 288°).

Answer 1

we have :
=(125*(cos288+isin288))^(1/3)
=5*(cos288+isin288)^(1/3)
and we know that cosx+isinx can be writeen as e^i(x) , thus :
=5*(e^(i288))^(1/3)
=5*e^(i*288*1/3) = 5*e^(i96)
=5*(cos96+isin96)

hope it helps

Answer 2

Hello,

let r = 125(cos(288°) + i sin(288°)). First, you can rewrite this complex number as

r = 125 * exp(i 288°)

This is called the polar representation with the complex exponential function. Now you want to find a number a that satisfies a^3 = r = 125 * exp(i 288°).

You know that by multiplying two complex numbers x and y, their radii multiply and their polar angles add. That means the number a you are looking for has radius 125^(1/3) = 5. For the polar angle b of the complex number a it holds 3b = 288°. It looks as this equation has only one solution but it has three, because 288° means the same on the complex plane as 288°+360° and 288°+720°. So, the polar angles of a can be 288°/3=96°, (288°+360°)/3 = 216° and (288° + 720°)/3 = 336°. Therefore,
all cube roots of your number are
5(cos 96° + i sin 96°),
5(cos 216° + i sin 216°) and
5(cos 336° + i sin 336°).

Answer 3

³√125 = 5 and 288/3 = 96 and 360/3 = 120, so
5(cos 96 + i sin 96)
5(cos 216 + i sin 216)
5(cos 336 + i sin 336)

Answer 4

5(cos( (288° + k360°)/3) + i sin ( (288° + k360°)/3 ) ) for k=0,1,2

z_0 = 5(cos 96° + i sin 96°)

z_1 = 5(cos 216° + i sin 216°)

z_2 = 5(cos 336° + i sin 336°)

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