# How do i write this in polar form? (cos(-2π/9)+isin(-2π/9))^3

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To write the expression ((\cos(-\frac{2\pi}{9}) + i\sin(-\frac{2\pi}{9}))^3) in polar form, you first convert the trigonometric expression to polar form using Euler's formula. Euler's formula states that (e^{i\theta} = \cos(\theta) + i\sin(\theta)).

Thus, (\cos(-\frac{2\pi}{9}) + i\sin(-\frac{2\pi}{9})) in polar form is (e^{-i\frac{2\pi}{9}}).

Now, raising (e^{-i\frac{2\pi}{9}}) to the power of 3, we get ((e^{-i\frac{2\pi}{9}})^3 = e^{-3i\frac{2\pi}{9}}).

So, the expression in polar form is (e^{-3i\frac{2\pi}{9}}).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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