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Let's break down this problem step by step, starting with the use of **De Moivre's Theorem** and then moving to the equation.

### Step 1: **Using De Moivre's Theorem to show** \( z^n - z^{-n} = 2i \sin(n\theta) \)

We are given \( z = \cos\theta + i\sin\theta \), which is the polar form of a complex number, where:
\[
z = e^{i\theta}
\]
From **De Moivre's Theorem**, we know that for any integer \( n \):
\[
z^n = \left( \cos \theta + i \sin \theta \right)^n = \cos(n\theta) + i \sin(n\theta)
\]
Similarly,
\[
z^{-n} = \left( \cos \theta + i \sin \theta \right)^{-n} = \cos(-n\theta) + i \sin(-n\theta) = \cos(n\theta) - i \sin(n\theta)
\]
Now, subtract the two expressions:
\[
z^n - z^{-n} = \left( \cos(n\theta) + i \sin(n\theta) \right) - \left( \cos(n\theta) - i \sin(n\theta) \right)
\]
This simplifies to:
\[
z^n - z^{-n} = 2i \sin(n\theta)
\]
Thus, we've shown the required expression.

### Step 2: **Deducing** \( 16 \sin^5 \theta = \sin 5\theta - 5 \sin 3\theta + 10 \sin \theta \)

We know that De Moivre's Theorem allows us to express powers of sine and cosine in terms of multiple angles. Expanding \( \sin(5\theta) \) using standard trigonometric identities, we get:
\[
\sin(5\theta) = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta
\]
Rearranging this gives:
\[
16\sin^5\theta = \sin(5\theta) - 5\sin(3\theta) + 10\sin(\theta)
\]
Thus, this identity has been derived.

### Step 3: **Solving the equation** \( 4 \sin^5 \theta + \sin(5\theta) = 0 \)

Using the result from Step 2:
\[
4\sin^5\theta + \sin(5\theta) = 0
\]
Substitute \( \sin(5\theta) = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta \):
\[
4\sin^5\theta + (5\sin\theta - 20\sin^3\theta + 16\sin^5\theta) = 0
\]
Simplify the equation:
\[
(4 + 16)\sin^5\theta - 20\sin^3\theta + 5\sin\theta = 0
\]
\[
20\sin^5\theta - 20\sin^3\theta + 5\sin\theta = 0
\]
Factor out \( 5\sin\theta \):
\[
5\sin\theta(4\sin^4\theta - 4\sin^2\theta + 1) = 0
\]
This gives two cases:
1. \( \sin\theta = 0 \), which gives \( \theta = 0, \pi, 2\pi \)
2. Solve \( 4\sin^4\theta - 4\sin^2\theta + 1 = 0 \)

Let \( x = \sin^2\theta \), so the equation becomes:
\[
4x^2 - 4x + 1 = 0
\]
This is a quadratic equation:
\[
\Delta = (-4)^2 - 4(4)(1) = 16 - 16 = 0
\]
Since the discriminant is zero, there is one repeated solution:
\[
x = \frac{4}{8} = \frac{1}{2}
\]
Thus, \( \sin^2\theta = \frac{1}{2} \), which gives \( \sin\theta = \pm \frac{\sqrt{2}}{2} \).

From this, \( \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).

### Solutions:
- \( \theta = 0, \pi, 2\pi \)
- \( \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \)

Would you like more details or any further clarification?

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Here are 5 related questions to expand your understanding:
1. What is the general form of De Moivre's Theorem?
2. How can De Moivre's Theorem be used to compute roots of complex numbers?
3. What are some other trigonometric identities involving powers of sine and cosine?
4. Can you deduce the general expression for \( \sin(7\theta) \) using similar methods?
5. How can you solve a general trigonometric equation like \( 4 \sin^5 \theta + \sin(5\theta) = 0 \) without substitution?

**Tip:** When solving trigonometric equations, it is often helpful to express higher powers of sine and cosine in terms of multiple angles using identities.

Given that z = cosθ + i sinθ, use De Moivre's Theorem to show that z^n - z^(-n) = 2i sin(nθ). Hence, deduce that 16 sin^5(θ) = sin(5θ) - 5 sin(3θ) + 10 sin(θ). Find all the solutions of the equation 4 sin^5(θ) + sin(5θ) = 0 which lie in the interval 0 ≤ θ ≤ 2π.

Learn how to use De Moivre's Theorem to solve complex equations involving trigonometric identities. This problem includes step-by-step instructions on deducing identities like 16 sin^5(θ) and solving the equation 4 sin^5(θ) + sin(5θ) = 0.

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