Let's solve this problem using De Moivre's Theorem.

Problem:

We need to show that: $\sin(5\theta) = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$

Solution:

De Moivre's Theorem states: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta).$ For $n = 5$, we expand $(\cos\theta + i\sin\theta)^5$: $(\cos\theta + i\sin\theta)^5 = \cos(5\theta) + i\sin(5\theta).$ Using the Binomial Theorem, expand $(\cos\theta + i\sin\theta)^5$: $(\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5\cos^4\theta(i\sin\theta) + 10\cos^3\theta(i\sin\theta)^2 + 10\cos^2\theta(i\sin\theta)^3 + 5\cos\theta(i\sin\theta)^4 + (i\sin\theta)^5.$ Simplify powers of $i$ (using $i^2 = -1, i^3 = -i, i^4 = 1$): $= \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta.$ Group real and imaginary terms:

• Real part: $\cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta.$
• Imaginary part: $5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta.$ From De Moivre's theorem, the imaginary part gives $\sin(5\theta)$: $\sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta.$

Step 2: Express $\cos^2\theta$ in terms of $\sin^2\theta$

Using $\cos^2\theta = 1 - \sin^2\theta$, substitute $\cos^2\theta$ and $\cos^4\theta = (1 - \sin^2\theta)^2$: $\cos^4\theta = (1 - \sin^2\theta)^2 = 1 - 2\sin^2\theta + \sin^4\theta.$ Now substitute back into the equation for $\sin(5\theta)$: $\sin(5\theta) = 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10(1 - \sin^2\theta)\sin^3\theta + \sin^5\theta.$

Step 3: Expand and simplify

Expand the terms:

1. First term: $5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta = 5\sin\theta - 10\sin^3\theta + 5\sin^5\theta.$
2. Second term: $-10(1 - \sin^2\theta)\sin^3\theta = -10\sin^3\theta + 10\sin^5\theta.$
3. Third term: $\sin^5\theta.$

Combine all terms: $\sin(5\theta) = (5\sin\theta) + (-10\sin^3\theta - 10\sin^3\theta) + (5\sin^5\theta + 10\sin^5\theta + \sin^5\theta).$ $\sin(5\theta) = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta.$

Final Result:

$\sin(5\theta) = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta.$

This verifies the given equation.

Would you like me to expand further or clarify any step?

Related Questions:

1. What is De Moivre's Theorem and how is it derived?
2. How can we use $\cos^2\theta + \sin^2\theta = 1$ in trigonometric expansions?
3. What is the binomial theorem, and how does it apply to $(a + b)^n$?
4. How can you derive $\cos(5\theta)$ using De Moivre’s Theorem?
5. What are some practical applications of De Moivre’s Theorem?

Tip:

Always break down the powers of $\sin\theta$ or $\cos\theta$ into smaller terms using fundamental identities like $\cos^2\theta = 1 - \sin^2\theta$ for simplifications.