The given equation is:

$y_n = a \cos(n\theta) + b \sin(n\theta),$

where $a$, $b$, and $\theta$ are constants. To eliminate the arbitrary constants $a$ and $b$, we need to form a difference equation involving $y_n$.

Steps:

1. Express $y_{n+1}$: Using the given equation: $y_{n+1} = a \cos((n+1)\theta) + b \sin((n+1)\theta).$ Expand using trigonometric angle addition formulas: $\cos((n+1)\theta) = \cos(n\theta)\cos\theta - \sin(n\theta)\sin\theta,$ $\sin((n+1)\theta) = \sin(n\theta)\cos\theta + \cos(n\theta)\sin\theta.$ Substitute into $y_{n+1}$: [ y_{n+1} = a[\cos(n\theta)\cos\theta - \sin(n\theta)\sin\theta] + b[\sin(n\theta)\cos\theta + \cos(n\theta)\sin\theta]. ] Simplify: $y_{n+1} = (a\cos\theta + b\sin\theta)\cos(n\theta) + (b\cos\theta - a\sin\theta)\sin(n\theta).$

2. Express $y_{n-1}$: Similarly, expand $y_{n-1}$: [ y_{n-1} = a[\cos(n\theta)\cos\theta + \sin(n\theta)\sin\theta] + b[\sin(n\theta)\cos\theta - \cos(n\theta)\sin\theta]. ] Simplify: $y_{n-1} = (a\cos\theta + b\sin\theta)\cos(n\theta) - (b\cos\theta - a\sin\theta)\sin(n\theta).$

3. Eliminate $a$ and $b$: From $y_n$, $y_{n+1}$, and $y_{n-1}$, observe that $a$ and $b$ are combined in terms involving $\cos(n\theta)$ and $\sin(n\theta)$. Use the fact that: $y_{n+1} + y_{n-1} = 2(a\cos\theta + b\sin\theta)\cos(n\theta),$ and $y_{n+1} - y_{n-1} = 2(b\cos\theta - a\sin\theta)\sin(n\theta).$ Combine these equations into a single second-order difference equation: $y_{n+1} - 2\cos\theta y_n + y_{n-1} = 0.$

Final Difference Equation:

$y_{n+1} - 2\cos\theta y_n + y_{n-1} = 0.$

This equation eliminates the constants $a$ and $b$, leaving a purely recursive relationship.

Would you like a detailed breakdown of any step?

Related Questions:

1. How can trigonometric identities simplify recursive equations?
2. What methods solve second-order linear difference equations?
3. How does the angle addition formula apply to recursive problems?
4. Can this approach be extended to equations with exponential or logarithmic terms?
5. What are practical applications of such difference equations?

Tip:

When dealing with trigonometric functions in recursive relations, leveraging symmetry and periodicity can greatly simplify the process.