Find a formula for the nth derivative of sin(ax+b) and cos(ax+b).

To find a formula for the nth derivative of \( \sin(ax + b) \) and \( \cos(ax + b) \), let's proceed step by step.

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### Part 1: nth Derivative of \( \sin(ax + b) \)
We start by observing the derivatives of \( \sin(ax + b) \):

1. First derivative:
\[
\frac{d}{dx}\sin(ax + b) = a\cos(ax + b)
\]

2. Second derivative:
\[
\frac{d^2}{dx^2}\sin(ax + b) = -a^2\sin(ax + b)
\]

3. Third derivative:
\[
\frac{d^3}{dx^3}\sin(ax + b) = -a^3\cos(ax + b)
\]

4. Fourth derivative:
\[
\frac{d^4}{dx^4}\sin(ax + b) = a^4\sin(ax + b)
\]

From these derivatives, we see a cyclic pattern:
- \( \sin(ax + b) \) alternates between \( \sin(ax + b) \) and \( \cos(ax + b) \), with alternating signs and increasing powers of \( a \).

Generalizing:
- When \( n \mod 4 = 0 \): \( \frac{d^n}{dx^n}\sin(ax + b) = a^n\sin(ax + b) \)
- When \( n \mod 4 = 1 \): \( \frac{d^n}{dx^n}\sin(ax + b) = a^n\cos(ax + b) \)
- When \( n \mod 4 = 2 \): \( \frac{d^n}{dx^n}\sin(ax + b) = -a^n\sin(ax + b) \)
- When \( n \mod 4 = 3 \): \( \frac{d^n}{dx^n}\sin(ax + b) = -a^n\cos(ax + b) \)

Thus, the formula is:
\[
\frac{d^n}{dx^n}\sin(ax + b) = a^n \times
\begin{cases}
\sin(ax + b), & \text{if } n \mod 4 = 0 \\
\cos(ax + b), & \text{if } n \mod 4 = 1 \\
-\sin(ax + b), & \text{if } n \mod 4 = 2 \\
-\cos(ax + b), & \text{if } n \mod 4 = 3
\end{cases}
\]

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### Part 2: nth Derivative of \( \cos(ax + b) \)
We now compute the derivatives of \( \cos(ax + b) \):

1. First derivative:
\[
\frac{d}{dx}\cos(ax + b) = -a\sin(ax + b)
\]

2. Second derivative:
\[
\frac{d^2}{dx^2}\cos(ax + b) = -a^2\cos(ax + b)
\]

3. Third derivative:
\[
\frac{d^3}{dx^3}\cos(ax + b) = a^3\sin(ax + b)
\]

4. Fourth derivative:
\[
\frac{d^4}{dx^4}\cos(ax + b) = a^4\cos(ax + b)
\]

Similarly, we see a cyclic pattern:
- \( \cos(ax + b) \) alternates between \( \cos(ax + b) \) and \( \sin(ax + b) \), with alternating signs and increasing powers of \( a \).

Generalizing:
- When \( n \mod 4 = 0 \): \( \frac{d^n}{dx^n}\cos(ax + b) = a^n\cos(ax + b) \)
- When \( n \mod 4 = 1 \): \( \frac{d^n}{dx^n}\cos(ax + b) = -a^n\sin(ax + b) \)
- When \( n \mod 4 = 2 \): \( \frac{d^n}{dx^n}\cos(ax + b) = -a^n\cos(ax + b) \)
- When \( n \mod 4 = 3 \): \( \frac{d^n}{dx^n}\cos(ax + b) = a^n\sin(ax + b) \)

Thus, the formula is:
\[
\frac{d^n}{dx^n}\cos(ax + b) = a^n \times
\begin{cases}
\cos(ax + b), & \text{if } n \mod 4 = 0 \\
-\sin(ax + b), & \text{if } n \mod 4 = 1 \\
-\cos(ax + b), & \text{if } n \mod 4 = 2 \\
\sin(ax + b), & \text{if } n \mod 4 = 3
\end{cases}
\]

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### Final Answer:
1. \( \frac{d^n}{dx^n}\sin(ax + b) \):
- Follows the formula above with cyclic behavior every 4 derivatives.

2. \( \frac{d^n}{dx^n}\cos(ax + b) \):
- Also follows the respective cyclic formula.

This provides the general formulas for the nth derivative of both \( \sin(ax + b) \) and \( \cos(ax + b) \).