What are the nth derivatives of sin(ax+b) and cos(ax+b), and why does nπ/2 appear?

In one line

\(D^n\sin(ax+b)=a^n\sin\left(ax+b+\dfrac{n\pi}2\right)\) and \(D^n\cos(ax+b)=a^n\cos\left(ax+b+\dfrac{n\pi}2\right)\): each derivative multiplies by \(a\) and shifts the angle by \(90^\circ\).

Explained simply

Differentiate \(\sin x\) four times: \(\cos x,-\sin x,-\cos x,\sin x\). It goes round in a cycle of four. Instead of remembering which sign comes when, note that \(\cos x=\sin(x+\pi/2)\). So every differentiation just adds \(\pi/2\) to the angle. After \(n\) steps the angle has grown by \(n\pi/2\). Think of turning a dial a quarter turn each time.

Technically, \(\dfrac{d}{dx}\sin(ax+b)=a\cos(ax+b)=a\sin\left(ax+b+\dfrac\pi2\right)\). Repeating, \(y_n=a^n\sin\left(ax+b+\dfrac{n\pi}2\right)\), which can be proved by induction.

Signal and vibration problems are full of sines and cosines, so this one formula is used constantly. The \(a^n\) factor also tells you something physical: a faster oscillation (larger \(a\)) has derivatives that grow much faster, which is why high-frequency noise is so strongly amplified by differentiation in electronic circuits.

How it works

1. Write \(y_1=a\cos(ax+b)\) as \(a\sin\left(ax+b+\frac\pi2\right)\).
2. Differentiate again: \(y_2=a^2\sin\left(ax+b+\frac{2\pi}2\right)\).
3. General: \(y_n=a^n\sin\left(ax+b+\frac{n\pi}2\right)\). Similarly for cosine.
4. For powers or products of sines and cosines, first convert to sums using identities like \(\sin^2x=\frac{1-\cos2x}2\) or \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\), then apply the formula to each term.

Key formula

\(D^n\sin(ax+b)=a^n\sin\left(ax+b+\dfrac{n\pi}{2}\right)\), \(\ D^n\cos(ax+b)=a^n\cos\left(ax+b+\dfrac{n\pi}{2}\right)\). Angles in radians.

Quick example

• \(y=\sin3x\): \(y_n=3^n\sin\left(3x+\frac{n\pi}2\right)\). For \(n=2\): \(9\sin(3x+\pi)=-9\sin3x\). Direct: \(y_1=3\cos3x\), \(y_2=-9\sin3x\). Correct.
• \(y=\cos x\cos3x=\frac12(\cos4x+\cos2x)\): \(y_n=\frac12\left[4^n\cos\left(4x+\frac{n\pi}2\right)+2^n\cos\left(2x+\frac{n\pi}2\right)\right]\).
• \(y=\sin^2x=\frac12-\frac12\cos2x\): for \(n\ge1\), \(y_n=-2^{n-1}\cos\left(2x+\frac{n\pi}2\right)\).

Common mistakes

• Adding \(n\pi\) instead of \(\dfrac{n\pi}{2}\).
• Forgetting \(a^n\), for example writing \(D^n\sin3x=\sin(3x+n\pi/2)\).
• Applying the formula to \(\sin^3x\) directly; first use \(\sin^3x=\frac{3\sin x-\sin3x}4\).

How MU asks it

3-mark parts: "Find \(y_n\) if \(y=\sin^3x\)" or "\(y=\cos x\cos2x\cos3x\)"; for the second, convert the product to a sum of four cosines first. For full marks, show the conversion identity, then apply the standard result to each term with its own \(a\).