Sine and Cosine Maclaurin Polynomials

Why Sine and Cosine Have Distinct Maclaurin Patterns

The derivatives of sine and cosine repeat in a short cycle. At zero, that cycle determines which powers appear in their Maclaurin polynomials, what the signs of the terms are, and which factorial belongs in each denominator.

A Maclaurin polynomial is a Taylor polynomial centered at \(x=0\). For sine and cosine, its terms can be constructed directly from the function values and derivatives at zero. The familiar derivative formulas \( (\sin x)'=\cos x \) and \( (\cos x)'=-\sin x \) assume that angles are measured in radians.

The factorial is essential: the coefficient of \(x^j\) is \(f^{(j)}(0)/j!\), not simply \(f^{(j)}(0)\). The repeating derivatives make these coefficients especially straightforward to find.

The Derivative Cycles at Zero

Differentiate sine repeatedly, remembering that the derivative of cosine is negative sine. The functions return to their starting point after four derivatives. Their values at zero follow from \(\sin 0=0\) and \(\cos 0=1\).

After the third derivative, the next derivative is sine again, so this pattern repeats. For cosine, the cycle begins at cosine and has a different alignment:

Thus, sine has nonzero derivatives at zero only at odd orders, while cosine has nonzero derivatives at zero only at even orders. The values of the nonzero derivatives alternate between \(1\) and \(-1\). Applying the Maclaurin coefficient formula converts each nonzero derivative into a term by dividing by the factorial of its order.

Building the Polynomials

For sine, the nonzero derivative orders are \(1,3,5,7,\ldots\). At these orders the derivative values are \(1,-1,1,-1,\ldots\). The resulting terms are \(x/1!\), \(-x^3/3!\), \(x^5/5!\), and so on.

For \(n=0\), the sine Maclaurin polynomial is \(P_0^{\sin}(x)=0\). In the formula, the largest allowed \(k\) is chosen so that \(2k+1\leq n\); terms above degree \(n\) are not included.

For cosine, the nonzero derivative orders are \(0,2,4,6,\ldots\), with values \(1,-1,1,-1,\ldots\). Its terms begin \(1\), \(-x^2/2!\), \(x^4/4!\), and \(-x^6/6!\).

The floor symbol \(\lfloor r\rfloor\) means the greatest integer less than or equal to \(r\). It sets the final index so that the largest power included does not exceed the requested degree. These formulas describe finite polynomials; constructing one does not require claiming that an infinite series converges.

Why the Factorial Denominators Appear

The factorial denominator ensures that differentiating a polynomial the required number of times recovers the prescribed derivative at zero. To see the key point, consider a term \(a_jx^j\). Differentiating it \(j\) times gives \(j!a_j\). If \(a_j=f^{(j)}(0)/j!\), that value is exactly \(f^{(j)}(0)\).

Terms of other degrees do not change this matching at zero. A term \(a_rx^r\) with \(r>j\), after \(j\) derivatives, still contains a positive power of \(x\), so its value at zero is zero. A term with \(r<j\) has become zero after \(j\) derivatives. Therefore the \(j\)th derivative of the Maclaurin polynomial, evaluated at zero, equals \(f^{(j)}(0)\) for each order included.

Worked Examples

Reading the Degree and the Missing Powers

“Degree \(n\)” sets an upper limit; it does not promise a nonzero term at every power up to \(n\). For instance, the degree-5 sine polynomial ends with an \(x^5\) term, while its \(x^2\) and \(x^4\) coefficients vanish. A degree-5 cosine polynomial instead ends at \(x^4\), because its \(x^5\) coefficient is zero. In fact, it is the same polynomial as the degree-4 cosine polynomial.

A useful way to construct either polynomial is to write the derivative values in order, including zeros, and then divide each value by the factorial for that order. This prevents two common oversights: skipping a derivative order and assigning a sign to the wrong power. Once the zero entries are visible, the missing powers are part of the calculation rather than gaps to fill.

A Common Sign and Factorial Error

A tempting but incorrect shortcut is to assign an alternating sign to every consecutive power. For example, one might write the degree-5 sine polynomial as

This is wrong: it puts nonzero coefficients on even powers, whose sine derivatives at zero are zero. It also assigns the wrong signs and denominators to several of the odd-power terms. The correction is to use the derivative cycle first:

Thus the correct result is \(x-x^3/3!+x^5/5!\). The signs alternate across the nonzero sine terms, whose powers differ by two; they do not alternate across a list that includes the missing even powers.