Maclaurin Polynomials for Sine

Why Build a Maclaurin Polynomial for Sine?

The sine function is familiar, but its values at many inputs are not simple to calculate exactly. A Maclaurin polynomial gives a polynomial model of sine near zero. Once its coefficients are known, evaluating that model uses only powers, multiplication, and addition.

The central task is to determine which powers appear and what their coefficients are. Sine’s derivatives repeat in a short cycle, so their values at zero determine the polynomial systematically. In particular, every even-power coefficient is zero.

Maclaurin Polynomial and the Derivative Cycle

For a function \(f\) with the required derivatives at zero, its Maclaurin polynomial of degree \(n\) is

Here \(f^{(0)}=f\), and \(k!\) is the factorial of \(k\), with \(0!=1\). The degree cutoff means that only powers \(x^k\) with \(k\le n\) are included. Some of their coefficients may be zero, so the polynomial’s actual degree can be less than \(n\).

For \(f(x)=\sin x\), differentiation gives a four-function cycle:

Evaluating this cycle at zero uses \(\sin 0=0\) and \(\cos 0=1\). The derivative values begin

Thus the coefficients of even powers vanish. The nonzero entries occur at odd derivative orders, with alternating signs. Dividing each derivative value by its factorial gives the coefficients of the polynomial.

Deriving the General Form

The cycle gives a direct rule for every derivative at zero. At an even order \(2m\), the derivative is either \(\sin x\) or \(-\sin x\), so its value at zero is zero. At an odd order \(2m+1\), the derivative is either \(\cos x\) or \(-\cos x\). Its value at zero alternates between \(1\) and \(-1\), beginning with \(1\) at order one.

More precisely, for every integer \(m\ge 0\),

Substituting these values into the definition of the Maclaurin polynomial proves the formula: even powers contribute zero, and the term at the odd power \(x^{2m+1}\) is \(\frac{(-1)^m}{(2m+1)!}x^{2m+1}\). Therefore, for \(n\ge 1\),

The index condition \(2m+1\le n\) enforces the degree cutoff. For \(n=0\), there are no odd powers to include, so \(P_0(x)=0\).

Worked Examples

Why the Degree Cutoff Matters

The notation \(P_n\) records the maximum allowed power, not a promise that the coefficient of \(x^n\) is nonzero. For sine, increasing the cutoff from an odd degree to the next even degree does not change the polynomial: the newly permitted even-power coefficient vanishes.

This pattern is useful when checking a calculation. If a proposed sine Maclaurin polynomial contains a nonzero even power, its coefficient calculation should be examined. Likewise, a degree-six cutoff should not include an \(x^7\) term, even though \(x^7\) is part of the next nonzero contribution.

Common Mistake: Losing the Alternating Sign

A tempting error is to assume the third derivative at zero is positive because the first derivative at zero is positive. That produces an incorrect cubic term. For example, the wrong calculation might be

But differentiating sine gives \(f'(x)=\cos x\), \(f''(x)=-\sin x\), and \(f'''(x)=-\cos x\). Hence \(f'''(0)=-1\), not \(1\). The correct cubic coefficient is

The correction is to write the derivative cycle before inserting coefficients. The cycle keeps both the zero values and the alternating signs visible.

Using the Polynomial as an Approximation

A sine Maclaurin polynomial is a local model centered at zero: its value and successive derivatives at zero agree with sine through the chosen cutoff. That makes it useful for estimating sine near zero using a finite polynomial. A higher cutoff includes additional terms when the next permitted odd power is reached.

The evaluation in a worked example is the value of the polynomial model, not an assertion that the polynomial equals \(\sin x\) at every input. For example, \(P_5(1/2)=1841/3840\) is the exact value of the degree-five polynomial at one-half. It is used as an approximation to \(\sin(1/2)\), not as an identity defining sine.