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Maclaurin Series for Sine and Cosine

Use the derivatives at zero to build the sine and cosine series, then verify that their Taylor remainders vanish for every real number.

Intermediate 9 min read

What You'll Learn

  • Generate sine and cosine Maclaurin coefficients from their repeating derivative patterns.
  • Use odd and even parity to identify which Taylor coefficients vanish.
  • Write the Maclaurin polynomials and infinite series for both functions.
  • Prove that the Taylor remainders tend to zero for every fixed real input.
  • Estimate sine and cosine at one using Taylor polynomials and explicit remainder bounds.

From Derivatives to Maclaurin Series

A Maclaurin series is a Taylor series centered at zero. For a function \(f\) with derivatives of all orders at zero, its degree-\(n\) Maclaurin polynomial is formed from the derivatives at zero:

$$ T_n(x)=\sum_{j=0}^{n}\frac{f^{(j)}(0)}{j!}x^j. $$

The earlier material on Taylor coefficients establishes this formula for the coefficients of the polynomial. For sine and cosine, the derivatives follow a particularly simple cycle, so their coefficients can be found directly. A further question is whether the resulting infinite series actually equals the function. We will answer that by bounding the Taylor remainder.

Definition: The Maclaurin series of \(f\) is the formal power series \(\sum_{j=0}^{\infty} f^{(j)}(0)x^j/j!\). It represents \(f(x)\) at a given \(x\) when the corresponding Taylor polynomials converge to \(f(x)\) there.

The Derivative Cycles and Parity

The derivatives of \(\sin x\) repeat every four differentiations:

$$ \sin x,\qquad \cos x,\qquad -\sin x,\qquad -\cos x,\qquad \sin x,\ldots $$

At zero, \(\sin 0=0\) and \(\cos 0=1\). Thus the derivative values at zero repeat as \(0,1,0,-1\). In particular, every even-order derivative of sine at zero vanishes, while the odd-order derivatives alternate between \(1\) and \(-1\).

For cosine, the derivatives also repeat every four differentiations, but their values at zero are \(1,0,-1,0\). Thus every odd-order derivative of cosine at zero vanishes, while its even-order derivatives alternate between \(1\) and \(-1\).

This pattern agrees with parity: sine is odd, meaning \(\sin(-x)=-\sin x\), and cosine is even, meaning \(\cos(-x)=\cos x\). The coefficients of even powers therefore disappear from the sine polynomial, and the coefficients of odd powers disappear from the cosine polynomial. The derivative cycles determine the signs of the remaining coefficients.

Takeaway: The nonzero sine coefficients occur at powers \(x^{2k+1}\), and the nonzero cosine coefficients occur at powers \(x^{2k}\). Their signs alternate, and each coefficient has the factorial of its power in the denominator.

Worked Example: Constructing a Sine Maclaurin Polynomial

Worked Example: Constructing a Sine Maclaurin Polynomial

Find the degree-nine Maclaurin polynomial for \(\sin x\).

Use \(T_9(x)=\sum_{j=0}^{9} \sin^{(j)}(0)x^j/j!\). The derivative values through order nine are:

$$ \begin{array}{c|rrrrrrrrrr} j&0&1&2&3&4&5&6&7&8&9\\ \hline \sin^{(j)}(0)&0&1&0&-1&0&1&0&-1&0&1 \end{array} $$

Substitute these values into the polynomial formula. The terms for even \(j\) vanish, and the nonzero terms give

$$ \begin{aligned} T_9(x) &=\frac{1}{1!}x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\frac{1}{9!}x^9\\ &=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}. \end{aligned} $$

To check the coefficients against the derivatives, differentiate at zero. The first, third, fifth, seventh, and ninth derivatives of this polynomial at zero are, respectively, \(1,-1,1,-1,1\); its even-order derivatives at zero vanish. These match the listed derivatives of \(\sin x\) through order nine, as required for the degree-nine Maclaurin polynomial.

Worked Example: Constructing a Cosine Maclaurin Polynomial

Worked Example: Constructing a Cosine Maclaurin Polynomial

Find the degree-ten Maclaurin polynomial for \(\cos x\).

The derivatives at zero through order ten follow the repeating pattern \(1,0,-1,0,\ldots\):

$$ \begin{array}{c|rrrrrrrrrrr} j&0&1&2&3&4&5&6&7&8&9&10\\ \hline \cos^{(j)}(0)&1&0&-1&0&1&0&-1&0&1&0&-1 \end{array} $$

Every odd-order term vanishes. The remaining terms are

$$ \begin{aligned} T_{10}(x) &=\frac{1}{0!}-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!} +\frac{x^8}{8!}-\frac{x^{10}}{10!}\\ &=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720} +\frac{x^8}{40320}-\frac{x^{10}}{3628800}. \end{aligned} $$

Here \(0!=1\), so the constant term is \(1\). Differentiating the polynomial and evaluating at zero gives the same even-order derivative values \(1,-1,1,-1,1,-1\) and zero odd-order values through order ten. This verifies the polynomial’s coefficients against the derivative cycle.

The Two Maclaurin Series

The same derivative patterns continue at every order. At a power \(x^m\), the Taylor coefficient is the \(m\)th derivative at zero divided by \(m!\). The sine and cosine series are therefore:

$$ \sin x =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots =\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!}, $$
$$ \cos x =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots =\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{(2k)!}. $$

The index \(k\) starts at zero in both sums. For sine, \(k=0\) gives \(x\); for cosine, it gives \(1\). The powers increase by two because the coefficients of the other parity vanish.

Why the Series Equal the Functions for Every Real \(x\)

Knowing the derivatives determines the Maclaurin polynomials, but it does not by itself show that those polynomials converge to the function. Taylor’s theorem supplies the missing step. For a function with \(n+1\) continuous derivatives between \(0\) and \(x\), the remainder after degree \(n\) has the form

$$ R_n(x)=f(x)-T_n(x) =\frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1} $$

for some \(\xi\) between \(0\) and \(x\). Every derivative of sine or cosine is one of \(\sin,\cos,-\sin,-\cos\). Since the absolute values of sine and cosine are at most \(1\) for real inputs, every derivative of either function has absolute value at most \(1\). Consequently, for either function,

$$ |R_n(x)|\leq \frac{|x|^{n+1}}{(n+1)!}. $$

Fix any real number \(x\). If \(x=0\), the remainder is zero. If \(x\ne 0\), the bound tends to zero as \(n\) increases: the ratio of consecutive quantities \(|x|^{n+1}/(n+1)!\) is \(|x|/(n+2)\), which is eventually less than, for example, \(1/2\). From that point onward the quantities decrease at least geometrically, so they approach zero. The remainder therefore approaches zero, and \(T_n(x)\) approaches \(f(x)\).

1
Form the Taylor polynomial.
Use the derivative values at zero to obtain the sine or cosine polynomial of degree \(n\).
2
Bound the remainder.
Every derivative of either function has absolute value at most \(1\), so the error is at most \(|x|^{n+1}/(n+1)!\).
3
Let the degree increase.
For each fixed real \(x\), the bound tends to zero; hence the Taylor polynomials converge to the function value.
Theorem: For every real \(x\), the Maclaurin series for sine and cosine converge to \(\sin x\) and \(\cos x\), respectively. The proof uses Taylor’s theorem and the fact that all derivatives of these functions are bounded in absolute value by \(1\).

Worked Example: Estimating Sine and Cosine at One

Worked Example: Estimating Sine and Cosine at One

Evaluate the degree-seven sine polynomial and degree-eight cosine polynomial at \(x=1\), and bound the errors.

For sine, substitute \(x=1\) into \(T_7(x)=x-x^3/3!+x^5/5!-x^7/7!\):

$$ \begin{aligned} T_7(1) &=1-\frac{1}{6}+\frac{1}{120}-\frac{1}{5040}\\ &=\frac{5040}{5040}-\frac{840}{5040}+\frac{42}{5040}-\frac{1}{5040}\\ &=\frac{4241}{5040}\\ &\approx 0.8414682540 \quad \text{(rounded to ten decimal places).} \end{aligned} $$

The Taylor remainder bound for degree seven gives \(|\sin(1)-T_7(1)|\leq 1/8!=1/40320\approx 0.0000248016\). Thus the displayed decimal is the rounded value of the polynomial; the bound, rather than the number of displayed decimal places, indicates its guaranteed accuracy as an estimate of \(\sin(1)\).

For cosine, substitute \(x=1\) into \(T_8(x)=1-x^2/2!+x^4/4!-x^6/6!+x^8/8!\):

$$ \begin{aligned} T_8(1) &=1-\frac{1}{2}+\frac{1}{24}-\frac{1}{720}+\frac{1}{40320}\\ &=\frac{40320}{40320}-\frac{20160}{40320}+\frac{1680}{40320} -\frac{56}{40320}+\frac{1}{40320}\\ &=\frac{21785}{40320} =\frac{4357}{8064}\\ &\approx 0.5403025794 \quad \text{(rounded to ten decimal places).} \end{aligned} $$

For degree eight, the remainder bound is \(|\cos(1)-T_8(1)|\leq 1/9!=1/362880\approx 0.0000027557\). Both estimates have been checked by substituting \(1\) into the stated polynomials and combining the fractions over common denominators.

A Sign Pitfall

A common error is to assume that the sine coefficients alternate starting with a positive cubic term. That misses the negative sign in the third derivative. For instance, the incorrect calculation \( \sin'''(0)=1\), and hence a term \(+x^3/3!\), does not agree with the derivative cycle: \(\sin'''x=-\cos x\), so \(\sin'''(0)=-1\). The correct cubic term is \(-x^3/3!\). As a direct check, the third derivative of \(+x^3/3!\) at zero is \(+1\), not \(-1\); the third derivative of \(-x^3/3!\) at zero is \(-1\).

It is also important not to confuse a Taylor polynomial with the infinite series. A polynomial uses only finitely many terms; the convergence argument is what establishes that taking terms of increasing degree approaches sine or cosine at every fixed real input.

Check Your Understanding

Use the derivative cycles and remainder argument to answer these questions.

  1. What are the first five nonzero terms of the Maclaurin series for \(\sin x\), including their signs and factorial denominators?
  2. Why do all odd powers have zero coefficients in the Maclaurin polynomial for \(\cos x\)?
  3. Write the degree-six Maclaurin polynomial for \(\cos x\). Which derivative values determine its coefficients?
  4. State the Taylor remainder bound used for sine and cosine, and explain why it tends to zero for a fixed real \(x\).
  5. For \(T_5(x)=x-x^3/3!+x^5/5!\), find the bound on \(|\sin(x)-T_5(x)|\) at \(x=1\).

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