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Taylor Polynomials · Tutorial 810 of 998

Sine and Cosine Polynomials at Nonzero Centers

Learn to build and verify sine and cosine Taylor polynomials about nonzero centers using angle addition and the functions’ repeating derivative patterns.

Beginner 10 min read

What You'll Learn

  • Write the Taylor polynomial for sine or cosine about a general center \(a\).
  • Use the repeating derivative values of sine and cosine to determine the coefficients.
  • Relate a nonzero-center polynomial to the Maclaurin polynomials for sine and cosine.
  • Track signs and powers of \(x-a\) through a chosen degree.
  • Verify a polynomial by checking its derivatives at the center.

Why the Center Changes the Coefficients

A Maclaurin polynomial describes a function near zero. A Taylor polynomial about a different center \(a\) instead describes it using powers of \(x-a\). For sine and cosine, shifting the center does not require a new derivative method: their derivatives repeat in a cycle, and angle-addition identities connect their values at \(a\) to their values near zero.

The important practical change is that the coefficients now depend on the center. For example, the slope of sine at \(a\) is \(\cos a\), not necessarily \(1\). The constant term is \(\sin a\), not necessarily \(0\). Keeping the center visible as \(x-a\) helps prevent both kinds of error.

Definition: For a function \(f\) with derivatives through order \(n\) at \(a\), its degree-\(n\) Taylor polynomial about \(a\) is $$ T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k. $$ Here \(f^{(0)}=f\), and \(0!=1\). The polynomial has degree at most \(n\); some of its highest-degree coefficients can be zero.

For sine and cosine, derivatives of every order exist. We use radians, as required for the standard derivative formulas \((\sin x)'=\cos x\) and \((\cos x)'=-\sin x\). Their repeating derivative values give the coefficients directly.

The Cyclic Derivative Values

Successive derivatives of sine repeat every four steps: sine, cosine, negative sine, negative cosine, and then sine again. Cosine has its own four-step cycle: cosine, negative sine, negative cosine, sine, and then cosine again.

$$ \begin{aligned} \sin x &: \quad \sin x,\ \cos x,\ -\sin x,\ -\cos x,\ \sin x,\ldots,\\ \cos x &: \quad \cos x,\ -\sin x,\ -\cos x,\ \sin x,\ \cos x,\ldots. \end{aligned} $$

Substituting these derivatives into the Taylor-polynomial definition produces the general formulas. In each summand, \(k\) is the derivative order, and the derivative is evaluated at the chosen center \(a\).

General formulas: The degree-\(n\) Taylor polynomials for sine and cosine about \(a\) are $$ T_n^{\sin,a}(x)=\sum_{k=0}^{n}\frac{\sin^{(k)}(a)}{k!}(x-a)^k $$ and $$ T_n^{\cos,a}(x)=\sum_{k=0}^{n}\frac{\cos^{(k)}(a)}{k!}(x-a)^k. $$ The coefficients repeat in the derivative cycles above. In particular, the first terms are $$ \begin{aligned} T_n^{\sin,a}(x) &=\sin a+\cos a(x-a)-\frac{\sin a}{2!}(x-a)^2 -\frac{\cos a}{3!}(x-a)^3+\cdots,\\ T_n^{\cos,a}(x) &=\cos a-\sin a(x-a)-\frac{\cos a}{2!}(x-a)^2 +\frac{\sin a}{3!}(x-a)^3+\cdots. \end{aligned} $$

The ellipses indicate that the same derivative cycles continue only as far as the requested degree. For instance, a degree-three polynomial stops at \((x-a)^3\); a degree-four polynomial also includes the term involving \((x-a)^4\).

Why the Formulas Work

The derivative formula gives one direct justification. To construct the degree-\(n\) Taylor polynomial, the coefficient of \((x-a)^k\) must be the \(k\)th derivative of the function at \(a\), divided by \(k!\). Sine and cosine have the cycles just listed, so inserting their values gives each displayed coefficient. The construction therefore uses the definition directly.

Angle addition gives a second perspective and explains how the nonzero-center formulas relate to Maclaurin polynomials. Write \(h=x-a\), so \(x=a+h\). The angle-addition identities give

$$ \sin(a+h)=\sin a\cos h+\cos a\sin h, \qquad \cos(a+h)=\cos a\cos h-\sin a\sin h. $$

The Maclaurin polynomials for \(\sin h\) and \(\cos h\) have already displayed their alternating odd-power and even-power patterns. Substitute those patterns into the identities, multiply by the constant values \(\sin a\) and \(\cos a\), and retain powers of \(h\) only through degree \(n\). For sine, the coefficient of \(h^k\) is precisely \(\sin^{(k)}(a)/k!\); for cosine, it is \(\cos^{(k)}(a)/k!\). Since \(h=x-a\), these are the Taylor polynomials about \(a\).

1
Choose the center and degree.
Set \(h=x-a\), and decide the largest power to include.
2
List derivatives in order.
Use the four-step cycle for the chosen function, evaluating each derivative at \(a\).
3
Divide by the factorial.
The coefficient of \(h^k\) is the \(k\)th derivative value divided by \(k!\).
4
Stop at the requested degree and check.
Confirm the polynomial’s derivatives at \(a\) match the function’s derivatives through order \(n\).

Worked Examples

Worked Example: A Degree-Four Sine Polynomial at \(\pi/6\)

Find the degree-four Taylor polynomial for \(\sin x\) about \(a=\pi/6\). Set \(h=x-\pi/6\). The derivative values at the center are \(\sin(\pi/6)=1/2\), \(\cos(\pi/6)=\sqrt{3}/2\), \(-\sin(\pi/6)=-1/2\), \(-\cos(\pi/6)=-\sqrt{3}/2\), and \(\sin(\pi/6)=1/2\). Use these values for orders zero through four:

$$ \begin{aligned} T_4^{\sin,\pi/6}(x) &=\frac12+\frac{\sqrt3}{2}h +\frac{-1/2}{2!}h^2 +\frac{-\sqrt3/2}{3!}h^3 +\frac{1/2}{4!}h^4\\ &=\frac12+\frac{\sqrt3}{2}h-\frac14h^2 -\frac{\sqrt3}{12}h^3+\frac1{48}h^4\\ &=\frac12+\frac{\sqrt3}{2}\left(x-\frac{\pi}{6}\right) -\frac14\left(x-\frac{\pi}{6}\right)^2 -\frac{\sqrt3}{12}\left(x-\frac{\pi}{6}\right)^3 +\frac1{48}\left(x-\frac{\pi}{6}\right)^4. \end{aligned} $$

Check by differentiating and evaluating at \(x=\pi/6\), where \(h=0\). The polynomial’s value is \(1/2\); its first derivative there is \(\sqrt3/2\); its second derivative is \(2!(-1/4)=-1/2\); its third derivative is \(3!(-\sqrt3/12)=-\sqrt3/2\); and its fourth derivative is \(4!(1/48)=1/2\). These are exactly \(\sin a,\cos a,-\sin a,-\cos a,\sin a\), respectively, so the Taylor coefficients are checked.

Worked Example: A Degree-Five Cosine Polynomial at \(\pi/3\)

Find the degree-five Taylor polynomial for \(\cos x\) about \(a=\pi/3\). Here \(h=x-\pi/3\). The derivative values at the center, from order zero to five, are \(1/2,-\sqrt3/2,-1/2,\sqrt3/2,1/2,-\sqrt3/2\). Dividing each by its factorial gives

$$ \begin{aligned} T_5^{\cos,\pi/3}(x) &=\frac12-\frac{\sqrt3}{2}h -\frac{1}{2!}\frac12h^2 +\frac{1}{3!}\frac{\sqrt3}{2}h^3 +\frac{1}{4!}\frac12h^4 -\frac{1}{5!}\frac{\sqrt3}{2}h^5\\ &=\frac12-\frac{\sqrt3}{2}h-\frac14h^2 +\frac{\sqrt3}{12}h^3+\frac1{48}h^4 -\frac{\sqrt3}{240}h^5\\ &=\frac12-\frac{\sqrt3}{2}\left(x-\frac{\pi}{3}\right) -\frac14\left(x-\frac{\pi}{3}\right)^2 +\frac{\sqrt3}{12}\left(x-\frac{\pi}{3}\right)^3\\ &\quad+\frac1{48}\left(x-\frac{\pi}{3}\right)^4 -\frac{\sqrt3}{240}\left(x-\frac{\pi}{3}\right)^5. \end{aligned} $$

For the check, evaluate successive derivatives at the center. The polynomial and its first five derivatives there are \(1/2,-\sqrt3/2,2!(-1/4)=-1/2,3!(\sqrt3/12)=\sqrt3/2,4!(1/48)=1/2,\) and \(5!(-\sqrt3/240)=-\sqrt3/2\). They match the cosine derivative cycle at \(\pi/3\), confirming every coefficient.

Worked Example: A Degree-Five Sine Polynomial at \(\pi/2\)

Find the degree-five Taylor polynomial for \(\sin x\) about \(a=\pi/2\). With \(h=x-\pi/2\), the derivatives of sine at the center, through order five, are \(1,0,-1,0,1,0\). Thus

$$ \begin{aligned} T_5^{\sin,\pi/2}(x) &=1+0h-\frac{1}{2!}h^2+0h^3+\frac{1}{4!}h^4+0h^5\\ &=1-\frac12\left(x-\frac{\pi}{2}\right)^2 +\frac1{24}\left(x-\frac{\pi}{2}\right)^4. \end{aligned} $$

The degree-five polynomial has no fifth-power term, so its actual degree is four. This is allowed: “degree five” here means the Taylor polynomial constructed through order five. Its value and first five derivatives at the center are \(1,0,-1,0,1,0\), matching the sine derivative values there. The same result follows from \(\sin(\pi/2+h)=\cos h\), whose degree-five Maclaurin polynomial is \(1-h^2/2+h^4/24\).

A Sign Error to Avoid

A common mistake is to use the wrong sign in the angle-addition identity for sine. For example, the incorrect calculation \(\sin(a+h)=\sin a\cos h-\cos a\sin h\) would give the coefficient of \(h\) as \(-\cos a\). But differentiating \(\sin(a+h)\) with respect to \(h\) at \(h=0\) gives \(\cos a\), so that sign cannot be correct. The correct identity is

$$ \sin(a+h)=\sin a\cos h+\cos a\sin h, $$

and its linear term is \(+\cos a\,h\). A quick check of the linear coefficient is often enough to catch a sign error before it spreads to higher powers.

Another frequent slip is to write powers of \(x\) rather than powers of \(x-a\). A Taylor polynomial about \(a\) must be organized in powers of \(x-a\). Its constant term is the function value at \(a\), and substituting \(x=a\) makes all higher-power terms vanish, as it should.

Why These Polynomials Are Useful

These formulas provide local polynomial models for sine and cosine near any chosen center. Choosing a center other than zero lets the constant term and slope match the function at the point of interest. This is useful when a calculation is naturally situated near a particular angle rather than near zero. For instance, near \(\pi/2\), the sine polynomial has a nonzero constant term and only even powers of \(x-\pi/2\) in the example above.

The coefficient pattern also makes the calculation systematic: evaluate sine and cosine at the center, follow the derivative cycle, divide by the relevant factorial, and stop at the requested order. The derivative check confirms that the resulting polynomial has the intended values and derivatives at the center. It does not, by itself, state an error bound for the approximation; that is a separate question.

Check Your Understanding

Use the derivative cycles and powers of \(x-a\) to answer these questions.

  1. Write the degree-three Taylor polynomial for \(\sin x\) about a general center \(a\).
  2. What are the first four derivatives of \(\cos x\), in order, and what pattern do they follow?
  3. For a degree-four Taylor polynomial about \(a\), what value does the polynomial take at \(x=a\), and why?
  4. Why can a degree-five Taylor polynomial for sine have no fifth-power term at a particular center?
  5. In constructing a Taylor polynomial about \(a\), why must the powers be \(x-a\) rather than \(x\)?

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