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Remainders and Error Estimates · Tutorial 869 of 998

Certifying Trigonometric Approximations

Use the structure of sine and cosine Taylor terms, or a derivative bound, to turn a polynomial approximation into a verified error guarantee.

Advanced 10 min read

What You'll Learn

  • Identify when the omitted sine or cosine terms satisfy the alternating-series remainder conditions
  • Use the first omitted term to bound an approximation error and determine its direction
  • Apply a Taylor remainder bound using a valid derivative bound on the interval
  • Compute certified approximations at arguments where early Taylor terms may not decrease
  • Express an error guarantee as an exact interval and round decimal endpoints outward

From a Trigonometric Polynomial to a Guarantee

A Taylor polynomial gives a convenient numerical approximation to sine or cosine, but the polynomial alone does not tell us how close it is to the function. A certified approximation includes a proven bound on the error. That bound lets us state an interval that is guaranteed to contain the true function value.

There are two useful ways to certify sine and cosine approximations near zero. The alternating-series method uses the sizes and signs of the omitted Taylor terms. The Taylor remainder method instead bounds a derivative on the interval from the expansion center to the argument. Both require checking their conditions; neither method should be applied just because the function is sine or cosine.

All trigonometric arguments in this tutorial are in radians. The Maclaurin series we use are

$$ \sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$ \cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots. $$

A truncation is a finite sum of initial terms. Its degree is the highest power retained, even when the coefficient of that power is zero. For example, the degree-eight Taylor polynomial for sine at zero is the same polynomial as its degree-seven Taylor polynomial, because the coefficient of \(x^8\) is zero.

Alternating-Series Certification

Definition: An alternating series has terms whose signs alternate. The alternating-series remainder bound applies when the magnitudes of the terms in the tail are nonincreasing and tend to zero. After truncating, the error has the sign of the first omitted term and its absolute value is at most the magnitude of that term.

For the sine series, the magnitude of the term indexed by \(k\) is \(a_k=|x|^{2k+1}/(2k+1)!\). For \(x\ne 0\), successive magnitudes have ratio

$$ \frac{a_{k+1}}{a_k} = \frac{|x|^2}{(2k+2)(2k+3)}. $$

For cosine, with \(a_k=|x|^{2k}/(2k)!\), the corresponding ratio is

$$ \frac{a_{k+1}}{a_k} = \frac{|x|^2}{(2k+1)(2k+2)}. $$

These ratios give a direct test of whether the terms are decreasing: a ratio at most one means the next term is no larger than the current term. It is enough for the terms to decrease from the first omitted term onward. Earlier terms in the series need not decrease if they are already included in the approximation.

Why the Alternating Remainder Bound Works

Consider a tail \(b_0-b_1+b_2-b_3+\cdots\), where \(b_j\ge 0\), the \(b_j\) are nonincreasing, and \(b_j\to 0\). Group consecutive terms in pairs. Each pair \(b_{2j}-b_{2j+1}\) is nonnegative, so the tail is nonnegative. Also, subtracting the tail from \(b_0\) gives \(b_1-b_2+b_3-\cdots\), which is nonnegative by the same pairing argument. Consequently, the tail lies between zero and \(b_0\). Reversing every sign gives the corresponding conclusion for a tail that starts negative. This proves both the direction and the first-omitted-term bound.

Worked Example: Certifying \(\sin(1)\) with an Alternating Tail

Retain the sine terms through degree five:

$$ P_5(1)=1-\frac{1}{3!}+\frac{1}{5!} =1-\frac{1}{6}+\frac{1}{120} =\frac{120-20+1}{120} =\frac{101}{120}. $$

The first omitted term is negative, with magnitude \(1/7!=1/5040\). For the sine-term magnitudes at \(|x|=1\), the ratio is \(1/((2k+2)(2k+3))<1\) for every \(k\ge 0\), so the omitted tail decreases to zero. Thus the error is negative and no larger in magnitude than \(1/5040\):

$$ \frac{101}{120}-\frac{1}{5040} \le \sin(1)\le \frac{101}{120}, \qquad \frac{4241}{5040}\le\sin(1)\le\frac{4242}{5040}. $$

The arithmetic checks because \(101/120=4242/5040\). The interval has decimal endpoints approximately \(0.8414682539\) and \(0.8416666667\); rounded outward to six decimal places, it is \(0.841468\le\sin(1)\le0.841667\). The approximation \(101/120\) is therefore an upper bound, not a two-sided estimate with an unspecified error direction.

Taylor Remainder Certification

Alternating terms are not the only route to a guarantee. Taylor’s theorem states that if \(f\) has continuous derivatives through order \(n+1\) on the interval joining \(0\) to \(x\), then

$$ f(x)=P_n(x)+R_n(x), \qquad R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} $$

for some \(c\) between \(0\) and \(x\). If \(|f^{(n+1)}(t)|\le M\) throughout that interval, it follows that

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

For sine and cosine, every derivative is one of \(\sin t\), \(\cos t\), \(-\sin t\), or \(-\cos t\). Since their absolute values are at most one for every real \(t\), we may take \(M=1\). This bound gives a symmetric interval around the polynomial value. It applies whether or not the Taylor terms decrease, provided the derivative condition holds.

Worked Example: Bounding \(\cos(1)\) with Taylor’s Theorem

Use the degree-four cosine polynomial. Substitution gives

$$ P_4(1)=1-\frac{1^2}{2!}+\frac{1^4}{4!} =1-\frac12+\frac1{24} =\frac{24-12+1}{24} =\frac{13}{24}. $$

For a degree-four Taylor polynomial, the remainder involves the fifth derivative. The fifth derivative of cosine is \(-\sin t\), whose absolute value is at most one on the entire interval \([0,1]\). Thus \(M=1\), and

$$ |R_4(1)|\le\frac{1\cdot |1|^5}{5!} =\frac1{120}. $$

The guaranteed interval is therefore

$$ \frac{13}{24}-\frac1{120} \le\cos(1)\le \frac{13}{24}+\frac1{120}, \qquad \frac{64}{120}\le\cos(1)\le\frac{66}{120}, $$

or \(8/15\le\cos(1)\le11/20\). To check the arithmetic, \(13/24=65/120\), so subtracting and adding \(1/120\) gives \(64/120\) and \(66/120\), respectively. The derivative bound supplies a symmetric error guarantee even though the alternating-series direction could also be used here.

Worked Example: Certifying \(\sin(3)\) Carefully

At \(x=3\), retain the sine terms through degree seven. Substitution and simplification give

$$ P_7(3) =3-\frac{3^3}{3!}+\frac{3^5}{5!}-\frac{3^7}{7!} =3-\frac{27}{6}+\frac{243}{120}-\frac{2187}{5040} =\frac{51}{560}. $$

For example, the sum can be checked over the common denominator \(5040\): its numerator is \(15120-22680+10206-2187=459\), so it equals \(459/5040=51/560\). The first omitted sine term is positive and has magnitude

$$ \frac{3^9}{9!} =\frac{19683}{362880} =\frac{243}{4480} \approx 0.0542410714. $$

The omitted tail begins with the term indexed by \(k=4\). From that term onward the ratio of successive magnitudes is \(9/((2k+2)(2k+3))<1\), so the alternating-series conditions hold for this tail. Its sign is positive, and its magnitude is at most \(243/4480\). In particular, the one-sided alternating-series certificate is

$$ \frac{51}{560}\le\sin(3)\le \frac{51}{560}+\frac{243}{4480} =\frac{651}{4480}. $$

The same polynomial can also be certified with Taylor’s theorem. Use the degree-eight Taylor polynomial, which equals \(P_7\) because sine has no \(x^8\) term. The ninth derivative of sine is \(\cos t\), bounded in absolute value by one on \([0,3]\). Therefore

$$ |R_8(3)|\le\frac{3^9}{9!} =\frac{243}{4480}. $$

This gives the symmetric interval

$$ \frac{51}{560}-\frac{19683}{362880} \le\sin(3)\le \frac{51}{560}+\frac{19683}{362880}. $$

To check its endpoints, write \(51/560=408/4480\). The lower endpoint is \((408-243)/4480=165/4480=33/896\), approximately \(0.0368303571\). The upper endpoint is \(651/4480=0.1453125\). Rounded outward to six decimal places, the symmetric certificate is \(0.036830\le\sin(3)\le0.145313\). The error bound is approximately \(0.0542410714\); retaining the exact fraction until the final decimal conversion avoids mistaking it for a nearby value.

Choosing and Checking a Method

The two methods answer slightly different questions. The alternating-series result can provide the sign of the error as well as its size, but only after the omitted tail has been checked for alternating signs, decreasing magnitudes, and terms tending to zero. A derivative bound usually provides a symmetric interval and works without an alternating series, but its usefulness depends on the size of the derivative bound and the power of \(|x|\) in the remainder estimate.

Method Condition to verify What the guarantee gives
Alternating-series remainder The omitted terms alternate, decrease in magnitude, and tend to zero. The error has the first omitted term’s sign and is bounded by its magnitude.
Taylor derivative remainder The required derivative is continuous and bounded on the interval from the center to the argument. A symmetric absolute-error bound \(M|x|^{n+1}/(n+1)!\).

A common mistake is to inspect only the first few terms of a trigonometric series and assume that the alternating bound applies to the whole series. For \(\sin(3)\), the first two term magnitudes are \(3\) and \(27/6=4.5\); they increase. Thus the terms do not decrease from the beginning, and one cannot use the alternating-series theorem for a tail that starts with those terms. The correction is to check the tail actually left out. After retaining terms through degree seven, its first omitted term is \(3^9/9!\), and the ratios from that point onward are less than one, so the alternating bound is valid. Alternatively, Taylor’s derivative bound can be used directly.

Reporting a Certified Approximation

A complete certification identifies the polynomial, the method, the verified condition, and the resulting error bound. If the method gives \(|f(x)-P_n(x)|\le E\), report the interval \(P_n(x)-E\le f(x)\le P_n(x)+E\). If the alternating tail has known sign, use that information to give the narrower one-sided interval when appropriate. When decimals are requested, calculate from the exact bound and round the lower endpoint downward and the upper endpoint upward. Ordinary rounding to nearest could accidentally exclude the true value.

These checks matter whenever a numerical value must be accompanied by a guarantee rather than a plausible estimate. A small polynomial value, by itself, is not evidence that the error is small; the certificate comes from the conditions and bound that connect the polynomial to the function.

Check Your Understanding

For each question, identify the condition that supports the claimed guarantee, not just the numerical calculation.

  1. For the cosine series at \(x=1/2\), approximate \(\cos(1/2)\) by \(1-x^2/2!+x^4/4!\). Find the first omitted term and state the alternating-series error direction.
  2. For a degree-six Taylor approximation to sine at \(x=2\), which derivative controls the Lagrange remainder, and what uniform bound can be used for its absolute value?
  3. Explain why checking that the terms alternate is not, by itself, enough to apply the alternating-series remainder bound.
  4. For the degree-eight sine polynomial at \(x=3\), state the exact Taylor derivative error bound and explain why the degree-eight polynomial equals the degree-seven one.
  5. If a certified interval has been calculated exactly, how should its endpoints be rounded to six decimal places so the decimal interval remains guaranteed?

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