Introduction
Trigonometric functions can change as their inputs change, but their derivatives place limits on how quickly they can do so. The Mean Value Theorem turns those limits into inequalities: knowing a derivative is bounded on an interval lets us bound the change in the function across that interval.
For sine and cosine, the derivative bounds are especially simple. Since both functions have absolute value at most \(1\), neither function can change by more than the distance between its inputs. A further argument gives a sharper, quadratic bound for how far cosine can lie below \(1\) near zero.
From a Derivative Bound to a Function-Change Bound
The Mean Value Theorem states that if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), where \(a<b\), then there is a point \(c\in(a,b)\) such that
In particular, if \(|f'(t)|\le M\) for every \(t\in(a,b)\), then taking absolute values gives
The same conclusion applies to any two distinct inputs \(x,y\): use the interval between them, with the smaller input as the left endpoint. If \(x=y\), the function change is zero and the inequality holds directly.
The assumptions matter. The function must satisfy the continuity and differentiability conditions on the interval, and the derivative bound must hold throughout the interval between the two inputs—not merely at the endpoints.
Global Bounds for Sine and Cosine
Throughout this tutorial, angles are measured in radians. The standard derivatives are \((\sin t)'=\cos t\) and \((\cos t)'=-\sin t\). Also, \(|\sin t|\le1\) and \(|\cos t|\le1\) for every real \(t\). Thus each derivative has absolute value at most \(1\).
For the first inequality, apply the Mean Value Theorem to \(f(t)=\sin t\) on the interval between \(x\) and \(y\). If the inputs differ, the theorem gives a point \(c\) between them such that
Since \(|\cos(c)|\le1\), taking absolute values proves \(|\sin x-\sin y|\le|x-y|\). If the inputs are equal, both sides are zero. For cosine, apply the theorem to \(f(t)=\cos t\); its derivative is \(-\sin t\), whose absolute value is also at most \(1\). This proves the second inequality.
These bounds are global: they hold for every pair of real inputs, no matter how far apart. On a particular interval, however, a tighter derivative bound can give a stronger estimate. For example, if \(|\cos t|\le M\) throughout the interval between \(x\) and \(y\), then the same Mean Value Theorem argument gives \(|\sin x-\sin y|\le M|x-y|\). Likewise, if \(|\sin t|\le M\) throughout that interval, then \(|\cos x-\cos y|\le M|x-y|\).
Worked Examples with the Global Bounds
Worked Example: Bounding a Change in Sine
Estimate the size of \(\sin(\pi/2)-\sin(0)\) using the global difference bound, then check the inequality.
The inputs are \(x=\pi/2\) and \(y=0\). The bound gives
The left side can also be evaluated exactly: \(\sin(\pi/2)=1\) and \(\sin(0)=0\), so it equals \(|1-0|=1\). The resulting inequality is \(1\le\pi/2\), which holds since \(\pi>2\). Thus the stated estimate is consistent with the exact function change.
Worked Example: Bounding a Change in Cosine
Use the global difference bound to estimate \(|\cos(\pi/3)-\cos(0)|\).
The input distance is \(|\pi/3-0|=\pi/3\). Therefore,
To check, \(\cos(\pi/3)=1/2\) and \(\cos(0)=1\), so the left side is \(|1/2-1|=1/2\). The inequality becomes \(1/2\le\pi/3\), true because \(\pi>3/2\). The derivative argument supplies the bound; the exact values verify it for these inputs.
A Sharper Bound for One Minus Cosine
The global cosine difference bound gives \(|1-\cos x|\le|x|\), by comparing \(\cos x\) with \(\cos 0=1\). Near zero, a quadratic estimate is more informative. We can derive it using derivative bounds and the Mean Value Theorem.
First consider \(t\ge0\). Applying the Mean Value Theorem to sine on \([0,t]\), when \(t>0\), gives a point \(c\in(0,t)\) such that
Since \(\cos(c)\le1\), this shows \(\sin t\le t\). For \(t=0\), the same inequality follows from \(\sin 0=0\). Now define \(q(t)=t^2/2-1+\cos t\) for \(t\ge0\). It has \(q(0)=0\), and
For any \(t>0\), the Mean Value Theorem applied to \(q\) on \([0,t]\) gives a point \(d\in(0,t)\) with \(q(t)-q(0)=q'(d)t\ge0\). Hence \(q(t)\ge0\), which is equivalent to \(1-\cos t\le t^2/2\). Also \(\cos t\le1\), so \(1-\cos t\ge0\).
For \(x<0\), cosine is even, so \(\cos x=\cos(-x)\), and \(x^2=(-x)^2\). Applying the result for the nonnegative input \(-x\) proves the upper bound for \(x\) as well. The lower bound again follows from \(\cos x\le1\). Together these establish the theorem for every real \(x\).
Worked Example: Bounding the Cosine Change at One Half
Use the quadratic bound to estimate \(1-\cos(1/2)\), then check the conditions and the resulting inequality.
Set \(x=1/2\) in the theorem:
The input is a real number, so the theorem applies; the intermediate arithmetic gives \((1/2)^2=1/4\), then \((1/4)/2=1/8\). Thus \(1-\cos(1/2)\) lies between \(0\) and \(1/8\). In particular, since \(\cos(1/2)=1-(1-\cos(1/2))\), the same result says \(7/8\le\cos(1/2)\le1\), which is the equivalent bound after rearranging.
When Local Derivative Bounds Improve the Estimate
A global derivative bound uses the largest permitted derivative magnitude, which may be much larger than the derivative values on a short interval. If the inputs are close to zero, sine itself is small there. That gives a sharper bound for changes in cosine.
For example, if \(x\) and \(y\) lie in \([-\delta,\delta]\), where \(\delta\ge0\), then \(|\sin t|\le|t|\le\delta\) for every \(t\) in this interval. The first inequality follows from the global sine difference bound by comparing \(t\) with \(0\). Applying the Mean Value Theorem to cosine now gives
This estimate is useful only when the derivative bound really holds on the entire interval between the inputs. Choosing a smaller constant based on the derivative at one endpoint, without controlling the derivative in between, does not justify the estimate.
Common Pitfall: Using Degrees with Radian Derivatives
The identities \((\sin t)'=\cos t\) and \((\cos t)'=-\sin t\) use radian inputs. If an angle is entered in degrees, it must first be converted to radians before applying these derivative-based bounds.
For instance, a tempting but incorrect argument is to use the global sine bound with inputs \(90\) and \(0\), obtaining \(|\sin 90-\sin 0|\le90\), and then treat the \(90\) as degrees. The bound's inputs are real numbers in radians, so that calculation instead refers to \(\sin(90)\) and \(\sin(0)\) with radian inputs. For the angle \(90^\circ\), convert first: \(90^\circ=\pi/2\) radians. The appropriate estimate is
with the exact change equal to \(1\). Keep the angle units consistent with the derivative formulas.
Why These Bounds Matter
The difference inequalities give direct control even when exact function values are difficult to compute. If two inputs differ by at most \(\varepsilon\), then their sine values differ by at most \(\varepsilon\), and their cosine values differ by at most \(\varepsilon\). Near zero, the quadratic estimate shows that \(1-\cos x\) is no larger than \(x^2/2\), a substantially smaller bound than \(|x|\) when \(|x|\) is small.
The common reasoning pattern is worth retaining: identify the function, identify a derivative bound valid over the full interval, check the Mean Value Theorem conditions, and multiply that derivative bound by the input distance. This tutorial's estimates concern changes in the functions themselves; estimates for errors in linear approximations are a separate application.
Check Your Understanding
For each question, identify which derivative bound is needed and check that it applies on the interval between the inputs.
- Use the global sine difference inequality to bound \(|\sin(2)-\sin(1)|\).
- Use the global cosine difference inequality to bound \(|\cos(0)-\cos(-3)|\).
- State the derivative bound that gives \(|\sin x-\sin y|\le M|x-y|\) on an interval.
- Use the quadratic theorem to bound \(1-\cos(-1/3)\).
- Why must an angle given in degrees be converted before applying the derivative-based inequalities?
