Proving the Local Sine Bound

Why a Local Bound Is Useful

A limit can be difficult to evaluate when a function oscillates or has a complicated expression. A useful strategy is to find a simple bound that controls the expression near the point of interest. For sine near zero, geometry gives just such a bound: the magnitude of the sine is no greater than the magnitude of its input.

This statement is about angles measured in radians. That qualification matters: the geometric comparison between an angle and lengths on the unit circle uses radian measure. We will first prove the comparison for small positive angles, extend it to negative angles, and then use it to bound limits and approximation errors.

The Geometric Sine Inequalities

Consider a unit circle centered at \(O\). Let \(A=(1,0)\), and let \(B=(\cos x,\sin x)\) lie on the circle, where \(0<x<\frac{\pi}{2}\). Draw the radius \(OB\), the chord \(AB\), and the tangent line to the circle at \(A\). The ray from \(O\) through \(B\) meets that tangent at a point \(T\).

Because \(0<x<\frac{\pi}{2}\), the point \(B\) is in the first quadrant, and the tangent intersection has coordinates \(T=(1,\tan x)\). The triangle \(OAB\) lies inside the circular sector \(OAB\), and the sector lies inside the larger triangle \(OAT\). Comparing their areas produces the needed inequalities.

Here is the area comparison in detail. The unit-circle triangle \(OAB\) has base \(OA=1\) and height \(\sin x\), so its area is \(\frac12\sin x\). The sector has radius \(1\) and angle \(x\) radians, so its area is \(\frac12x\). The triangle \(OAT\) has base \(OA=1\) and height \(AT=\tan x\), so its area is \(\frac12\tan x\). The containments are strict for an angle strictly between \(0\) and \(\frac{\pi}{2}\), giving

Multiplying each part by \(2\) proves the theorem. The use of radians is essential in the sector-area formula: for a unit circle, the sector area is \(x/2\) when the angle is \(x\) radians.

From Geometry to the Absolute-Value Bound

The first part of the geometric inequality gives \(\sin x<x\) when \(0<x<\frac{\pi}{2}\). At \(x=0\), both sides are zero. For negative \(x\), use the oddness of sine: \(\sin(-u)=-\sin u\). If \(-\frac{\pi}{2}<x<0\), then \(u=-x\) satisfies \(0<u<\frac{\pi}{2}\), so \(|\sin x|=|\sin u|=\sin u<u=|x|\).

This is a local bound: it is proved on an interval around zero, which is exactly what is needed for limits as \(x\) approaches zero. A proof of a local statement need not establish the same inequality for every real input.

A Useful Consequence: Bounding the Sine Approximation Error

The bound controls sine itself. It also lets us control how far \(\sin x\) is from \(x\). For \(0<x<\frac{\pi}{2}\), the other geometric inequality, \(x<\tan x\), can be multiplied by the positive number \(\cos x\). Since \(\tan x\cos x=\sin x\), this gives \(x\cos x<\sin x\). Combining this with \(\sin x<x\) yields

To bound the remaining cosine term, use the identity \(1-\cos x=2\sin^2(x/2)\). Because \(|x/2|<\frac{\pi}{2}\), the local sine bound applies to \(x/2\): \(|\sin(x/2)|\le |x|/2\). Therefore

It follows for positive \(x\) that \(0\le x-\sin x\le x^3/2\). For negative \(x\), the function \(x-\sin x\) is odd, so its absolute value at \(-x\) equals its absolute value at \(x\). Thus the error estimate holds on both sides of zero.

This estimate is sufficient for limit arguments and numerical error control. It is not a claim that the constant \(1/2\) is the sharpest possible constant; the point is that the error is bounded by a constant times \(|x|^3\).

Using a Bound to Prove a Limit

Suppose an expression \(g(x)\), for \(x\) near zero, satisfies \(|g(x)|\le C|x|^p\), where \(C>0\) and \(p>0\). Since \(C|x|^p\) tends to zero as \(x\to0\), the squeeze theorem gives \(g(x)\to0\). In an epsilon-delta form, given \(\varepsilon>0\), choose \(\delta=(\varepsilon/C)^{1/p}\), possibly reduced to stay within the interval where the bound holds. Then \(0<|x|<\delta\) implies \(|g(x)|\le C|x|^p<\varepsilon\).

For sine itself, \(|\sin x|\le|x|\) near zero immediately gives \(\lim_{x\to0}\sin x=0\). The same reasoning works when sine is evaluated at a function of \(x\), provided that input tends to zero and lies in the interval where the local bound applies.

Using the Bound to Estimate an Error

The same inequality can give a numerical guarantee without requiring a long decimal expansion of sine. The error estimate applies to radian inputs with magnitude less than \(\frac{\pi}{2}\). It compares the sine value with the input itself.

Common Mistakes

A frequent error is to apply the positive-input inequality \(\sin x\le x\) unchanged to negative \(x\). For example, at \(x=-0.2\), the local sine bound gives \(\sin(-0.2)\approx-0.1987\), but \(-0.1987\le-0.2\) is false. The correct statement for both signs is \(|\sin x|\le|x|\). In signed form, the comparison reverses for negative inputs: oddness gives \(\sin(-u)=-\sin u>-u\) when \(u>0\) is small.

Another mistake is to insert a degree measure into a radian inequality. The expression \(\sin x\) in calculus uses radian input, and the sector-area formula used above gives area \(x/2\) only when \(x\) is in radians. Keep the angle unit consistent throughout both the proof and any numerical estimate.