Introduction
A small positive angle on the unit circle has three closely related measurements: the height of a point on the circle, the angle itself, and the slope of a ray from the origin. These measurements are expressed by the sine, the radian measure, and the tangent.
For an acute angle measured in radians, geometry places them in a strict order:
The comparison comes from placing three shapes in the same part of the unit circle: a triangle inside a circular sector, and that sector inside a triangle formed using a tangent line. The angle must be in radians for the area comparison to produce the formula above.
The angle and the unit circle
Work in the coordinate plane with a circle of radius \(1\) centered at the origin \(O\). Let \(A=(1,0)\), and choose an angle \(\theta\) with \(0<\theta<\frac{\pi}{2}\). The ray from \(O\) at angle \(\theta\) meets the circle at \(B=(\cos\theta,\sin\theta)\). Because the angle is acute, both coordinates of \(B\) are positive.
The arc from \(A\) to \(B\) has length \(\theta\), because on a unit circle the radian measure of an angle equals the length of its intercepted arc. Thus \(\theta\) here is a number of radians, not a number of degrees. For example, a quarter-turn has radian measure \(\frac{\pi}{2}\), not \(90\).
The sine and tangent in the comparison also have direct geometric meanings in this diagram. The height of \(B\) above the horizontal diameter is \(\sin\theta\). The ray from \(O\) through \(B\) meets the vertical tangent line at \(A\), whose equation is \(x=1\), at a point with height \(\tan\theta\).
Three areas to compare
The relevant shapes share the same two rays from \(O\): the horizontal ray \(OA\) and the ray \(OB\). The small triangle \(OAB\) lies inside the sector between the two rays. The sector, in turn, lies inside the larger triangle formed by \(O\), \(A\), and the intersection of \(OB\) with the tangent at \(A\).
The area of the inner triangle is \(\frac12\sin\theta\): its base \(OA\) has length \(1\), and its height is the vertical coordinate \(\sin\theta\) of \(B\). The area of the outer triangle is \(\frac12\tan\theta\): it has base \(OA=1\) and height \(\tan\theta\).
The area of the sector is \(\frac12\theta\). To see why, the whole unit disk has area \(\pi\), and a sector’s share of the disk is the same fraction as its angle’s share of a full turn. A full turn has \(2\pi\) radians, so the sector’s fraction is \(\frac{\theta}{2\pi}\). Therefore,
This area formula is where the radian convention enters. If the angle were recorded in degrees, the fraction of a full turn would use \(360\) in the denominator instead.
Derivation of the geometric bounds
We now justify the containment of the three regions and translate it into inequalities. The point where \(OB\) meets the tangent \(x=1\) is \(C=(1,\tan\theta)\): along a ray at angle \(\theta\), the ratio of vertical to horizontal coordinates is \(\tan\theta\), so setting the horizontal coordinate equal to \(1\) gives the stated height.
The straight chord \(AB\) lies inside the circular disk. Together with the radii \(OA\) and \(OB\), it encloses triangle \(OAB\), which is contained in the sector bounded by those radii and the arc \(AB\).
The triangle \(OAC\) is bounded by \(OA\), the tangent segment \(AC\), and the ray \(OC=OB\). The unit-circle arc from \(A\) to \(B\) stays on or to the left of \(x=1\), and the sector lies between \(OA\) and \(OB\). It is therefore contained in triangle \(OAC\).
For \(0<\theta<\frac{\pi}{2}\), the chord does not follow the arc, so the sector has more area than triangle \(OAB\). The arc is not the tangent segment \(AC\), so triangle \(OAC\) has more area than the sector.
The strict area comparison is \(\frac12\sin\theta<\frac12\theta<\frac12\tan\theta\). Multiplying every term by the positive number \(2\) gives the claimed inequalities.
The argument uses strict containment in area, not merely a comparison of the boundaries. The circle’s curved arc differs from its chord and from its tangent segment on a region of positive area, which is why both inequalities are strict.
Worked examples
Worked Example: Comparing the three quantities at \(\frac{\pi}{6}\)
Take \(\theta=\frac{\pi}{6}\), which lies between \(0\) and \(\frac{\pi}{2}\). The standard unit-circle values are \(\sin\frac{\pi}{6}=\frac12\) and \(\tan\frac{\pi}{6}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}\). Substituting gives
As decimals to six places, these are \(0.500000\), \(0.523599\), and \(0.577350\), respectively. Thus the order is \(\frac12<\frac{\pi}{6}<\frac{\sqrt3}{3}\), consistent with the geometric theorem. The theorem supplies the exact comparison; the decimals are only a numerical check.
Worked Example: Comparing the quantities at \(\frac{\pi}{4}\)
For \(\theta=\frac{\pi}{4}\), the angle is again in the required interval. The unit-circle values are \(\sin\frac{\pi}{4}=\frac{\sqrt2}{2}\) and \(\tan\frac{\pi}{4}=1\). Hence the three quantities are
Their six-place decimal values are \(0.707107\), \(0.785398\), and \(1.000000\). Therefore \(\frac{\sqrt2}{2}<\frac{\pi}{4}<1\). This is an instance of the strict area comparison, with the radian value \(\frac{\pi}{4}\) used for the middle quantity.
Worked Example: Comparing the quantities at \(\frac{\pi}{3}\)
Let \(\theta=\frac{\pi}{3}\). This angle is acute, so the theorem applies even though it is not especially close to zero. Its sine and tangent are \(\frac{\sqrt3}{2}\) and \(\sqrt3\). The quantities in order are
To check the ordering numerically to six decimal places, \(\frac{\sqrt3}{2}\approx0.866025\), \(\frac{\pi}{3}\approx1.047198\), and \(\sqrt3\approx1.732051\). Thus \(\frac{\sqrt3}{2}<\frac{\pi}{3}<\sqrt3\), as predicted. The example also illustrates that the geometric inequalities hold throughout the acute-angle interval, not only for angles very near zero.
Conditions and common pitfalls
The restrictions \(0<\theta<\frac{\pi}{2}\) are part of the result. At \(\theta=0\), all three quantities are \(0\), so the inequalities are equalities rather than strict. At \(\theta=\frac{\pi}{2}\), the ray \(OB\) is parallel to the tangent \(x=1\); it does not meet that tangent at a finite point, and \(\tan\theta\) is undefined. The tangent-triangle comparison therefore cannot be used there.
For example, the degree numeral \(30\) does not fit between \(\sin 30^\circ=\frac12\) and \(\tan 30^\circ=\frac{1}{\sqrt3}\). The angle’s radian measure is \(\frac{\pi}{6}\), and the correct comparison is \(\frac12<\frac{\pi}{6}<\frac{1}{\sqrt3}\). Radians are not a cosmetic choice here: they make the arc length on the unit circle equal to the angle’s numerical measure, which makes the sector area \(\frac12\theta\).
A second mistake is to treat a diagram as if its apparent sizes were a proof. A sketch may help identify which regions are being compared, but the reasoning must specify the containment and compute each area. Here, the relevant areas are exactly \(\frac12\sin\theta\), \(\frac12\theta\), and \(\frac12\tan\theta\); the positive area gaps establish the strict comparisons.
Why these bounds matter
The inequalities give a geometric way to control the sine of a positive acute angle using the angle and its tangent. Their strength comes from comparing a curved boundary with straight boundaries without needing to approximate the curve. The chord supplies an inscribed triangle, while the tangent supplies a circumscribed triangle.
This comparison is particularly useful when studying behavior for angles near zero. It provides precise bounds that can be used in later limit arguments. The bounds themselves do not evaluate a limit: they establish the geometric estimates needed before a limit argument is made.
When using the result, check three things: the angle is positive, it is less than \(\frac{\pi}{2}\), and it is expressed in radians. With those conditions in place, the area comparison yields
Check Your Understanding
Answer these questions using the geometry and conditions developed above.
- Why is the area of a sector in a unit circle with radian angle \(\theta\) equal to \(\frac12\theta\)?
- What are the areas of triangle \(OAB\), the sector, and triangle \(OAC\), respectively?
- Which geometric containments produce the strict inequalities, and why are they strict for \(0<\theta<\frac{\pi}{2}\)?
- Why does the tangent-triangle construction fail at \(\theta=\frac{\pi}{2}\)?
- What goes wrong if an angle measured in degrees is substituted directly for \(\theta\)?
