The Weierstrass Substitution and Its Derivation

Why Use the Half-Angle Parameter?

The substitution \(t=\tan(x/2)\) is useful because it expresses both \(\sin x\) and \(\cos x\) as rational functions of one variable. The key idea is geometric: a line of slope \(t\) through a fixed point on the unit circle meets the circle at another point whose coordinates are rational functions of \(t\).

This is not an arbitrary algebraic trick. The parameter \(t\) is the slope of a line, and the unit circle gives the trigonometric coordinates. Once those coordinates have been derived, differentiating the relation between \(t\) and \(x\) gives the corresponding formula for \(dx\).

Deriving the Circle Parametrization

Use \(X\) and \(Y\) for coordinates in the plane, so that \(x\) can continue to denote the angle. The unit circle has equation \(X^2+Y^2=1\), and the point corresponding to angle \(x\) has coordinates \((\cos x,\sin x)\).

Choose the fixed point \((-1,0)\) on the circle. A nonvertical line through this point with slope \(t\) has equation \(Y=t(X+1)\). Find its second intersection with the circle by substituting this equation into \(X^2+Y^2=1\):

Expanding and collecting terms gives

The known intersection \(X=-1\) is one root. Factoring the quadratic makes the other root visible:

The second intersection therefore has

Substitute this value into the equation of the line:

For the point at angle \(x\), \(X=\cos x\) and \(Y=\sin x\). The line’s slope is consequently \(Y/(X+1)\), which, using the half-angle identities or the geometry of the angle, is \(\tan(x/2)\). Thus setting \(t=\tan(x/2)\) yields the two rational formulas.

Relating the Slope to the Half-Angle

The geometric slope relation can also be seen directly. Let \(\theta=x/2\). The standard half-angle identities give

provided \(\cos(x/2)\ne 0\). On the unit circle, \(\sin x/(1+\cos x)\) is exactly \(Y/(X+1)\), the slope of the line from \((-1,0)\) to \((X,Y)\). Hence this slope is \(t=\tan(x/2)\). Combining this with the coordinate calculation establishes

Deriving the Differential

On any interval where \(x\) avoids odd multiples of \(\pi\), differentiate \(t=\tan(x/2)\) with respect to \(x\). The chain rule gives

Since \(1+t^2\) is positive for real \(t\), this derivative is nonzero. We can solve for the differential:

The same result follows by using the inverse relation \(x=2\arctan t\) on the interval \(-\pi<x<\pi\). Differentiating gives \(dx/dt=2/(1+t^2)\). The two derivations express the same local change of variable.

Worked Examples

Domain and Interpretation

The parameter \(t\) is real whenever \(x/2\) is not an angle at which tangent is undefined. Thus the excluded angles are \(x=(2k+1)\pi\), where \(k\) is an integer. Geometrically, these angles correspond to the point \((-1,0)\), the fixed point used to draw the line. There is no finite slope for a line from that point to itself, so it is not represented by a finite value of \(t\).

On the interval \(-\pi<x<\pi\), the angle and parameter correspond one-to-one: as \(x\) ranges over this interval, \(t\) ranges over all real numbers, and \(x=2\arctan t\). Outside this interval, the tangent has period \(\pi\), so \(t\) records the angle modulo \(2\pi\). This is why the rational formulas for sine and cosine remain valid on other intervals but the angle must be interpreted with the appropriate branch.

Why the Substitution Rationalizes Trigonometric Expressions

A rational function of \(t\) is a quotient of polynomials in \(t\). The formulas for sine and cosine are rational functions, and the differential factor \(2/(1+t^2)\) is rational as well. Consequently, expressions built from sums, products, and quotients of sine and cosine can be rewritten using rational operations in \(t\), wherever their denominators are nonzero.

For example, \(\sin x+\cos x\) becomes

This observation explains the method’s importance in integral calculus: after substituting the expressions for sine, cosine, and \(dx\), a trigonometric integrand may become a rational function of \(t\). Whether a particular integral is simpler after substitution depends on its form; rationalization is a tool, not a guarantee of shorter work.

A Common Mistake

A frequent error is to set \(t=\tan x\) instead of \(t=\tan(x/2)\), then use the half-angle formulas. For \(x=\pi/3\), that incorrect choice gives \(t=\sqrt{3}\), and the formulas would produce

The sine value happens to be correct, but the cosine value is not: \(\cos(\pi/3)=1/2\). The error is using the wrong parameter. The correct value is \(t=\tan(\pi/6)=1/\sqrt{3}\), which gives \(\cos x=1/2\). Always define \(t\) as the tangent of the half-angle before applying these formulas.

Summary

The Weierstrass substitution comes from parametrizing the unit circle by the slope of a line through \((-1,0)\). The second intersection has coordinates \(((1-t^2)/(1+t^2),\,2t/(1+t^2))\), and identifying those coordinates with \((\cos x,\sin x)\) gives the rational formulas. Differentiating \(t=\tan(x/2)\) supplies the differential factor.