Deriving the Sine Derivative from First Principles

Why Derive the Sine Derivative?

A derivative describes the instantaneous rate of change of a function. For sine, the familiar rule is that its derivative is cosine. First principles reveal why this relationship holds and which assumptions are built into it.

The calculation depends on the change in sine between two nearby inputs. The sine subtraction identity rewrites that change in terms of sine and cosine of the small increment. Two standard trigonometric limits then determine what happens as the increment approaches zero.

The Difference Quotient

For a function \(f\), its derivative at \(x\), when the limit exists, is defined by

Here \(h\) is a nonzero increment that approaches zero. For \(f(x)=\sin x\), the quotient becomes

The numerator is not yet in a form where the limit is immediate. The sine subtraction identity, written as an addition identity, states

Using this identity in the difference quotient gives

The factor \(x\) is fixed while \(h\) approaches zero. Thus the key is to determine the limits of the two expressions involving \(h\).

The Trigonometric Limits Needed

The standard small-angle limit, with angles measured in radians, is

One justification comes from the unit-circle geometric inequality. For \(0<h<\frac{\pi}{2}\), comparing areas in the unit circle gives \(\sin h<h<\tan h\). Since all quantities are positive, division yields

As \(h\to0^+\), continuity of cosine gives \(\cos h\to1\). The squeeze theorem therefore gives \(\frac{\sin h}{h}\to1\) from the right. For negative \(h\), the quotient is even:

So the same limit holds from the left, establishing the two-sided limit. The geometric inequality uses radian measure: on a unit circle, the angle in radians equals the length of its intercepted arc.

We also need the limit of \(\frac{\cos h-1}{h}\). The half-angle identity \(\cos h-1=-2\sin^2(h/2)\) gives, for \(h\ne0\),

The last equality follows because \(h=2(h/2)\): multiplying \(\frac{\sin(h/2)}{h/2}\) by \(\sin(h/2)\) gives \(\frac{2\sin^2(h/2)}{h}\). As \(h\to0\), the first factor approaches \(1\) by the standard sine limit, and the second approaches \(0\). Hence

This factorization is useful to check carefully: the coefficient multiplying the product is \(-1\), not \(-\frac12\).

Deriving the Derivative

We can now take the limit of the difference quotient. The sine subtraction identity has already expressed it as a sum of a fixed multiple of each standard limit:

The limit laws apply because \(\sin x\) and \(\cos x\) are constants with respect to \(h\), and both limits in the sum exist. This proves the derivative formula at every fixed real \(x\).

Worked Examples

Why Radian Measure Matters

The limit \(\frac{\sin h}{h}\to1\) is stated for a numerical input \(h\) measured in radians. If the input is instead a number of degrees, an increment of \(h\) degrees corresponds to an increment of \(\frac{\pi h}{180}\) radians. The degree-based difference quotient consequently includes the conversion factor \(\frac{\pi}{180}\), as the preceding example showed.

This is not a change in the geometry of sine. It is a change in the unit used to measure the input. A rate of change per degree and a rate of change per radian have different numerical values because one radian and one degree represent different angular increments.

A Common Pitfall

A frequent error is to apply \(\lim_{h\to0}\frac{\sin h}{h}=1\) without checking that \(h\) is measured in radians. For degree inputs, writing that limit as \(1\) skips the conversion.

For example, at zero, if \(x\) is measured in degrees, then

As \(h\to0\), the quotient on the right approaches \(1\), so the limit is \(\frac{\pi}{180}\), not \(1\). The correction is to convert the increment to radians before using the standard limit.

Another algebraic pitfall is losing a factor when rewriting the cosine limit. In particular,

not \(-\frac12\left(\frac{\sin(h/2)}{h/2}\right)\sin(h/2)\). The displayed identity can be checked by multiplying: \(\frac{\sin(h/2)}{h/2}\sin(h/2)=\frac{2\sin^2(h/2)}{h}\).