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Trigonometric and Exponential Derivatives · Tutorial 142 of 999

Deriving the Cosine Derivative

Use the difference quotient and radian-based trigonometric limits to derive the cosine derivative and understand why its sign is negative.

Intermediate 9 min read

What You'll Learn

  • Set up the derivative of cosine from its limit definition.
  • Use the cosine angle-addition identity to simplify the difference quotient.
  • Apply the standard trigonometric limits needed for the derivation.
  • Explain why the cosine derivative has a negative sign.
  • Compute cosine slopes at selected angles directly from the definition.
  • Recognize a common sign error in the cosine difference quotient.

Why Derive the Cosine Derivative?

The familiar formula for the derivative of cosine is short, but it has a sign that is easy to misremember: the derivative is negative sine, not sine. Starting from the definition of a derivative makes the sign understandable. It also shows exactly which trigonometric identities and limits are needed.

In this derivation, angles are measured in radians. The standard trigonometric limits used below apply with radian measure. The derivative definition asks us to examine the change in a function over a small change \(h\) in its input and then take the limit as \(h\) approaches zero.

Goal: Starting from the difference quotient for \(f(x)=\cos x\), show that for every real \(x\), \(f'(x)=-\sin x\).

The Difference Quotient for Cosine

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

$$ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}. $$

The value \(h\) must be nonzero inside the quotient; the limit describes what happens as such nonzero values approach zero. For the cosine function, substitute \(f(x)=\cos x\):

$$ \frac{d}{dx}\cos x = \lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h}. $$

At this stage, the quotient still depends on both \(x\) and \(h\). The cosine angle-addition identity will separate those roles:

$$ \cos(x+h)=\cos x\cos h-\sin x\sin h. $$

The minus sign in this identity is crucial. It will be carried into the difference quotient and ultimately into the derivative.

The Limits Needed

The standard trigonometric limit is

$$ \lim_{u\to 0}\frac{\sin u}{u}=1. $$

We also need the limit of \((\cos u-1)/u\). It equals zero. This follows from the identity \(1-\cos u=2\sin^2(u/2)\) and the sine limit above. For nonzero \(u\), rewrite the expression as

$$ \frac{\cos u-1}{u} = -\frac{2\sin^2(u/2)}{u} = -\frac{u}{2} \left(\frac{\sin(u/2)}{u/2}\right)^2. $$

As \(u\) approaches zero, the ratio \(\sin(u/2)/(u/2)\) approaches \(1\), while \(u/2\) approaches zero. Therefore, the entire product approaches zero:

$$ \lim_{u\to 0}\frac{\cos u-1}{u}=0. $$

This second limit says that near zero, the change in cosine contributes no term proportional to \(h\) in the difference quotient. The sine term, in contrast, has a quotient that approaches \(1\). That distinction is what allows the derivative to simplify.

Deriving the Formula

Apply the angle-addition identity to the numerator of the difference quotient. Then collect the terms involving \(\cos x\) and \(\sin x\):

$$ \begin{aligned} \frac{\cos(x+h)-\cos x}{h} &= \frac{\cos x\cos h-\sin x\sin h-\cos x}{h}\\ &= \cos x\frac{\cos h-1}{h} - \sin x\frac{\sin h}{h}. \end{aligned} $$

For a fixed real number \(x\), the values \(\cos x\) and \(\sin x\) do not vary as \(h\) approaches zero. The two standard limits can therefore be applied to the two quotients separately.

1
Start with the definition.
Write the difference quotient \(\bigl(\cos(x+h)-\cos x\bigr)/h\).
2
Use the angle-addition identity.
Replace \(\cos(x+h)\) with \(\cos x\cos h-\sin x\sin h\), preserving the minus sign.
3
Separate the small-angle quotients.
The expression becomes \(\cos x(\cos h-1)/h-\sin x\sin h/h\).
4
Take the limit.
The first quotient approaches \(0\); the second approaches \(1\). The result is \(-\sin x\).

In full, the limit is

$$ \begin{aligned} \lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h} &= \cos x\lim_{h\to 0}\frac{\cos h-1}{h} - \sin x\lim_{h\to 0}\frac{\sin h}{h}\\ &= \cos x(0)-\sin x(1)\\ &=-\sin x. \end{aligned} $$
Theorem: For every real \(x\), the derivative of cosine is $$ \frac{d}{dx}\cos x=-\sin x. $$ The derivation uses the cosine angle-addition identity and the radian-based limits \(\lim_{h\to0}(\cos h-1)/h=0\) and \(\lim_{h\to0}\sin h/h=1\).

Where the Negative Sign Comes From

The negative sign is not an extra rule added at the end. It is already present when the angle-addition identity is substituted into the difference quotient. The identity contains the term \(-\sin x\sin h\). After division by \(h\), that term becomes \(-\sin x(\sin h/h)\), and the sine quotient approaches positive \(1\). Thus the surviving contribution is \(-\sin x\).

The other contribution, \(\cos x(\cos h-1)/h\), approaches zero. It does not change the sign or contribute another term to the derivative. In short, the minus sign comes from the cosine addition identity, while the standard sine limit determines that the factor multiplying it approaches \(1\).

Worked Examples from the Definition

Worked Example: Find the Slope of Cosine at \(x=0\)

Use the difference quotient at zero, not just the general derivative formula:

$$ \begin{aligned} \lim_{h\to0}\frac{\cos(0+h)-\cos 0}{h} &=\lim_{h\to0}\frac{\cos h-1}{h}\\ &=0. \end{aligned} $$

The limit is zero by the cosine limit derived above. Now check against the general result: \(-\sin 0=0\). The direct calculation and the formula agree, so the slope at \(x=0\) is \(0\).

Worked Example: Find the Slope of Cosine at \(x=\pi/2\)

Use the angle-addition identity to evaluate the difference quotient at \(x=\pi/2\):

$$ \begin{aligned} \lim_{h\to0} \frac{\cos(\pi/2+h)-\cos(\pi/2)}{h} &= \lim_{h\to0}\frac{-\sin h-0}{h}\\ &= -\lim_{h\to0}\frac{\sin h}{h}\\ &=-1. \end{aligned} $$

Here, \(\cos(\pi/2+h)=-\sin h\) follows from the cosine addition identity, and \(\cos(\pi/2)=0\). Check with the derived formula: \(-\sin(\pi/2)=-1\). Therefore, the slope at \(x=\pi/2\) is \(-1\), as the direct limit confirms.

Worked Example: Find the Slope of Cosine at \(x=\pi/3\)

Let \(x=\pi/3\) in the general difference quotient and use the exact values \(\cos(\pi/3)=1/2\) and \(\sin(\pi/3)=\sqrt{3}/2\):

$$ \begin{aligned} \lim_{h\to0} \frac{\cos(\pi/3+h)-\cos(\pi/3)}{h} &= \lim_{h\to0} \left[ \frac{1}{2}\frac{\cos h-1}{h} - \frac{\sqrt{3}}{2}\frac{\sin h}{h} \right]\\ &= \frac{1}{2}(0)-\frac{\sqrt{3}}{2}(1)\\ &=-\frac{\sqrt{3}}{2}. \end{aligned} $$

Check by substituting \(x=\pi/3\) into the derivative formula: \(-\sin(\pi/3)=-\sqrt{3}/2\). The direct difference-quotient limit gives the same exact slope.

Common Sign Error

A frequent mistake is to write the cosine addition identity with a plus sign, as if \(\cos(x+h)=\cos x\cos h+\sin x\sin h\). That is the wrong identity. Using it would lead to this incorrect calculation:

$$ \frac{\cos(x+h)-\cos x}{h} \ \overset{\text{wrong}}{=}\ \cos x\frac{\cos h-1}{h} + \sin x\frac{\sin h}{h} \ \longrightarrow\ \sin x. $$

The error occurs before taking the limit: the cosine angle-addition formula has a minus sign between its two terms. Correcting that identity corrects the quotient and the result:

$$ \cos(x+h)=\cos x\cos h-\sin x\sin h \quad\Longrightarrow\quad \frac{\cos(x+h)-\cos x}{h} = \cos x\frac{\cos h-1}{h} - \sin x\frac{\sin h}{h} \longrightarrow -\sin x. $$

A useful check is \(x=\pi/2\): the incorrect result would give \(\sin(\pi/2)=1\), while the direct difference quotient gives \(-1\). Keeping the addition identity visible until the quotient is simplified helps prevent this sign error.

Why the Derivation Matters

Memorizing a derivative formula can make routine differentiation quick, but the difference-quotient derivation explains what the formula records. A derivative is a limiting rate of change: here, it measures how cosine changes when its input changes by a small amount. The angle-addition identity expresses that change in terms of the small angle \(h\), and the standard limits identify which terms remain as \(h\) approaches zero.

The derivation also provides a reliable way to check the sign. At an angle such as \(\pi/2\), the direct quotient is \(-\sin h/h\), which approaches \(-1\). A claimed positive sign cannot agree with that computation. When applying the derivative later, the formula is concise; when the sign is uncertain, the derivation shows where it comes from.

Check Your Understanding

Answer each question using the derivation and examples above.

  1. Write the difference quotient used to define the derivative of \(\cos x\).
  2. Which angle-addition identity is used, and which term in it produces the negative sign?
  3. Why does \((\cos h-1)/h\) approach zero as \(h\) approaches zero?
  4. Use the standard trigonometric limits to find the slope of cosine at \(x=\pi\).
  5. What incorrect derivative results if the cosine addition identity is written with a plus sign in place of its minus sign?

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