3.5 Derivatives of Trigonometric Functions

Big idea

Sine and cosine are the first functions in this chapter that are not built from arithmetic on $x$. The power rule has nothing to say about them, so their derivatives have to come from the limit definition, and the limit definition needs two facts about small angles that were established with the squeeze theorem: the ratio of $\sin\theta$ to $\theta$ tends to $1$, and the ratio of $\cos\theta - 1$ to $\theta$ tends to $0$.

Those two limits are the entire content of this section. Once the angle addition formula splits $\sin(x+h)$ into pieces, the difference quotient sorts itself into one term containing $(\cos h - 1)/h$ and another containing $(\sin h)/h$, and the two special limits finish the job. The answer, that the derivative of sine is cosine, then propagates: cosine by the same argument, and the other four functions by the quotient rule, since each is a ratio of sines and cosines.

The first of those limits is also the reason radians are not optional. The statement that $\sin\theta$ behaves like $\theta$ for small $\theta$ is false in degrees, where the ratio approaches $\pi/180$ instead. Every derivative formula in this section silently assumes radian measure, and using degrees introduces a constant factor into all of them.

What you get is a second family of building blocks. From here on, a differentiation problem can mix polynomials and trigonometric functions in products and quotients, and the work is deciding which structural rule governs the outermost operation before reaching for any formula.

Decoder

The derivative of $\sin x$ is $\cos x$, the derivative of $\cos x$ is $-\sin x$, and both statements require the angle to be measured in radians.

The minus sign on the cosine derivative is not decoration. Cosine starts at its maximum at $x = 0$ and decreases through the first quadrant, so its rate of change there must be negative, and $-\sin x$ is negative exactly on the intervals where cosine falls. Reading the pair off the graphs is the fastest way to recover which one picks up the sign if you forget.

Differentiating four times returns you to where you started, since sine goes to cosine to $-\sin x$ to $-\cos x$ and back. Higher derivatives of sine and cosine therefore depend only on the remainder of the order after division by four, which makes a question about the fiftieth derivative a question about arithmetic.

The classic mistake is treating $\sin^2 x$ and $\sin(x^2)$ as if they were the same object, or applying a rule to $\sin(3x)$ that only covers $\sin x$. Until the composition rule arrives, keep the arguments simple: the formulas here are for the functions of $x$ itself, and anything nested needs the tool of the next section.

Definitions and results

The two special limits. For $\theta$ in radians,

$$ \lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1, \qquad \lim_{\theta \to 0}\frac{\cos\theta - 1}{\theta} = 0 $$

Derivatives of sine and cosine.

$$ \frac{d}{dx}\sin x = \cos x, \qquad \frac{d}{dx}\cos x = -\sin x $$

Derivatives of the other four. Each follows from the quotient rule applied to a ratio of sine and cosine:

$$ \frac{d}{dx}\tan x = \sec^2 x, \qquad \frac{d}{dx}\cot x = -\csc^2 x $$

$$ \frac{d}{dx}\sec x = \sec x \tan x, \qquad \frac{d}{dx}\csc x = -\csc x \cot x $$

The co- pattern. Every function whose name begins with “co” has a minus sign in its derivative. That one observation covers half the table.

Domains. The formulas hold wherever the function itself is defined. Tangent and secant are undefined at odd multiples of $\pi/2$, and cotangent and cosecant at multiples of $\pi$, so their derivatives are as well.

Higher derivatives cycle. Since $\frac{d^4}{dx^4}\sin x = \sin x$, the $n$th derivative of sine is determined by the remainder of $n$ divided by $4$. The same holds for cosine.

Combining with the structural rules. A product such as $x^2 \sin x$ needs the product rule with a trigonometric factor; a ratio such as $\frac{\sin x}{1 + \cos x}$ needs the quotient rule. The trigonometric formulas supply $f'$ and $g'$; the structural rule decides how they are assembled.

Identities simplify the answer. After differentiating, look for $\sin^2 x + \cos^2 x$, which collapses to $1$, and for common factors. The unsimplified answer is correct but often three times as long.

Worked examples

Where the derivative of sine comes from

Start from the definition and expand with the angle addition formula, $\sin(x+h) = \sin x \cos h + \cos x \sin h$:

$$ \frac{\sin(x+h) - \sin x}{h} = \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h} $$

For fixed $x$, the factors $\sin x$ and $\cos x$ are constants as $h \to 0$. The first fraction tends to $0$ and the second to $1$, so the whole expression tends to $\cos x$. The same argument with $\cos(x+h) = \cos x \cos h - \sin x \sin h$ produces $-\sin x$, the minus sign coming from the minus in the identity.

Tangent from the quotient rule

Write $\tan x = \dfrac{\sin x}{\cos x}$ and differentiate:

$$ \frac{d}{dx}\tan x = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x $$

The Pythagorean identity is what turns a three-term answer into one term. The result is positive everywhere tangent is defined, which matches the fact that tangent increases across every branch.

A product with a tangent line

Differentiate $y = x^2 \sin x$ and find the tangent line at $x = \pi$. The product rule gives

$$ \frac{dy}{dx} = 2x\sin x + x^2\cos x $$

At $x = \pi$, $\sin \pi = 0$ and $\cos \pi = -1$, so the point is $(\pi, 0)$ and the slope is $2\pi(0) + \pi^2(-1) = -\pi^2$. The tangent line is

$$ y = -\pi^2 (x - \pi) $$

A sign check: just past $x = \pi$, sine is negative and $x^2$ is positive, so $y$ is negative and the curve is falling through the axis. A negative slope is what that requires.

A quotient that collapses

Differentiate $y = \dfrac{\sin x}{1 + \cos x}$. With $f = \sin x$ and $g = 1 + \cos x$:

$$ \frac{dy}{dx} = \frac{\cos x (1 + \cos x) - \sin x(-\sin x)}{(1+\cos x)^2} = \frac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2} $$

The identity turns the numerator into $\cos x + 1$, which cancels one factor from the denominator:

$$ \frac{dy}{dx} = \frac{1}{1 + \cos x} $$

Check at $x = 0$: the formula gives $1/2$. Near the origin the function itself is close to $x/2$, since $\sin x \approx x$ and $1 + \cos x \approx 2$, and a function behaving like $x/2$ has slope $1/2$. The two agree.

Practice

Start with the six formulas on their own, evaluated at specific angles and combined with constant multiples and sums. Keep every angle in radians.

Then mix them with the structural rules. These place trigonometric functions inside products and quotients, so identify the outermost operation first, then supply the trigonometric derivatives as the pieces.

Quiz

Five items on the derivatives of the six trigonometric functions and on products and quotients built from them.