Differentiating trigonometric functions

The three basic derivatives

$$\frac{d}{dx}(\sin x) = \cos x$$

$$\frac{d}{dx}(\cos x) = -\sin x$$

$$\frac{d}{dx}(\tan x) = \sec^2 x$$

• The minus sign belongs to cosine. Differentiating $\cos$ introduces it; differentiating $\sin$ does not.
• These results are only true in radians. In degrees they pick up a factor of $\frac{\pi}{180}$, which is why Level 3 Calculus works in radians throughout.

The reciprocal trigonometric functions

• The standard explicitly includes reciprocal trigonometric functions:

  • $$\sec x = \dfrac{1}{\cos x}$$
  • $\csc x = \dfrac{1}{\sin x}$ (also written $\text{cosec}\,x$)
  • $$\cot x = \dfrac{1}{\tan x} = \dfrac{\cos x}{\sin x}$$

• Note the pairing: sec goes with cos, and cosec goes with sin. It is deliberately counterintuitive, and it is the first thing to check when one of these appears.

• Their derivatives:

$$\frac{d}{dx}(\sec x) = \sec x \tan x$$

$$\frac{d}{dx}(\csc x) = -\csc x \cot x$$

$$\frac{d}{dx}(\cot x) = -\csc^2 x$$

• The pattern: every function beginning with "co" — cos, cosec, cot — differentiates to something negative. That one observation carries all six signs.

With a function inside

• Combined with the chain rule, the working forms are:

$$\frac{d}{dx}\left(\sin(ax + b)\right) = a\cos(ax+b)$$

$$\frac{d}{dx}\left(\cos(ax+b)\right) = -a\sin(ax+b)$$

$$\frac{d}{dx}\left(\tan(ax+b)\right) = a\sec^2(ax+b)$$

• Generally, for any inner function $u$:

$$\frac{d}{dx}\left(\sin u\right) = u'\cos u \qquad \frac{d}{dx}\left(\cos u\right) = -u'\sin u$$

• Worked through:
  • $$\dfrac{d}{dx}\left(\sin 5x\right) = 5\cos 5x$$
  • $$\dfrac{d}{dx}\left(\cos(3x - 1)\right) = -3\sin(3x-1)$$
  • $$\dfrac{d}{dx}\left(\tan(x^2)\right) = 2x\sec^2(x^2)$$

Powers of trigonometric functions

• $\sin^3 x$ means $(\sin x)^3$ — the power is on the whole function, and it needs the chain rule:

$$\frac{d}{dx}\left(\sin^n x\right) = n\sin^{n-1}x \times \cos x$$

• Worked through: $\dfrac{d}{dx}\left(\cos^4 x\right) = 4\cos^3 x \times (-\sin x) = -4\cos^3 x\sin x$.

• Distinguish carefully:

  • $\sin^2 x = (\sin x)^2$ — square the output. Chain rule.
  • $\sin(x^2)$ — square the input first. Also chain rule, but a different one.
  • These give completely different derivatives: $2\sin x\cos x$ versus $2x\cos(x^2)$.

Identities worth having ready

• Questions often need an identity before or after differentiating:

  • $$\sin^2 x + \cos^2 x = 1$$
  • $$\sin 2x = 2\sin x\cos x$$
  • $$\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$$
  • $$1 + \tan^2 x = \sec^2 x$$

• The 2025 report notes that many candidates "tried to use trigonometric identities to differentiate, but lacked the skills to apply these correctly" — so know the few above properly rather than half-remembering many.

Solving the equations that result

• Setting a trigonometric derivative to zero gives a trigonometric equation, which has infinitely many solutions.
• Read the domain given in the question — usually something like $0 \leq x \leq 2\pi$ — and give every solution inside it.
• $\cos x = 0$ on $0 \leq x \leq 2\pi$ gives $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. Giving only one is a half-answer.

Exact values

• Keep answers exact where you can. The values worth knowing:

$x$	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$
$\sin x$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$
$\cos x$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$0$