Differentiating Trigonometric Functions

The six trigonometric functions all have derivatives that are themselves trigonometric, so differentiating them keeps you inside a family you already know: $\sin$ becomes $\cos$, $\tan$ becomes $\sec^2$, and so on. Combined with the chain, product and quotient rules and the identities from the trigonometry unit, these give the stationary-point and tangent questions that appear on almost every P3 paper, usually with exact answers in terms of $\pi$ and surds.

Why the derivative of sin x is cos x

Look at the graph of $y = \sin x$ with $x$ in radians. At $x = 0$ it climbs most steeply. At $x = \tfrac{\pi}{2}$ it is momentarily flat. At $x = \pi$ it is falling as steeply as it ever falls. Plot those gradients against $x$ and you get $1, 0, -1$ at $0, \tfrac{\pi}{2}, \pi$, which is exactly the graph of $\cos x$.

$y = \sin x$ (solid) and its gradient function $y = \cos x$ (dashed). Where $\sin x$ peaks, $\cos x$ crosses zero.

The precise argument uses the small-angle fact $\dfrac{\sin h}{h} \to 1$ as $h \to 0$, which is true only in radians. In degrees, $\dfrac{\sin h^\circ}{h} \to \dfrac{\pi}{180}$, so every derivative would carry an ugly factor of $\dfrac{\pi}{180}$. Radians are the units in which trigonometric calculus is clean, and that is why P3 calculus always uses them.

The sign pattern is worth fixing in memory: differentiating $\sin \to \cos \to -\sin \to -\cos \to \sin$ cycles every four steps. The functions starting with "co" ($\cos$, $\operatorname{cosec}$, $\cot$) all have a minus sign in their derivative.

The reciprocal functions

$\sec x = \dfrac{1}{\cos x}$, $\operatorname{cosec} x = \dfrac{1}{\sin x}$ and $\cot x = \dfrac{\cos x}{\sin x}$, as in reciprocal trigonometric functions. Their derivatives follow from the chain rule or the quotient rule.

The same methods give $\operatorname{cosec} x$ and $\cot x$ (see the practice questions).

These are in the list of formulae, but you may be asked to prove any of them, and proofs of $\tan x$ and $\sec x$ are common opening parts.

Composites: powers and functions of trig functions

Read the notation carefully, because it decides the layers.

Expression	Means	Derivative
$\sin 3x$	$\sin(3x)$	$3\cos 3x$
$\sin^3 x$	$(\sin x)^3$	$3\sin^2 x\cos x$
$\sin x^3$	$\sin(x^3)$	$3x^2\cos(x^3)$
$\tan^2 x$	$(\tan x)^2$	$2\tan x\sec^2 x$
$\sec 3x$	$\sec(3x)$	$3\sec 3x\tan 3x$
$\ln(\sin x)$	$\ln$ of $\sin x$	$\dfrac{\cos x}{\sin x} = \cot x$
$\ln(\sec x)$	$\ln$ of $\sec x$	$\dfrac{\sec x\tan x}{\sec x} = \tan x$
$e^{\cos x}$	$e$ to the $\cos x$	$-\sin x\, e^{\cos x}$

Note that $\tan^{-1} x$ is the inverse function, not $(\tan x)^{-1} = \cot x$. It has its own note: differentiating the inverse tangent.

Use identities before or after differentiating

Trigonometric identities often make a derivative simpler or make $\dfrac{dy}{dx} = 0$ solvable.

Before: $2\sin x\cos x = \sin 2x$, so its derivative is $2\cos 2x$ with no product rule. Similarly $\cos^2 x - \sin^2 x = \cos 2x$.

After: if $\dfrac{dy}{dx}$ contains both $\cos 2x$ and $\sin x$, rewrite $\cos 2x = 1 - 2\sin^2 x$ to get a quadratic in $\sin x$. If it contains $\sin 2x$ and $\cos x$, use $\sin 2x = 2\sin x\cos x$ and factorise out $\cos x$.

The standard identities are in double angles and compound angles. In the solving step, a common factor such as $\cos x$ must be factorised out, never cancelled, or you lose the solutions with $\cos x = 0$.

Worked examples

$y = 2\sin x + \cos 2x$ on $0 \le x \le \pi$: two equal maxima and a minimum between them.

Practice

Answers

1. (a) $-15\sin(3x - 1)$. (b) $4\sin^3 x\cos x$. (c) $-2\operatorname{cosec} 2x\cot 2x$. (d) $2x\sec^2(x^2)$.

2. (a) $\dfrac{\sec x\tan x}{\sec x} = \tan x$. (b) $\dfrac{2\cos 2x}{\sin 2x} = 2\cot 2x$.

3. $\operatorname{cosec} x = (\sin x)^{-1}$, so the derivative is $-(\sin x)^{-2}\cos x = -\dfrac{1}{\sin x}\cdot\dfrac{\cos x}{\sin x} = -\operatorname{cosec} x\cot x$.

4. $\dfrac{dy}{dx} = \sin x + x\cos x$. At $\tfrac{\pi}{3}$: $\dfrac{\sqrt{3}}{2} + \dfrac{\pi}{3}\cdot\dfrac{1}{2} = \dfrac{\sqrt{3}}{2} + \dfrac{\pi}{6}$.

5. $y = \sec\tfrac{\pi}{3} = 2$. Gradient $\sec\tfrac{\pi}{3}\tan\tfrac{\pi}{3} = 2\sqrt{3}$. Tangent: $y - 2 = 2\sqrt{3}\left(x - \tfrac{\pi}{3}\right)$.

6. $\dfrac{dy}{dx} = e^x\cos x - e^x\sin x = e^x(\cos x - \sin x)$. Then $\dfrac{d^2y}{dx^2} = e^x(\cos x - \sin x) + e^x(-\sin x - \cos x) = -2e^x\sin x$.

7. $\dfrac{dy}{dx} = \sec^2 x - \operatorname{cosec}^2 x = \dfrac{\sin^2 x - \cos^2 x}{\sin^2 x\cos^2 x} = \dfrac{-\cos 2x}{\tfrac{1}{4}\sin^2 2x} = -\dfrac{4\cos 2x}{\sin^2 2x}$, using $\sin x\cos x = \tfrac{1}{2}\sin 2x$. Stationary when $\cos 2x = 0$: $2x = \tfrac{\pi}{2}$, $x = \tfrac{\pi}{4}$, where $y = 1 + 1 = 2$. Point $\left(\tfrac{\pi}{4}, 2\right)$.

8. $\dfrac{dy}{dx} = \cos x(1 + \cos x) + \sin x(-\sin x) = \cos x + \cos^2 x - \sin^2 x = 2\cos^2 x + \cos x - 1 = (2\cos x - 1)(\cos x + 1)$. In $0 < x < \pi$, $\cos x \ne -1$, so $\cos x = \tfrac{1}{2}$, $x = \tfrac{\pi}{3}$, $y = \tfrac{\sqrt{3}}{2}\cdot\tfrac{3}{2} = \tfrac{3\sqrt{3}}{4}$. Sign test: at $x = \tfrac{\pi}{6}$, $(2\cos x - 1) > 0$; at $x = \tfrac{\pi}{2}$, $2\cos x - 1 = -1 < 0$; and $\cos x + 1 > 0$ throughout. The gradient goes from positive to negative: a maximum at $\left(\tfrac{\pi}{3}, \tfrac{3\sqrt{3}}{4}\right)$.

9. Quotient rule: $\dfrac{dy}{dx} = \dfrac{(1 + \sin^2 x)(-\sin x) - \cos x(2\sin x\cos x)}{(1 + \sin^2 x)^2} = \dfrac{-\sin x\left(1 + \sin^2 x + 2\cos^2 x\right)}{(1 + \sin^2 x)^2}$. Since $2\cos^2 x = 2 - 2\sin^2 x$, the bracket is $3 - \sin^2 x$, giving $\dfrac{\sin x(\sin^2 x - 3)}{(1 + \sin^2 x)^2}$. Now $\sin^2 x - 3 \le 1 - 3 < 0$, never zero, so stationary points need $\sin x = 0$: $x = 0, \pi, 2\pi$.