17. Differentiation of Trigonometric Functions

Adapted from: Classical single-variable calculus (continuity of sin/cos, key limits, trig derivatives, wave graphs, angles between curves). Rewritten for teaching — not a verbatim textbook dump.

After Review of Trigonometry, we prove the derivative formulas and read the graphs of the six trig functions.

---

Continuity of $\cos x$ and $\sin x$

For any fixed $\theta$,

$$\lim_{h\to 0}\cos(\theta+h)=\cos\theta \qquad\text{and}\qquad \lim_{h\to 0}\sin(\theta+h)=\sin\theta.$$

On the unit circle (Fig. 17-1), as $h\to 0$ the terminal point $C$ for angle $\theta+h$ approaches the point $B$ for angle $\theta$. Hence the $x$-coordinate $\cos(\theta+h)$ approaches $\cos\theta$, and the $y$-coordinate $\sin(\theta+h)$ approaches $\sin\theta$.

Fig. 17-1 — continuity of sin and cos on the unit circle

---

Two key limits

To differentiate $\sin$ and $\cos$ we need

$$\begin{align} \lim_{\theta\to 0}\frac{\sin\theta}{\theta}&=1 \tag{17.1}\\[0.4em] \lim_{\theta\to 0}\frac{1-\cos\theta}{\theta}&=0 \tag{17.2} \end{align}$$

(Use radians — these limits fail in degrees.)

From (17.1) we get (17.2):

$$\begin{aligned} \frac{1-\cos\theta}{\theta} &=\frac{1-\cos\theta}{\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta} =\frac{\sin^{2}\theta}{\theta(1+\cos\theta)} =\frac{\sin\theta}{\theta}\cdot\frac{\sin\theta}{1+\cos\theta}. \end{aligned}$$

Hence

$$\lim_{\theta\to 0}\frac{1-\cos\theta}{\theta} =\lim_{\theta\to 0}\frac{\sin\theta}{\theta}\cdot\lim_{\theta\to 0}\frac{\sin\theta}{1+\cos\theta} =1\cdot\frac{0}{1+1}=0.$$

Check: key limits

Quiz

Key trig limits

1 / 2

lim θ→0 (sin θ)/θ equals…

Pick an answer to see if you are right.

---

Derivatives of sine and cosine

$$\begin{align} D_{x}(\sin x)&=\cos x \tag{17.3}\\[0.35em] D_{x}(\cos x)&=-\sin x \tag{17.4} \end{align}$$

The proof of (17.3) expands $\sin(x+h)$ with an addition formula, then applies (17.1)–(17.2). From (17.3) and the cofunction identity $\cos x=\sin(\pi/2-x)$, the Chain Rule gives

$$D_{x}(\cos x) =D_{x}\!\left(\sin\!\left(\frac{\pi}{2}-x\right)\right) =\cos\!\left(\frac{\pi}{2}-x\right)\cdot(-1) =-\sin x.$$

Hands On: Numerical check of $(\sin)'$

---

Graph of $\sin x$

Since $\sin(x+2\pi)=\sin x$, it is enough to sketch one period $[0,2\pi]$.

Critical numbers where $D_{x}(\sin x)=\cos x=0$ in $[0,2\pi]$: $x=\pi/2$ and $x=3\pi/2$.

Second derivative: $D_{x}^{2}(\sin x)=-\sin x$.

• At $\pi/2$: $-\sin(\pi/2)=-1<0$ → relative maximum $(\pi/2,1)$
• At $3\pi/2$: $-\sin(3\pi/2)=1>0$ → relative minimum $(3\pi/2,-1)$

Increasing where $\cos x>0$ (first and fourth quadrants on $[0,2\pi]$): $0<x<\pi/2$ and $3\pi/2<x<2\pi$.

Concave up where $-\sin x>0$ (third and fourth): $\pi<x<2\pi$. Inflection at $(\pi,0)$, and also at $(0,0)$ and $(2\pi,0)$ by periodicity.

Fig. 17-2 — y = sin x

---

Graph of $\cos x$

Cofunction: $\sin(\pi/2+x)=\cos x$. So the graph of $\cos x$ is the graph of $\sin x$ shifted left by $\pi/2$.

Fig. 17-3 — y = cos x (sin shifted left by π/2)

Both graphs are repeating waves of length $2\pi$. Multiplying the argument by a constant changes the period/frequency; multiplying the value changes the amplitude.

Example 17.1 — $y=\cos 3x$

$\cos 3(x+2\pi/3)=\cos(3x+2\pi)=\cos 3x$, so the period is $p=2\pi/3$. Over an interval of length $2\pi$ there are 3 waves — the frequency $f=3$. In general $pf=2\pi$, so $f=2\pi/p$.

Fig. 17-4 — y = cos 3x (period 2π/3, frequency 3)

For $b>0$, $\sin bx$ and $\cos bx$ have frequency $b$ and period $2\pi/b$.

Example 17.2 — $y=2\sin x$

Double the $y$-values of $\sin x$. Period and frequency unchanged ($p=2\pi$, $f=1$); amplitude (max height) is $2$.

Fig. 17-5 — y = 2 sin x (amplitude 2)

Example 17.3 — general $A\sin bx$

If $b>0$, then $y=A\sin bx$ and $y=A\cos bx$ have period $2\pi/b$, frequency $b$, and amplitude $|A|$.

Fig. 17-6 — y = 1.5 sin 4x

Check: amplitude / period

Quiz

Amplitude and period

1 / 2

Period of y = 3 cos(2x) is…

Pick an answer to see if you are right.

---

Other trigonometric functions

$$\begin{aligned} \tan x&=\frac{\sin x}{\cos x},& \cot x&=\frac{\cos x}{\sin x}=\frac{1}{\tan x},\\ \sec x&=\frac{1}{\cos x},& \csc x&=\frac{1}{\sin x}. \end{aligned}$$

Derivatives

$$\begin{align} D_{x}(\tan x)&=\sec^{2} x \tag{17.5}\\ D_{x}(\cot x)&=-\csc^{2} x \tag{17.6}\\ D_{x}(\sec x)&=\tan x\,\sec x \tag{17.7}\\ D_{x}(\csc x)&=-\cot x\,\csc x \tag{17.8} \end{align}$$

(Quotient / reciprocal rules from $\sin$ and $\cos$.)

Other relationships

$$\tan^{2} x+1=\sec^{2} x \tag{17.9}$$

Proof: $\dfrac{\sin^{2} x}{\cos^{2} x}+1=\dfrac{\sin^{2} x+\cos^{2} x}{\cos^{2} x}=\sec^{2} x$.

$$\tan(x+\pi)=\tan x,\qquad \cot(x+\pi)=\cot x \tag{17.10}$$

So $\tan$ and $\cot$ have period $\pi$ (not $2\pi$).

$$\tan(-x)=-\tan x,\qquad \cot(-x)=-\cot x \tag{17.11}$$

(Odd functions: $\sin$ odd, $\cos$ even.)

---

Graph of $y=\tan x$

Period $\pi$: sketch on $(-\pi/2,\pi/2)$. Oddness: draw $(0,\pi/2)$ and reflect through the origin.

Vertical asymptotes where $\cos x=0$: $x=\pm\pi/2$ (and every odd multiple of $\pi/2$ by periodicity).

$D_{x}(\tan x)=\sec^{2} x>0$ wherever defined → strictly increasing on each interval between asymptotes.

$$D_{x}^{2}(\tan x)=2\sec x\cdot(\tan x\sec x)=2\tan x\,\sec^{2} x,$$

so concave up when $\tan x>0$ (e.g. $0<x<\pi/2$), with an inflection at $(0,0)$.

$x$	$\tan x$
$0$	$0$
$\pi/6$	$\sqrt{3}/3\approx 0.58$
$\pi/4$	$1$
$\pi/3$	$\sqrt{3}\approx 1.73$

Fig. 17-7 — y = tan x on (−π/2, π/2)

For an acute angle in a right triangle: $\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}$.

---

Graph of $y=\sec x$

Asymptotes where $\cos x=0$: $x=(2n+1)\pi/2$. Period $2\pi$ like $\cos$. Also $|\sec x|\ge 1$ because $|\cos x|\le 1$.

Critical numbers from $D_{x}(\sec x)=\tan x\sec x=0$: $x=0$ (relative min) and $x=\pi$ (relative max) on $(-\pi,\pi)$, by the first-derivative test.

$$D_{x}^{2}(\sec x)=\sec x\,(\tan^{2} x+\sec^{2} x),$$

so no inflection on $(-\pi/2,\pi/2)$ (second derivative keeps the sign of $\sec x>0$ there) — concave upward.

Fig. 17-8 — y = sec x

Check: tan and sec

Quiz

Tan and sec

1 / 2

Period of tan x is…

Pick an answer to see if you are right.

---

Angles between curves

The angle of inclination of a nonvertical line $\mathscr{L}$ is the smaller counterclockwise angle $\alpha$ from the positive $x$-axis to $\mathscr{L}$. If $m$ is the slope, then

$$m=\tan\alpha.$$

Fig. 17-9 — inclination α

Reason (Fig. 17-10): a parallel unit-direction ray meets the circle at $(\cos\alpha,\sin\alpha)$, so slope $=\sin\alpha/\cos\alpha=\tan\alpha$.

Fig. 17-10 — m = tan α

The angle between two curves at an intersection $P$ is the smaller of the two angles between their tangent lines at $P$. If the slopes are $m_1$ and $m_2$,

$$\tan\phi=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|$$

when $1+m_1 m_2\neq 0$. If $1+m_1 m_2=0$, the tangents are perpendicular ($\phi=\pi/2$).

Check: inclination

Quiz

Inclination

1 / 2

A line of slope 1 has inclination…

Pick an answer to see if you are right.

---

Key takeaways

• Continuity of $\sin$ and $\cos$ from the unit circle; key limits $(\sin\theta)/\theta\to 1$ and $(1-\cos\theta)/\theta\to 0$ (radians).
• $D_x(\sin x)=\cos x$, $D_x(\cos x)=-\sin x$; then $\tan,\cot,\sec,\csc$ by quotient/reciprocal.
• $A\sin bx$: amplitude $|A|$, period $2\pi/b$, frequency $b$.
• $\tan$ has period $\pi$ and vertical asymptotes at odd multiples of $\pi/2$; $\sec=1/\cos$ shares those poles.
• Inclination: $m=\tan\alpha$; angle between curves = acute angle between tangents.

Further Reading

• 17b. Trig Diff — Practice — proofs and differentiation drills
• 18. Inverse Trig Functions — arcsin, arccos, arctan and derivatives
• 16. Review of Trigonometry — radians and unit-circle definitions
• 16b. Trig Review — Practice — conversion and identity drills
• 15. Curve Sketching — extrema, concavity, asymptotes used above