Graphs of sec x, cosec x and cot x: asymptotes where the reciprocated function is zero, turning points at y = 1 and y = -1, period, domain, range, symmetry, the identities 1 + tan squared = sec squared and 1 + cot squared = cosec squared, and solving equations (moved into Extension 1 from Mathematics Advanced in the 2024 syllabus)

Domain

$\sec x = \frac{1}{\cos x}$ is undefined where $\cos x = 0$, so the domain is all real $x$ except $x = \frac{\pi}{2} + k\pi$ for integers $k$.

Range

Since $-1 \leq \cos x \leq 1$ and $\cos x \neq 0$, the reciprocal satisfies $\sec x \leq -1$ or $\sec x \geq 1$. The range is $(-\infty, -1] \cup [1, \infty)$.

Period

The same as cosine, $2\pi$.

Symmetry

$\sec(-x) = \frac{1}{\cos(-x)} = \frac{1}{\cos x} = \sec x$, so the function is even and the graph is symmetric about the $y$-axis.

Marker's note: one mark for the domain and range, one for the period and the symmetry with a reason.

Asymptotes

$\cot x = \frac{\cos x}{\sin x}$ is undefined where $\sin x = 0$, so the vertical asymptotes are $x = k\pi$ for integers $k$.

Intercepts

$\cot x = 0$ where $\cos x = 0$, that is at $x = \frac{\pi}{2} + k\pi$.

Period

$\pi$, the same as $\tan x$. The range is all real numbers and the function is odd, because $\cot(-x) = -\cot x$.

Marker's note: one mark for the asymptotes, one for the intercepts and the period.

Rewrite in cosine. $\sec x = 2$ means $\cos x = \frac{1}{2}$.

Solve in the given domain. The cosine is positive in the first and fourth quadrants, with reference angle $\frac{\pi}{3}$:

Marker's note: one mark for $\cos x = \frac{1}{2}$, one for each solution. $\sec x = 2$ does not mean $\cos x = 2$; the reciprocal flips.

Rewrite in sine. $\operatorname{cosec} x = -2$ gives $\sin x = -\frac{1}{2}$.

Solve. Sine is negative in the third and fourth quadrants, with reference angle $\frac{\pi}{6}$:

Marker's note: one mark for $\sin x = -\frac{1}{2}$, one for each solution.

Write $\sec x$ as $\frac{1}{\cos x}$ (valid when $\cos x \neq 0$):

Take roots. $\cos x = \pm\frac{1}{\sqrt{2}}$, which has one solution in each quadrant:

None of these makes $\cos x = 0$, so no solution is lost or invalid.

Marker's note: one mark for $\cos^2 x = \frac{1}{2}$, one for the two signs, one for all four solutions. Taking only the positive root is the typical slip.

Use $\tan^2 x = \sec^2 x - 1$. Then $2\left( \sec^2 x - 1 \right) + \sec x = 1$, which rearranges to

Case 1: $\sec x = 1$ gives $\cos x = 1$, so $x = 0$ or $x = 2\pi$.

Case 2: $\sec x = -\frac{3}{2}$ gives $\cos x = -\frac{2}{3}$. The reference angle is $\cos^{-1}\left( \frac{2}{3} \right) \approx 0.84$ rad, and cosine is negative in the second and third quadrants:

Check one value. At $x = 2.30$, $\tan^2 x \approx 1.25$ and $\sec x = -1.5$, so $2(1.25) - 1.5 = 1$.

Marker's note: one mark for the identity, one for the factorisation, one for the solutions from $\sec x = 1$, one for the two decimal solutions.

Compare with $y = \sin 2x$. The sine curve $y = \sin 2x$ has period $\frac{2\pi}{2} = \pi$ and zeroes at $x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$.

Vertical asymptotes occur at these zeroes: $x = 0$, $\frac{\pi}{2}$, $\pi$, $\frac{3\pi}{2}$ and $2\pi$.

Turning points occur where $\sin 2x = \pm 1$, so $y = \pm 1$:

• $\left( \frac{\pi}{4}, 1 \right)$ and $\left( \frac{5\pi}{4}, 1 \right)$ are minimum points on upward-opening branches,
• $\left( \frac{3\pi}{4}, -1 \right)$ and $\left( \frac{7\pi}{4}, -1 \right)$ are maximum points on downward-opening branches.

Shape. Between consecutive asymptotes there is one branch, alternately above $y = 1$ and below $y = -1$, so there are four branches in the interval.

Period and range. The period is $\pi$ and the range is $y \leq -1$ or $y \geq 1$.

Marker's note: one mark for the asymptotes, one for the turning points, one for the alternating branches, one for the period and range. The graph never enters the strip $-1 < y < 1$.

Use the cofunction identity

$\cot x = \tan\left( \frac{\pi}{2} - x \right) = -\tan\left( x - \frac{\pi}{2} \right)$.

Describe the transformations

Translate the graph of $y = \tan x$ right by $\frac{\pi}{2}$ to obtain $y = \tan\left( x - \frac{\pi}{2} \right)$, then reflect it in the $x$-axis to obtain $y = -\tan\left( x - \frac{\pi}{2} \right) = \cot x$.

Asymptotes

The asymptotes of $y = \tan x$ are $x = \frac{\pi}{2} + k\pi$. Translating right by $\frac{\pi}{2}$ moves them to $x = \pi + k\pi = k\pi$ (reflection in the $x$-axis does not move vertical lines), which matches the zeroes of $\sin x$.

Marker's note: one mark for the identity, one for the transformations in a valid order, one for the asymptotes.