This note covers the three reciprocal trigonometric functions (secant, cosecant and cotangent) and the three inverse functions (arcsin, arccos and arctan). It also covers the two Pythagorean-type identities that involve them. They let you simplify expressions, solve equations and describe angles from known ratios.
Key ideas
The reciprocal functions are built from the three basic ones. Each is undefined wherever its denominator is zero, so each graph has vertical asymptotes.
- \(\sec\theta\) is \(1/\cos\theta\). It is undefined when \(\cos\theta=0\), for example at \(\theta=\tfrac{\pi}{2}\).
- \(\csc\theta\) is \(1/\sin\theta\). It is undefined when \(\sin\theta=0\), for example at \(\theta=0,\ \pi\).
- \(\cot\theta\) is \(1/\tan\theta=\cos\theta/\sin\theta\). It is undefined when \(\sin\theta=0\).
The graphs of \(\sec\) and \(\csc\) never enter the band between \(-1\) and \(1\), because the reciprocal of a number whose size is at most 1 has size at least 1. The range of both is \(y\le-1\) or \(y\ge1\). The range of \(\cot\) is all real numbers. The period of \(\sec\) and \(\csc\) is \(2\pi\) and the period of \(\cot\) is \(\pi\).
Dividing the identity \(\sin^2\theta+\cos^2\theta=1\) by \(\cos^2\theta\) gives \(\tan^2\theta+1=\sec^2\theta\). Dividing by \(\sin^2\theta\) gives \(1+\cot^2\theta=\csc^2\theta\). These let you switch between \(\tan\) and \(\sec\), or \(\cot\) and \(\csc\), without knowing the angle.
The inverse functions reverse the basic ones. Because \(\sin\), \(\cos\) and \(\tan\) repeat, each is first restricted to an interval where it is one-to-one. Then the inverse is a proper function.
- \(\arcsin x\): domain \(-1\le x\le1\), range \(-\tfrac{\pi}{2}\le y\le\tfrac{\pi}{2}\).
- \(\arccos x\): domain \(-1\le x\le1\), range \(0\le y\le\pi\).
- \(\arctan x\): domain all real numbers, range \(-\tfrac{\pi}{2}\lt y\lt\tfrac{\pi}{2}\). The lines \(y=\pm\tfrac{\pi}{2}\) are horizontal asymptotes.
Each inverse graph is the reflection of the restricted original graph in the line \(y=x\). The notation \(\sin^{-1}x\) means the same as \(\arcsin x\). It does not mean \(1/\sin x\).
The two-way relationship only holds in one direction without care. For \(x\) in the domain, \(\sin(\arcsin x)=x\). But \(\arcsin(\sin x)=x\) only if \(x\) is inside the restricted range.
When you sketch the reciprocal graphs, use the original graph as a guide. Draw the original curve lightly first. Wherever it crosses the \(x\)-axis, the reciprocal has a vertical asymptote. Wherever the original has a maximum of 1, the reciprocal has a minimum of 1 at the same place, and the other way round. Where the original is positive the reciprocal is positive, and where it is negative the reciprocal is negative. So \(\csc x\) is made of upward-opening and downward-opening U-shaped branches, each sitting on a peak or trough of \(\sin x\).
In general, \(\sec\) is an even function, so its graph is symmetric about the \(y\)-axis. \(\csc\) and \(\cot\) are odd functions, so their graphs have rotational symmetry about the origin. The graph of \(\cot x\) falls from left to right between each pair of asymptotes, which is the opposite of the way \(\tan x\) rises.
The identities are often used to prove other identities or to turn an equation into a quadratic. For example, replacing \(\sec^2x\) by \(1+\tan^2x\) turns an equation containing both \(\sec\) and \(\tan\) into one containing only \(\tan\). A calculator gives only one solution for an inverse function, so use the symmetry of the graph to find the others in the interval.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\sec\theta=\dfrac{1}{\cos\theta}\) | Reciprocal of cosine; asymptotes where \(\cos\theta=0\). |
| \(\csc\theta=\dfrac{1}{\sin\theta}\) | Reciprocal of sine; asymptotes where \(\sin\theta=0\). |
| \(\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{\cos\theta}{\sin\theta}\) | Reciprocal of tangent; asymptotes where \(\sin\theta=0\). |
| \(1+\tan^2\theta=\sec^2\theta\) | Links \(\tan\) and \(\sec\); useful for finding one from the other. |
| \(1+\cot^2\theta=\csc^2\theta\) | Links \(\cot\) and \(\csc\). |
| \(\arcsin x\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\) | The angle in that interval whose sine is \(x\). |
| \(\arccos x\in[0,\pi]\) | The angle in that interval whose cosine is \(x\). |
| \(\arctan x\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})\) | The angle in that interval whose tangent is \(x\). |
Worked examples
Example 1 The angle \(\theta\) satisfies \(\tan\theta=-2\) and \(\tfrac{\pi}{2}\lt\theta\lt\pi\). Find the exact values of \(\sec\theta\) and \(\csc\theta\).
Use \(\sec^2\theta=1+\tan^2\theta=1+(-2)^2=5\).
So \(\sec\theta=\pm\sqrt5\). In the second quadrant cosine is negative, so secant is negative: \(\sec\theta=-\sqrt5\).
Then \(\cos\theta=-\tfrac{1}{\sqrt5}\). Since \(\tan\theta=\sin\theta/\cos\theta\), \(\sin\theta=\tan\theta\cos\theta=(-2)\left(-\tfrac{1}{\sqrt5}\right)=\tfrac{2}{\sqrt5}\). This is positive, as it should be in the second quadrant.
So \(\csc\theta=\tfrac{1}{\sin\theta}=\tfrac{\sqrt5}{2}\).
Answer: \(\sec\theta=-\sqrt5\), \(\csc\theta=\dfrac{\sqrt5}{2}\).
Example 2 Solve \(\sec x=-2\) for \(0\le x\le2\pi\).
Rewrite in terms of cosine: \(\dfrac{1}{\cos x}=-2\), so \(\cos x=-\tfrac12\).
The acute reference angle is \(\arccos\tfrac12=\tfrac{\pi}{3}\).
Cosine is negative in the second and third quadrants, so \(x=\pi-\tfrac{\pi}{3}\) or \(x=\pi+\tfrac{\pi}{3}\).
Answer: \(x=\dfrac{2\pi}{3}\) or \(x=\dfrac{4\pi}{3}\).
Example 3 (a) Find the exact value of \(\arccos\left(\cos\tfrac{7\pi}{6}\right)\). (b) State the domain and range of \(f(x)=\arcsin(2x-1)\).
(a) First evaluate the inside: \(\cos\tfrac{7\pi}{6}=-\cos\tfrac{\pi}{6}=-\tfrac{\sqrt3}{2}\).
Now find the angle in \([0,\pi]\) with cosine \(-\tfrac{\sqrt3}{2}\). That is \(\pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}\).
Notice that the answer is not \(\tfrac{7\pi}{6}\), because \(\tfrac{7\pi}{6}\) lies outside the range of \(\arccos\).
(b) The input to \(\arcsin\) must lie between \(-1\) and \(1\): \(-1\le2x-1\le1\). Adding 1 gives \(0\le2x\le2\), so \(0\le x\le1\).
The output of \(\arcsin\) lies between \(-\tfrac{\pi}{2}\) and \(\tfrac{\pi}{2}\), and every value in that interval is reached as \(2x-1\) runs over \([-1,1]\).
Answer: (a) \(\dfrac{5\pi}{6}\). (b) Domain \(0\le x\le1\); range \(-\dfrac{\pi}{2}\le f(x)\le\dfrac{\pi}{2}\).
Common errors
- Pairing the wrong reciprocal with the wrong function. Secant goes with cosine, not sine; cosecant goes with sine.
- Reading \(\sin^{-1}x\) as \(\dfrac{1}{\sin x}\). The reciprocal is \(\csc x\).
- Writing \(\arccos(\cos x)=x\) for every \(x\). This is only true when \(x\) is in \([0,\pi]\).
- Forgetting to choose the sign when taking a square root in \(\sec^2\theta=1+\tan^2\theta\). The quadrant decides it.
- Giving a range of \([-1,1]\) for \(y=\sec x\). Its values are never strictly between \(-1\) and \(1\).
Exam tip: When an equation contains \(\sec\), \(\csc\) or \(\cot\), rewrite it in sine and cosine first. Then solve it as usual, and check each answer against the stated interval.
Check yourself
- Solve \(\csc x=2\) for \(0\le x\le2\pi\).
Answer
\(\sin x=\tfrac12\). The reference angle is \(\tfrac{\pi}{6}\), and sine is positive in quadrants 1 and 2. So \(x=\tfrac{\pi}{6}\) or \(x=\tfrac{5\pi}{6}\). - Given \(\cot\theta=\tfrac34\) with \(\theta\) acute, find \(\csc\theta\).
Answer
\(\csc^2\theta=1+\cot^2\theta=1+\tfrac{9}{16}=\tfrac{25}{16}\). Since \(\theta\) is acute, \(\csc\theta=\tfrac54\). - Find the exact value of \(\arcsin\left(-\tfrac12\right)\).
Answer
The angle in \([-\tfrac{\pi}{2},\tfrac{\pi}{2}]\) with sine \(-\tfrac12\) is \(-\tfrac{\pi}{6}\). - Find the exact value of \(\arcsin\left(\sin\tfrac{5\pi}{6}\right)\).
Answer
\(\sin\tfrac{5\pi}{6}=\tfrac12\). The angle in \([-\tfrac{\pi}{2},\tfrac{\pi}{2}]\) with sine \(\tfrac12\) is \(\tfrac{\pi}{6}\). So the answer is \(\tfrac{\pi}{6}\), not \(\tfrac{5\pi}{6}\).