Reciprocal and inverse trigonometric functions

Reciprocal and inverse trigonometric functions are different ideas. This distinction is essential:

$$\boxed{\sec x=\frac1{\cos x}} \qquad\text{but}\qquad \boxed{\cos^{-1}x=\arccos x}.$$

The first takes the reciprocal of a trigonometric value. The second returns an angle whose cosine is $x$. In particular,

$$\cos^{-1}x\ne \frac1{\cos x}.$$

Prerequisites

You should be able to:

• use exact trigonometric values and identify signs in each quadrant;
• interpret domains, ranges and inverse functions from composite and inverse functions;
• sketch transformations and recognise asymptotes;
• work deliberately in degrees or radians.

Reciprocal trigonometric functions

The three reciprocal functions are defined by

$$\boxed{\sec x=\frac1{\cos x}}, \qquad \boxed{\csc x=\frac1{\sin x}}, \qquad \boxed{\cot x=\frac1{\tan x}=\frac{\cos x}{\sin x}}.$$

The notation $\operatorname{cosec}x$ is also used for $\csc x$. Calculator buttons for these functions are not needed: evaluate the corresponding sine, cosine or tangent, then take the reciprocal.

Exact values

Reciprocating an exact value gives an exact reciprocal value. Rationalise denominators where appropriate.

Worked example 1: evaluate reciprocal functions exactly

Find $\sec60^\circ$, $\csc45^\circ$ and $\cot(5\pi/6)$.

Since $\cos60^\circ=1/2$,

$$\sec60^\circ=\frac1{1/2}=2.$$

Since $\sin45^\circ=\sqrt2/2$,

$$\csc45^\circ =\frac1{\sqrt2/2} =\frac2{\sqrt2} =\sqrt2.$$

The angle $5\pi/6$ is in quadrant II with reference angle $\pi/6$, so

$$\tan\left(\frac{5\pi}{6}\right)=-\frac1{\sqrt3}.$$

Therefore

$$\cot\left(\frac{5\pi}{6}\right)=-\sqrt3.$$

Hence

$$\boxed{\sec60^\circ=2,\quad \csc45^\circ=\sqrt2,\quad \cot(5\pi/6)=-\sqrt3}.$$

Undefined values

A reciprocal is undefined when its denominator is zero:

$$\begin{array}{c|c|c} \text{function} & \text{undefined when} & \text{radian values} \\ \hline \sec x & \cos x=0 & x=\frac\pi2+n\pi \\ \csc x & \sin x=0 & x=n\pi \\ \cot x & \sin x=0 & x=n\pi \end{array} \qquad n\in\mathbb Z.$$

For example, $\sec(\pi/2)$ is undefined because it would require division by zero. It is not equal to zero or infinity.

Graphs of $\sec x$, $\csc x$ and $\cot x$

The reciprocal graphs follow directly from the familiar trigonometric graphs.

For any non-zero number $u$:

• if $u=1$ or $u=-1$, then $1/u=u$;
• if $0<|u|<1$, then $|1/u|>1$;
• as $u$ approaches $0$, the magnitude of $1/u$ increases without bound;
• $u$ and $1/u$ always have the same sign.

Since $-1\leq\sin x\leq1$ and $-1\leq\cos x\leq1$, reciprocal values can never lie strictly between $-1$ and $1$. Thus

$$\boxed{\text{range of }\sec x=(-\infty,-1]\cup[1,\infty)}$$

and

$$\boxed{\text{range of }\csc x=(-\infty,-1]\cup[1,\infty)}.$$

The zeros of cosine become vertical asymptotes of secant. The zeros of sine become vertical asymptotes of cosecant. Neither $\sec x$ nor $\csc x$ has a zero, because a fraction with numerator $1$ cannot equal zero.

Key graph facts are:

Function	Period	Vertical asymptotes	Range
$\sec x$	$2\pi$	$x=\pi/2+n\pi$	$y\leq-1$ or $y\geq1$
$\csc x$	$2\pi$	$x=n\pi$	$y\leq-1$ or $y\geq1$
$\cot x$	$\pi$	$x=n\pi$	all real numbers

Between consecutive asymptotes, $\cot x$ decreases from positive values to negative values and crosses the $x$ axis where $\cos x=0$, namely at $x=\pi/2+n\pi$.

Worked example 2: features of a transformed reciprocal graph

State the period, vertical asymptotes and range of

$$y=2\sec(3x)-1.$$

Secant has period $2\pi$. Replacing $x$ by $3x$ divides the period by $3$:

$$\boxed{\text{period}=\frac{2\pi}{3}}.$$

Asymptotes occur where $\cos(3x)=0$:

$$3x=\frac\pi2+n\pi,$$

so

$$\boxed{x=\frac\pi6+\frac{n\pi}{3}},\qquad n\in\mathbb Z.$$

For $u=\sec(3x)$, either $u\leq-1$ or $u\geq1$. Multiplying by $2$ and subtracting $1$ gives

$$2u-1\leq-3 \qquad\text{or}\qquad 2u-1\geq1.$$

Therefore

$$\boxed{\text{range}=(-\infty,-3]\cup[1,\infty)}.$$

Self check 1

1. Evaluate $\sec(2\pi/3)$, $\csc(3\pi/2)$ and $\cot(\pi/4)$ exactly.
2. Where is $\csc(2x)$ undefined?
3. State the period and range of $y=3\csc x+2$.
4. Explain why the equation $\sec x=1/2$ has no real solutions.

Answers

1. $\sec(2\pi/3)=-2$, $\csc(3\pi/2)=-1$ and $\cot(\pi/4)=1$.
2. It is undefined when $\sin(2x)=0$, so $2x=n\pi$ and $x=n\pi/2$, where $n\in\mathbb Z$.
3. The period is $2\pi$. Since $\csc x\leq-1$ or $\csc x\geq1$, the range is $y\leq-1$ or $y\geq5$.
4. Every real secant value satisfies $|\sec x|\geq1$, so $1/2$ is outside its range.

Identities involving reciprocal functions

Dividing the identity $\sin^2x+\cos^2x\equiv1$ by $\cos^2x$ gives

$$\tan^2x+1\equiv\sec^2x.$$

Dividing instead by $\sin^2x$ gives

$$1+\cot^2x\equiv\csc^2x.$$

Thus the three Pythagorean identities are

$$\boxed{\sin^2x+\cos^2x\equiv1},$$ $$\boxed{1+\tan^2x\equiv\sec^2x}, \qquad \boxed{1+\cot^2x\equiv\csc^2x}.$$

The notation $\sec^2x$ means $(\sec x)^2$. It does not mean $\sec(\sec x)$.

Worked example 3: find reciprocal values from one ratio

Given that $\tan\theta=-3/4$ and $\theta$ lies in quadrant II, find $\sec\theta$ exactly.

Use

$$1+\tan^2\theta=\sec^2\theta.$$

Then

$$\sec^2\theta =1+\left(-\frac34\right)^2 =1+\frac9{16} =\frac{25}{16}.$$

Therefore

$$\sec\theta=\pm\frac54.$$

In quadrant II, cosine is negative, so its reciprocal secant is also negative:

$$\boxed{\sec\theta=-\frac54}.$$

The identity determines the magnitude but not the sign. The quadrant supplies the missing information.

Inverse trigonometric functions

An inverse trigonometric function reverses a trigonometric function:

$$\sin^{-1}x=\arcsin x, \qquad \cos^{-1}x=\arccos x, \qquad \tan^{-1}x=\arctan x.$$

For example,

$$\arcsin\left(\frac12\right)=30^\circ$$

because $\sin30^\circ=1/2$.

However, sine and cosine repeat every $360^\circ$, and tangent repeats every $180^\circ$. There are infinitely many angles with the same trigonometric value. A function must produce exactly one output, so each original function is restricted before it is inverted.

Principal values, domains and ranges

The standard restrictions are:

Inverse function	Input domain	Principal value range in radians	Principal value range in degrees
$y=\arcsin x$	$-1\leq x\leq1$	$-\pi/2\leq y\leq\pi/2$	$-90^\circ\leq y\leq90^\circ$
$y=\arccos x$	$-1\leq x\leq1$	$0\leq y\leq\pi$	$0^\circ\leq y\leq180^\circ$
$y=\arctan x$	all real $x$	$-\pi/2<y<\pi/2$	$-90^\circ<y<90^\circ$

The endpoints $\pm\pi/2$ are excluded from the range of arctangent because tangent is undefined there.

These ranges explain calculator outputs. Although $\sin150^\circ=1/2$,

$$\arcsin(1/2)=30^\circ,$$

not $150^\circ$, because $150^\circ$ is outside the principal range of arcsine.

The graph of an inverse is the reflection of the restricted original graph in the line $y=x$. Consequently, the original range becomes the inverse domain, and the original restricted domain becomes the inverse range.

Worked example 4: find principal values exactly

Evaluate

$$\arcsin\left(-\frac{\sqrt3}{2}\right), \qquad \arccos\left(-\frac{\sqrt2}{2}\right), \qquad \arctan(-1)$$

in radians.

The reference angles are $\pi/3$, $\pi/4$ and $\pi/4$ respectively.

Arcsine outputs an angle in $[-\pi/2,\pi/2]$. The required sine is negative, so

$$\arcsin\left(-\frac{\sqrt3}{2}\right)=-\frac\pi3.$$

Arccosine outputs an angle in $[0,\pi]$. Cosine is negative in the relevant part of quadrant II, so

$$\arccos\left(-\frac{\sqrt2}{2}\right)=\frac{3\pi}{4}.$$

Arctangent outputs an angle in $(-\pi/2,\pi/2)$. The required tangent is negative, so

$$\arctan(-1)=-\frac\pi4.$$

Therefore

$$\boxed{-\frac\pi3,\quad\frac{3\pi}{4},\quad-\frac\pi4}.$$

Worked example 5: use an inverse function in a model

A straight path rises $7.4$ m over a horizontal distance of $32$ m. Find its angle of elevation to the nearest tenth of a degree.

If the angle is $\theta$, then

$$\tan\theta=\frac{7.4}{32}.$$

Apply arctangent to both sides:

$$\theta=\arctan\left(\frac{7.4}{32}\right).$$

In degree mode,

$$\theta\approx13.018^\circ.$$

Hence

$$\boxed{\theta\approx13.0^\circ}.$$

Do not round $7.4/32$ before applying arctangent. Keeping the full calculator value avoids avoidable error.

Compositions and the order of operations

Within their stated domains,

$$\sin(\arcsin x)=x \qquad (-1\leq x\leq1),$$ $$\cos(\arccos x)=x \qquad (-1\leq x\leq1),$$

and

$$\tan(\arctan x)=x \qquad (x\in\mathbb R).$$

In the opposite order, the answer is the original angle only when that angle lies in the inverse function’s principal range.

Worked example 6: why cancellation can fail

Evaluate $\arcsin(\sin(5\pi/6))$.

First evaluate the inner function:

$$\sin\left(\frac{5\pi}{6}\right)=\frac12.$$

Then apply arcsine:

$$\arcsin\left(\frac12\right)=\frac\pi6.$$

Therefore

$$\boxed{\arcsin(\sin(5\pi/6))=\pi/6},$$

not $5\pi/6$. Arcsine must return its principal value in $[-\pi/2,\pi/2]$.

Worked example 7: exact composition without finding the angle

Find

$$\cos\left(\arcsin\frac35\right)$$

exactly.

Let

$$\theta=\arcsin\frac35.$$

Then $\sin\theta=3/5$ and $-\pi/2\leq\theta\leq\pi/2$. Since $3/5>0$, $\theta$ lies in quadrant I, where cosine is positive.

Using $\sin^2\theta+\cos^2\theta=1$,

$$\cos^2\theta=1-\frac9{25}=\frac{16}{25}.$$

Therefore $\cos\theta=4/5$, and hence

$$\boxed{\cos\left(\arcsin\frac35\right)=\frac45}.$$

Calculator discipline

Before evaluating an inverse function, check the angle mode.

$$\arctan(1)=45^\circ$$

in degree mode, but

$$\arctan(1)=\frac\pi4\approx0.7854$$

in radian mode. These describe the same angle in different units.

Also check that an arcsine or arccosine input lies in $[-1,1]$. A calculator error for $\arccos(1.2)$ reflects a mathematical fact: no real angle has cosine $1.2$.

Common misconceptions

• Confusing reciprocal and inverse notation: $\sin^{-1}x$ conventionally means $\arcsin x$, while $(\sin x)^{-1}$ means $1/\sin x=\csc x$. Use $\arcsin$ when ambiguity is possible.
• Cancelling without checking a range: $\arccos(\cos x)=x$ is guaranteed only for $0\leq x\leq\pi$.
• Treating a principal value as every solution: $\arcsin(1/2)=30^\circ$ is one chosen output. The equation $\sin x=1/2$ has further solutions.
• Ignoring undefined points: a reciprocal graph has an asymptote wherever its denominator is zero.
• Forgetting the sign after taking a square root: identities such as $\sec^2x=1+\tan^2x$ produce two possible signs until a quadrant or interval is used.
• Mixing degrees and radians: write a degree symbol when using degrees. A bare angle is normally interpreted in radians in advanced mathematics.

Self check 2

1. Evaluate $\arcsin(-1/2)$, $\arccos(-1/2)$ and $\arctan(\sqrt3)$ exactly in radians.
2. State the domain and range of $y=\arccos x$.
3. Evaluate $\cos(\arccos(-0.3))$.
4. Evaluate $\arccos(\cos(5\pi/3))$.
5. Find $\sin(\arctan(3/4))$ exactly.
6. Explain why $\arcsin 2$ has no real value.

Answers

1. $-\pi/6$, $2\pi/3$ and $\pi/3$ respectively.
2. Domain $[-1,1]$ and range $[0,\pi]$.
3. $-0.3$, because $\cos(\arccos x)=x$ for every $x\in[-1,1]$.
4. $\cos(5\pi/3)=1/2$, so the principal value is $\arccos(1/2)=\pi/3$.
5. Let $\theta=\arctan(3/4)$. Then $\tan\theta=3/4$ and $\theta$ is in quadrant I. A right triangle has opposite side $3$, adjacent side $4$ and hypotenuse $5$, so $\sin\theta=3/5$.
6. The range of real sine is $[-1,1]$, so no real angle has sine $2$.

Summary

$$\sec x=\frac1{\cos x}, \qquad \csc x=\frac1{\sin x}, \qquad \cot x=\frac1{\tan x}.$$

Reciprocal functions inherit undefined points from zeros in their denominators. Secant and cosecant have range $(-\infty,-1]\cup[1,\infty)$.

Inverse trigonometric functions return principal angles:

$$\arcsin x\in\left[-\frac\pi2,\frac\pi2\right], \quad \arccos x\in[0,\pi], \quad \arctan x\in\left(-\frac\pi2,\frac\pi2\right).$$

These restrictions make the inverses single valued. They must be used when interpreting calculator outputs and simplifying compositions.

Next, develop reciprocal and Pythagorean relationships in trigonometric identities, then use principal values alongside symmetry and periodicity in trigonometric equations.