Mt9, Mt13–Mt15 - Reciprocal and inverse trigonometric functions | OCR A-Level Mathematics B (MEI)

Use the inverse functions in Mt9 and the reciprocal definitions in Mt13 without confusing their notation. Mt14 connects their graphs, domains and ranges to sine, cosine and tangent; Mt15 supplies two squared identities for exact values and algebra.

Reciprocal functions and their domains

Can cotangent exist where tangent does not? At $90^\circ$ , sine is $1$ and cosine is $0$ . Tangent divides by zero, but cosine divided by sine is $0$ . That is the value of cotangent. A careful definition reveals a valid value that a careless reciprocal button sequence would miss.

You already know

A quotient is defined only when its denominator is nonzero. Sine and cosine are signed coordinates on the unit circle, and $\tan\theta=\sin\theta/\cos\theta$ on its domain.

The three reciprocal functions are defined using those coordinates.

Secant, cosecant and cotangent

\sec\theta=\frac1{\cos\theta}\qquad(\cos\theta\ne0)

\operatorname{cosec}\theta=\frac1{\sin\theta}\qquad \cot\theta=\frac{\cos\theta}{\sin\theta} \qquad(\sin\theta\ne0)

For integer $k$ , secant excludes $\pi/2+k\pi$ , while cosecant and cotangent exclude $k\pi$ . In degrees the exclusions are $90^\circ+180^\circ k$ and $180^\circ k$ . These definitions are not in the formula sheet: learn them.

On the common domain where tangent is nonzero and defined, cotangent also equals $1/\tan\theta$ . That shorter expression has a smaller domain than the cosine-over-sine definition: at $\pi/2$ , cotangent equals zero. Do not throw that point away.

Worked example 1: exact reciprocals in quadrant II

Question. Find $\sec(3\pi/4)$ , $\operatorname{cosec}(3\pi/4)$ and $\cot(3\pi/4)$ exactly.

Plan. Use the signed sine and cosine values before taking reciprocals or their ratio.

\begin{aligned} \sin(3\pi/4)&=\frac{\sqrt2}{2},\quad\cos(3\pi/4)=-\frac{\sqrt2}{2} && \text{unit-circle values}\\ \sec(3\pi/4)&=-\frac2{\sqrt2}=-\sqrt2 && \text{invert cosine}\\ \operatorname{cosec}(3\pi/4)&=\sqrt2 && \text{invert sine}\\ \cot(3\pi/4)&=-1 && \text{cosine over sine} \end{aligned}

Why this step? The reciprocal retains the sign of the value inverted; the reference angle alone does not determine that sign.

Answer. The three values are $-\sqrt2$ , $\sqrt2$ and $-1$ .

Check. The reciprocal products with cosine and sine equal $1$ ; cotangent times sine equals cosine.

Keep the reference angle but move to quadrant III. This changes sine's sign while cosine stays negative.

Worked example 2: changing the quadrant

Question. Find the three values for $\theta=5\pi/4$ .

Plan. Apply the definitions to two negative coordinates of equal magnitude.

\begin{aligned} \sin\theta&=-\frac{\sqrt2}{2},\quad\cos\theta=-\frac{\sqrt2}{2} && \text{quadrant III}\\ \sec\theta&=-\sqrt2 && \text{invert cosine}\\ \operatorname{cosec}\theta&=-\sqrt2 && \text{invert sine}\\ \cot\theta&=1 && \text{equal negative coordinates} \end{aligned}

Why this step? A quotient of two negative values is positive, even though both reciprocal values are negative.

Answer. $-\sqrt2$ , $-\sqrt2$ and $1$ , respectively.

Check. Neither coordinate is zero, and each result meets its defining equation.

Another notation needs separating: $\sin^2\theta$ means $(\sin\theta)^2$ , whereas $\sin^{-1}x$ conventionally means an inverse function, not $1/\sin x$ .

Watch out

Wrong: use $1/\tan(\pi/2)$ to conclude that cotangent is undefined. Right: evaluate $\cos(\pi/2)/\sin(\pi/2)=0$ . The defining denominator is sine.

Move the same reference angle into quadrant IV.

Reading reciprocal graphs

A reciprocal graph keeps the sign of the denominator, becomes large near its zeros and retains its values $1$ and $-1$ .

Secant and cosecant follow reciprocal values, while cotangent decreases between its excluded arguments

For secant and cosecant the range is $( -\infty,-1]\cup[1,\infty)$ , because reciprocating a nonzero value with magnitude at most $1$ gives magnitude at least $1$ . Both have period $2\pi$ and no zero. Secant is even; cosecant is odd. Their excluded inputs and vertical asymptotes are $\pi/2+k\pi$ and $k\pi$ , respectively.

Cosecant has a minimum $1$ at $\pi/2$ on $(0,\pi)$ , and a maximum $-1$ at $3\pi/2$ on $(\pi,2\pi)$ . Secant has $\sec0=1$ and $\sec\pi=-1$ , with branches separated by the zeros of cosine.

Cotangent has domain $x\ne k\pi$ , range $\mathbb R$ , period $\pi$ and odd symmetry. On $(0,\pi)$ it decreases from arbitrarily large positive to arbitrarily large negative values and crosses zero at $\pi/2$ . Its zeros are $\pi/2+k\pi$ . Where both are defined, tangent and cotangent are reciprocals, but their excluded inputs differ. These graph properties are not in the formula sheet: rebuild them from the definitions.

Worked example 3: using the denominator's graph

Question. Describe the asymptotes, period and branch vertices of $y=\operatorname{cosec}(2x)$ on $0\le x\le\pi$ .

Plan. Map zeros and extrema of $\sin(2x)$ to reciprocal features.

\begin{aligned} \sin(2x)=0&\iff x=k\pi/2 && \text{vertical asymptotes}\\ 2T=2\pi&\implies T=\pi && \text{period}\\ \sin(2x)=1,-1&\text{ at }x=\pi/4,3\pi/4 && \text{fixed reciprocal values} \end{aligned}

Why this step? Zeros of the denominator locate missing inputs; the reciprocal function itself never crosses zero.

Answer. Asymptotes occur at $0,\pi/2,\pi$ . The branches have vertices $(\pi/4,1)$ and $(3\pi/4,-1)$ , with period $\pi$ .

Check. The signs are positive on $(0,\pi/2)$ and negative on $(\pi/2,\pi)$ , just as for $\sin(2x)$ .

Change only the reciprocal family, from cosecant to secant. The argument and interval stay fixed, but the denominator's zeros move.

Worked example 4: changing sine to cosine

Question. Give the corresponding features of $y=\sec(2x)$ on $0\le x\le\pi$ .

Plan. Use zeros and extrema of $\cos(2x)$ .

\begin{aligned} 2x=\pi/2+k\pi&\implies x=\pi/4+k\pi/2 && \text{excluded inputs}\\ 2T=2\pi&\implies T=\pi && \text{same period}\\ \cos(2x)=1,-1,1&\text{ at }x=0,\pi/2,\pi && \text{reciprocal values} \end{aligned}

Why this step? The period stays the same, but using the sine's zeros would place every asymptote incorrectly.

Answer. The asymptotes are $x=\pi/4,3\pi/4$ . There are positive branch portions at the included endpoints $(0,1)$ and $(\pi,1)$ , and a negative branch with vertex $(\pi/2,-1)$ . The period is $\pi$ .

Check. At the centre, $\cos\pi=-1$ , so its reciprocal is $-1$ .

When sketching, draw separate branches beside dashed asymptotes. A finite line joining across an asymptote would assert values at excluded inputs.

Watch out

Wrong: reciprocal sine and reciprocal cosine still have range $[-1,1]$ . Right: their values have magnitude at least $1$ ; the gap between $-1$ and $1$ contains no curve.

Apply the same denominator reasoning to cotangent, whose period differs.

Identities with reciprocal functions

Divide the unit-circle identity by a nonzero square to turn it into a reciprocal identity, retaining the division condition.

For example, dividing $\sin^2\theta+\cos^2\theta=1$ by $\sin^2\theta$ gives $1+\cos^2\theta/\sin^2\theta=1/\sin^2\theta$ . Division by cosine squared gives the other form.

Squared reciprocal identities

\tan^2\theta+1=\sec^2\theta\qquad(\cos\theta\ne0)

\cot^2\theta+1=\operatorname{cosec}^2\theta\qquad(\sin\theta\ne0)

Not in the formula sheet: learn it.

The squares determine magnitudes. A quadrant or interval is still needed to determine an unsquared value. They also show why $\sec\theta=1/2$ is impossible for real $\theta$ : its square would be below $1$ .

Worked example 5: recovering a positive cosecant

Question. Given $\cot\theta=-3$ and $\pi/2<\theta<\pi$ , find $\operatorname{cosec}\theta$ exactly.

Plan. Use the cotangent identity and choose the sign from sine's quadrant.

\begin{aligned} \operatorname{cosec}^2\theta&=1+(-3)^2=10 && \text{squared identity}\\ \operatorname{cosec}\theta&=\sqrt{10} && \text{sine positive in II} \end{aligned}

Why this step? Negative cotangent does not mean negative cosecant; cotangent is the ratio of two signed values.

Answer. $\operatorname{cosec}\theta=\sqrt{10}$ .

Check. Take $\sin\theta=1/\sqrt{10}$ , $\cos\theta=-3/\sqrt{10}$ . Their squares sum to $1$ and their ratio is $-3$ .

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Mt9, Mt13–Mt15 - Reciprocal and inverse trigonometric functions

Reciprocal functions and their domains

Reading reciprocal graphs

Identities with reciprocal functions

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