International A Level (IAL)MathsEdexcelPure 3FlashcardsTrigonometryReciprocal & Inverse Trigonometric Functions

Reciprocal & Inverse Trigonometric Functions (Edexcel International A Level (IAL) Maths: Pure 3): Flashcards

Exam code: YMA01

1/23

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FrontDefinitions of Reciprocal Trigonometric Functions
Complete the three reciprocal trigonometric functions:
$\sec x = \frac{1}{______}$
$cosec x = \frac{1}{______}$
$\cot x = \frac{1}{______}$

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Cards in this collection (23)

Complete the three reciprocal trigonometric functions:
$\sec x = \frac{1}{______}$
$cosec x = \frac{1}{______}$
$\cot x = \frac{1}{______}$
The completed definitions are:
$\sec x = \frac{1}{\cos x}$
$cosec x = \frac{1}{\sin x}$
$\cot x = \frac{1}{\tan x}$
Cosecant is sometimes written $\csc x$ instead.
True or False?
$\sec x$ is the reciprocal of $\sin x$ .
False.
$\sec x$ is the reciprocal of $\cos x$ , and it is $cosec x$ that goes with $\sin x$ .
The pairing is the opposite way round from what the first three letters of each name suggest.
How can $\cot x$ be written in terms of $\sin x$ and $\cos x$ ?
$\cot x = \frac{\cos x}{\sin x}$ .
It is $\tan x$ turned upside down, so the sine and the cosine simply swap places.
Where does $\cot x$ take the value zero?
Wherever $\tan x$ is undefined.
A reciprocal is zero only where the function it came from grows without limit, so cotangent's zeros sit exactly where tangent has its asymptotes.
How do you solve an equation containing $\sec$ , $cosec$ or $\cot$ ?
Convert each one into the ordinary trigonometric function it is the reciprocal of, then solve in the usual way.
So $\sec x = 2$ becomes $\cos x = \frac{1}{2}$ , which is an equation you already know how to handle.
Why is it often useful to rewrite an expression entirely in $\sin$ and $\cos$ ?
Because every one of the six trigonometric functions can be written that way, so a mixture of them stops being a problem.
Once everything is in $\sin$ and $\cos$ , the parts can be combined and cancelled like ordinary algebraic fractions.
Where does a reciprocal trigonometric graph have its vertical asymptotes?
Wherever the original function is zero, since you cannot divide by zero.
So $\sec x$ has them where $\cos x = 0$ , and $cosec x$ where $\sin x = 0$ .
The range of both $\sec x$ and $cosec x$ is $y \leq______$ or $y \geq______$ .
The range of both is $y \leq - 1$ or $y \geq 1$ .
Since $\sin$ and $\cos$ never exceed $1$ in size, their reciprocals can never be smaller than $1$ in size.
What are the periods of $\sec x$ , $cosec x$ and $\cot x$ ?
$\sec x$ and $cosec x$ both repeat every $360 °$ , or $2 π$ radians.
$\cot x$ repeats every $180 °$ , or $π$ radians, just as $\tan x$ does.
True or False?
$\cot x$ , like $\sec x$ , can never take a value between $- 1$ and $1$ .
False.
$\cot x$ takes every real value, because $\tan x$ does too.
It is $\sec$ and $cosec$ that are restricted, not all three.
How do you sketch a reciprocal trigonometric graph?
Sketch the original function first, then take the reciprocal of every value on it.
Where the original is large the reciprocal is close to zero, and where the original reaches $\pm 1$ the two graphs touch.
Which reciprocal trigonometric graph is symmetrical about the $y$ -axis?
$\sec x$ , because $\cos x$ is.
Taking reciprocals does not disturb a symmetry the original graph already has.
The two reciprocal identities are:
$\tan^{2} x + 1 \equiv______$ and $1 + \cot^{2} x \equiv______$
$\tan^{2} x + 1 \equiv \sec^{2} x$ and $1 + \cot^{2} x \equiv {cosec}^{2} x$
Both follow from $\sin^{2} x + \cos^{2} x \equiv 1$ , so neither has to be memorised separately.
How do you derive $\tan^{2} x + 1 \equiv \sec^{2} x$ ?
Divide every term of $\sin^{2} x + \cos^{2} x \equiv 1$ by $\cos^{2} x$ .
That works because $\frac{\sin x}{\cos x} = \tan x$ and $\frac{1}{\cos x} = \sec x$ .
What do you divide $\sin^{2} x + \cos^{2} x \equiv 1$ by to reach the $cosec$ identity?
By $\sin^{2} x$ .
That turns the first term into $1$ , the second into $\cot^{2} x$ and the right-hand side into ${cosec}^{2} x$ .
True or False?
$\sec^{2} x - \tan^{2} x = 1$ wherever both are defined.
True.
It is $\tan^{2} x + 1 \equiv \sec^{2} x$ with the $\tan^{2} x$ moved across.
Spotting the rearranged forms inside a longer expression is what the identity is actually for.
When are the reciprocal trigonometric identities needed?
When an expression mixes $\sec$ , $cosec$ or $\cot$ with $\tan$ , or with each other.
Substituting one of them removes a squared reciprocal term, which often collapses the whole expression.
Why must the domain of $\sin x$ be restricted before $\arcsin x$ can exist?
Because $\sin x$ is many-to-one over all real $x$ , and only a one-to-one function has an inverse.
Restricting it to $- \frac{π}{2} \leq x \leq \frac{π}{2}$ makes it one-to-one while still producing every output from $- 1$ to $1$ .
The domains are restricted to $- \frac{π}{2} \leq x \leq \frac{π}{2}$ for $\sin$ , $______$ for $\cos$ , and $- \frac{π}{2} < x < \frac{π}{2}$ for $\tan$ .
The domains are restricted to $- \frac{π}{2} \leq x \leq \frac{π}{2}$ for $\sin$ , $0 \leq x \leq π$ for $\cos$ , and $- \frac{π}{2} < x < \frac{π}{2}$ for $\tan$ .
Cosine gets a different interval because it is one-to-one from $0$ to $π$ rather than symmetrically about zero.
What are the ranges of $\arcsin x$ , $\arccos x$ and $\arctan x$ ?
$- \frac{π}{2} \leq \arcsin x \leq \frac{π}{2}$ and $- \frac{π}{2} < \arctan x < \frac{π}{2}$ .
$0 \leq \arccos x \leq π$ , each range matching the restricted domain it came from.
What is the domain of $\arcsin x$ , and why is $\arctan x$ different?
$\arcsin x$ is defined only for $- 1 \leq x \leq 1$ , because those are the only values sine ever produces.
$\arctan x$ is defined for all real $x$ , because tangent produces every real value.
True or False?
$\sin^{- 1} x$ means $\frac{1}{\sin x}$ .
False.
It means $\arcsin x$ , the inverse function, and the $- 1$ is not a power.
$\frac{1}{\sin x}$ is $cosec x$ , which is a completely different thing.
What happens to the graph of $\arctan x$ for large values of $x$ ?
It flattens out towards the horizontal asymptotes $y = \frac{π}{2}$ and $y = - \frac{π}{2}$ .
It never reaches them, because $\tan x$ never actually attains those angles.

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