A LevelMathsEdexcelPureFlashcardsTrigonometryReciprocal & Inverse Trigonometric Functions

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

Exam code: 9MA0

1/22

0Still learning

Know0

FrontDefinitions of Reciprocal Trigonometric Functions
$\sec x = \frac{1}{______}$ , $cosec x = \frac{1}{______}$ , $\cot x = \frac{1}{______}$

Was this flashcard helpful?

Cards in this collection (22)

$\sec x = \frac{1}{______}$ , $cosec x = \frac{1}{______}$ , $\cot x = \frac{1}{______}$
$\sec x = \frac{1}{\cos x}$ , $cosec x = \frac{1}{\sin x}$ , $\cot x = \frac{1}{\tan x}$
Note that the "co" in cosec goes with sine, not with cosine, which is the pairing most often reversed.
How can $\cot x$ be written in terms of $\sin$ and $\cos$ ?
$\cot x = \frac{\cos x}{\sin x}$ .
It is the reciprocal of $\tan x = \frac{\sin x}{\cos x}$ , so the fraction simply turns over.
How do you solve an equation containing $\sec$ , $cosec$ or $\cot$ ?
Convert them into $\sin$ , $\cos$ or $\tan$ first, then solve in the usual way.
The reciprocal functions have no solving techniques of their own.
True or False?
$cosec x$ is sometimes written as $\csc x$ .
True.
$\csc$ is simply an alternative abbreviation for cosecant and means exactly the same thing.
It turns up in textbooks and on calculators, so it is worth recognising.
Why is $\sec x$ not the same thing as $\cos^{- 1} x$ ?
$\sec x$ is the reciprocal of $\cos x$ , that is $\frac{1}{\cos x}$ .
$\cos^{- 1} x$ is the inverse function, which returns an angle, and the $- 1$ there is not a power at all.
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.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Radian MeasureNext:Compound & Double Angle Formulae