A LevelMathsCambridge (CIE)Pure 3FlashcardsTrigonometryReciprocal Trigonometric Functions

Reciprocal Trigonometric Functions (Cambridge (CIE) A Level Maths: Pure 3): Flashcards

Exam code: 9709

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

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Cards in this collection (18)

Complete the three reciprocal trigonometric functions:
$\sec x =______$
$cosec x =______$
$\cot x =______$
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 of $cosec x$ .
Which regular function does $cosec$ pair with?
$cosec$ is built from sine, and $\sec$ from cosine, which is the opposite of what the names suggest.
A reliable check is the third letter of each name: the s in cosec points to sine, and the c in sec points to cosine.
How can $\cot x$ be written in terms of sine and cosine?
It can be written as $\cot x = \frac{\cos x}{\sin x}$ .
That form is often more useful than the reciprocal one, because it lets you combine $\cot$ with other sine and cosine terms over a common denominator.
True or False?
$\sec x$ is another name for the inverse function $\cos^{- 1} x$ .
False.
$\sec x$ is the reciprocal of the cosine, whereas $\cos^{- 1} x$ is the angle whose cosine is $x$ .
The two are quite different: $\sec 60 ° = 2$ , while $\cos^{- 1} 0.5 = 60 °$ .
How do you start solving an equation containing $\sec$ , $cosec$ or $\cot$ ?
Rewrite every reciprocal function in terms of $\sin$ , $\cos$ or $\tan$ , then solve it as an ordinary trigonometric equation.
For an expression such as $\frac{1 + \cot x}{1 + \tan x}$ , putting everything over sine and cosine lets the whole thing cancel down.
True or False?
The equation $\cot x = 10$ becomes $\tan x = 10$ .
False.
Taking the reciprocal of both sides gives $\tan x = \frac{1}{10}$ , not $10$ .
Swapping a reciprocal function for its regular one always flips the number on the other side as well.
True or False?
$\cot x$ can be equal to zero.
True.
At $x = 90 °$ the cosine is $0$ and the sine is $1$ , so $\cot x = 0$ .
Those are exactly the values where $\tan x$ is undefined, which is why the reciprocal form is misleading here.
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.

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