Define the CAST diagram .

A diagram of the four quadrants of a full turn, used to decide which quadrants the solutions of a trigonometric equation lie in. Each quadrant is labelled with the function or functions that are positive in it.

True or False? The CAST diagram and a sketch of the graph both work by using the symmetry of the trigonometric functions.

True. Each takes one known angle and produces the rest from where the curve repeats or reflects, so neither can find a solution the other misses.

You have a list of angles from a trigonometric equation. What three checks should they pass before you give them as the solutions?

That every branch you solved actually had solutions; that each answer lies inside the interval you were given; and that you have found all of them, not just the principal value. If you changed the interval at the start, change the solutions back before you check any of this.

ASMathsAQAPureFlashcardsTrigonometryTrigonometric Equations

Trigonometric Equations (AQA AS Maths: Pure): Flashcards

Exam code: 7356

1/21

0Still learning

Know0

FrontLinear Trigonometric Equations
Define the CAST diagram.

Was this flashcard helpful?

Cards in this collection (21)

Define the CAST diagram.
A diagram of the four quadrants of a full turn, used to decide which quadrants the solutions of a trigonometric equation lie in.
Each quadrant is labelled with the function or functions that are positive in it.
Between $0 °$ and $90 °$ all three functions are positive. Complete the other three quadrants:
Only $______$ is positive between $90 °$ and $180 °$
Only $______$ is positive between $180 °$ and $270 °$
Only $______$ is positive between $270 °$ and $360 °$
The completed quadrants are:
Only $\sin$ is positive between $90 °$ and $180 °$
Only $\tan$ is positive between $180 °$ and $270 °$
Only $\cos$ is positive between $270 °$ and $360 °$
Read anticlockwise from the first quadrant, the four labels spell A, S, T, C.
Define the principal value when solving a trigonometric equation.
The single angle your calculator gives when you use $\sin^{- 1}$ , $\cos^{- 1}$ or $\tan^{- 1}$ .
It is one solution out of infinitely many, and it need not lie inside the interval you were given.
After finding the principal value of $\cos x = 0.5$ , how does the CAST diagram give you the other solutions between $0 °$ and $360 °$ ?
Draw the principal value $60 °$ as a line from the origin, then draw a line making the same acute angle with the horizontal in each of the other three quadrants.
Keep only the lines in the quadrants where $\cos$ is positive, and read each one as an angle measured anticlockwise from $0 °$ , giving $60 °$ and $300 °$ .
True or False?
$\cos x = 0.5$ and $\cos x = - 0.5$ have their solutions in different quadrants.
True.
$\cos x$ is positive in the first and fourth quadrants, so $\cos x = 0.5$ gives $60 °$ and $300 °$ .
It is negative in the second and third, so $\cos x = - 0.5$ gives $120 °$ and $240 °$ .
You have every solution of $\cos x = 0.5$ between $0 °$ and $360 °$ , but the interval given is $- 360 ° \leq x \leq 720 °$ . How do you find the rest?
Add and subtract $360 °$ from each solution you already have, as often as you like, keeping every result that lands inside the interval.
From $60 °$ and $300 °$ this gives $- 300 °$ , $- 60 °$ , $60 °$ , $300 °$ , $420 °$ and $660 °$ .
To solve $\sin 2 x = \frac{1}{\sqrt{2}}$ for $0 ° \leq x \leq 360 °$ , what has to happen to the interval before you start?
Whatever the equation does to the angle, do to the interval.
The angle here is $2 x$ , so double every part to get $0 ° \leq 2 x \leq 720 °$ , and solve $\sin Z = \frac{1}{\sqrt{2}}$ over that wider interval instead.
Solving $\cos (θ - 30 °) = 0.5$ gives $Z = 60 °$ and $Z = 300 °$ , where $Z = θ - 30 °$ . Complete the solutions for $θ$ :
$θ =______,______$
The completed solutions are:
$θ = 90 °, 330 °$
Each $Z$ is $θ - 30 °$ , so add $30 °$ to undo the substitution, and both results lie inside the original interval $0 ° \leq θ \leq 360 °$ .
Define a quadratic trigonometric equation.
An equation in which a trigonometric function appears squared, such as $2 \sin^{2} θ = - 3 \cos θ$ or $\tan^{2} x - 2 \tan x = 0$ .
Treat it as an ordinary quadratic in that function, and expect two branches to solve rather than one.
An identity has already been used on $11 \sin x = 5 \cos^{2} x + 7$ . Complete the rearrangement into a quadratic in $\sin x$ :
$11 \sin x = 5 - 5 \sin^{2} x + 7$
$______+ 11 \sin x - 12 = 0$
The completed quadratic is:
$5 \sin^{2} x + 11 \sin x - 12 = 0$
Collect everything on the side that leaves the squared term positive, which keeps the factorising straightforward.
Why does replacing $\cos θ$ with a single letter help you factorise $2 \cos^{2} θ - 3 \cos θ - 2 = 0$ ?
It turns the equation into the ordinary quadratic $2 c^{2} - 3 c - 2 = 0$ , which factorises as $(2 c + 1) (c - 2) = 0$ .
Putting $\cos θ$ back gives $(2 \cos θ + 1) (\cos θ - 2) = 0$ , so nothing is lost by the substitution.
Factorising gives $\cos θ = - \frac{1}{2}$ and $\cos θ = 2$ . What do you do with each branch?
$\cos θ = 2$ has no solutions, because $\cos θ$ never rises above $1$ , so that branch is discarded.
Every solution comes from $\cos θ = - \frac{1}{2}$ , which gives $θ = 120 °$ and $θ = 240 °$ in $0 ° \leq θ \leq 360 °$ .
True or False?
To solve $\tan^{2} x - 2 \tan x = 0$ you can divide both sides by $\tan x$ .
False.
Dividing by $\tan x$ throws away every solution of $\tan x = 0$ .
Factorising instead gives $\tan x (\tan x - 2) = 0$ , so in $0 ° \leq x \leq 360 °$ the solutions are $0 °$ , $63.4 °$ , $180 °$ , $243.4 °$ and $360 °$ .
You have factorised into two branches and found one solution from each. Why is that not the finished answer?
Each branch is a trigonometric equation in its own right, so each has its own full set of solutions inside the interval.
$6 \sin^{2} x - \sin x - 1 = 0$ on $0 ° \leq x < 360 °$ gives $\sin x = \frac{1}{2}$ and $\sin x = - \frac{1}{3}$ , and between them four solutions: $30 °$ , $150 °$ , $199.47 °$ and $340.53 °$ .
A quadratic gives $\sin x = - \frac{1}{4}$ , and $\sin^{- 1} (- \frac{1}{4}) = - 14.5 °$ , outside the interval $0 ° \leq x \leq 360 °$ . Is that principal value any use?
Yes: a principal value outside the interval still locates the solutions that are inside it.
Sketching $y = \sin x$ shows the two in range are $180 ° + 14.5 ° = 194.5 °$ and $360 ° - 14.5 ° = 345.5 °$ .
Before you solve a trigonometric equation, what do you check about the angle, and why?
Whether it is simply $x$ or $θ$ , or a function of it such as $2 x$ or $θ - 30 °$ .
A function of the angle means the interval has to be changed before you start, and the solutions changed back at the end.
A trigonometric equation can be attacked in more than one way. Complete the list of four methods:
- sketching the graph
- using trigonometric $______$
- using the $______$ diagram
- $______$ a quadratic equation
The completed list is:
- sketching the graph
- using trigonometric identities
- using the CAST diagram
- factorising a quadratic equation
Which one is quickest depends on the equation, and more than one will often work.
When can you solve a trigonometric equation without using an identity at all?
When rearranging, and factorising if it is needed, already leaves you with simple $\sin$ , $\cos$ or $\tan$ terms.
$3 + 5 \cos 2 x = 1$ is one of these: it rearranges straight to $\cos 2 x = - \frac{2}{5}$ , with no identity anywhere.
An equation contains both $\sin x$ and $\cos x$ . How do you decide which identity to use?
Look at whether the equation is linear or quadratic in the trigonometric functions: the square is the tell.
A linear one usually needs $\tan x \equiv \frac{\sin x}{\cos x}$ , and a quadratic one usually needs $\sin^{2} x + \cos^{2} x \equiv 1$ .
True or False?
The CAST diagram and a sketch of the graph both work by using the symmetry of the trigonometric functions.
True.
Each takes one known angle and produces the rest from where the curve repeats or reflects, so neither can find a solution the other misses.
You have a list of angles from a trigonometric equation. What three checks should they pass before you give them as the solutions?
That every branch you solved actually had solutions; that each answer lies inside the interval you were given; and that you have found all of them, not just the principal value.
If you changed the interval at the start, change the solutions back before you check any of this.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Trigonometric FunctionsNext:Exponential & Logarithms