Do you check that a solution exists before or after looking for more solutions?

Check that it exists first . A root that no angle can produce contributes nothing at all to the solution set, so there is nothing there to go looking for.

True or False? Each trigonometric equation has one correct method that has to be used to solve it.

False. A sketch, the CAST diagram and a substituted identity can all reach the same answers. The shape of the equation is what makes one route shorter than another, not any rule about which one is allowed.

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Trigonometric Equations (Cambridge (CIE) A Level Maths: Pure 1): Flashcards

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FrontSimple Trigonometric Identities
Define trigonometric identity.

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Cards in this collection (24)

Define trigonometric identity.
A trigonometric identity is a statement that is true for every value of the angle, not just for particular ones.
The sign $\equiv$ is used in place of $=$ to show this, and is read as "is identical to".
Complete the two identities you need to know:
$\tan θ \equiv \frac{______}{______} and \sin^{2} θ +______\equiv 1$
The completed identities are:
$\tan θ \equiv \frac{\sin θ}{\cos θ} and \sin^{2} θ + \cos^{2} θ \equiv 1$
Note that $\sin^{2} θ$ means ${(\sin θ)}^{2}$ , the sine of the angle squared, and not the sine of $θ^{2}$ .
Where does $\tan θ \equiv \frac{\sin θ}{\cos θ}$ come from?
From the right-angled triangle definitions.
Dividing $\frac{opposite}{hypotenuse}$ by $\frac{adjacent}{hypotenuse}$ cancels the hypotenuse and leaves $\frac{opposite}{adjacent}$ , which is the tangent.
Where does $\sin^{2} θ + \cos^{2} θ \equiv 1$ come from?
From Pythagoras' theorem.
In a right-angled triangle $a^{2} + b^{2} = c^{2}$ , and dividing every term by $c^{2}$ turns $\frac{a}{c}$ into $\sin θ$ and $\frac{b}{c}$ into $\cos θ$ .
True or False?
$\sin^{2} θ + \cos^{2} θ \equiv 1$ holds for obtuse and negative angles as well as acute ones.
True.
An identity is true for every value of the angle, which is exactly what the $\equiv$ sign is claiming.
The right-angled triangle is only where the result is most easily seen, not a limit on where it applies.
How does dividing an equation through by $\cos x$ help?
It turns $\frac{\sin x}{\cos x}$ into $\tan x$ , leaving a single trigonometric function to solve for.
For example $5 \cos x = 2 \sin x$ becomes $5 = 2 \tan x$ , so $\tan x = \frac{5}{2}$ .
You have an equation in $\sin x$ and $\cos^{2} x$ . Which way round do you use the Pythagorean identity?
Replace the squared term, writing $\cos^{2} x$ as $1 - \sin^{2} x$ .
Substituting for the squared function is what leaves the whole equation in $\sin x$ ; going the other way would reintroduce the function you were trying to remove.
What does the CAST diagram tell you?
Which of the three functions are positive in each $90 °$ quadrant, going anticlockwise from $0 °$ .
All three are positive from $0 °$ to $90 °$ , then only Sine, then only Tangent, then only Cosine.
How do you use the CAST diagram to find every solution between $0 °$ and $360 °$ ?
Draw the principal value into the diagram measured from $0 °$ , then draw a line making that same angle in all four quadrants.
Read off the angles in the quadrants where the function takes the sign you need, so $\cos x = \frac{1}{2}$ gives $60 °$ and $300 °$ .
You have the solutions between $0 °$ and $360 °$ . How do you find any others the interval asks for?
Add or subtract $360 °$ to each of them, as many times as the interval allows.
For $\cos x = \frac{1}{2}$ over $- 360 ° \leq x \leq 720 °$ that turns two solutions into six: $- 300 °$ , $- 60 °$ , $60 °$ , $300 °$ , $420 °$ and $660 °$ .
True or False?
To find all the solutions of $\tan x = 2$ , you add $360 °$ to the principal value each time.
False.
Tangent repeats every $180 °$ , so it is $180 °$ that you add each time.
From $\tan^{- 1} 2 = 63.4 °$ the next solution is $243.4 °$ , not $423.4 °$ .
How do you solve an equation such as $\cos (θ - 30 °) = 0.5$ ?
Substitute $Z = θ - 30 °$ and transform the interval in the same way, so $0 ° \leq θ \leq 360 °$ becomes $- 30 ° \leq Z \leq 330 °$ .
Solve for $Z$ inside that new interval, then convert each answer back, here by adding $30 °$ to give $θ = 90 °$ and $330 °$ .
To solve $\sin 2 x = \frac{1}{\sqrt{2}}$ for $0 ° \leq x \leq 360 °$ , the interval for $2 x$ becomes:
$______\leq 2 x \leq______$
The interval becomes:
$0 ° \leq 2 x \leq 720 °$
Both ends are multiplied by $2$ , so the equation has to be solved over an interval twice as wide before the answers are halved back.
What makes a trigonometric equation quadratic?
It contains the square of a trigonometric function, such as $\sin^{2} θ$ or $\cos^{2} θ$ .
$2 \cos^{2} θ - 3 \cos θ - 2 = 0$ is a quadratic in $\cos θ$ in exactly the way $2 y^{2} - 3 y - 2 = 0$ is a quadratic in $y$ .
Why is it worth replacing $\cos θ$ with a single letter?
Because the equation then looks like an ordinary quadratic, which makes the factorisation far easier to spot.
Swap the letter back afterwards, so that each bracket gives you an equation in $\cos θ$ rather than in the letter.
Replacing $\cos θ$ with $C$ turns $2 \cos^{2} θ - 3 \cos θ - 2 = 0$ into $2 C^{2} - 3 C - 2 = 0$ , which factorises to:
$(2 C +______) (C -______) = 0$
The factorised form is:
$(2 C + 1) (C - 2) = 0$
Rewritten with the function back in place, that is $(2 \cos θ + 1) (\cos θ - 2) = 0$ .
Factorising leaves $\cos θ = - \frac{1}{2}$ or $\cos θ = 2$ . What do you do next?
Discard $\cos θ = 2$ , because no angle has a cosine of $2$ , and solve only $\cos θ = - \frac{1}{2}$ .
A quadratic always offers two roots, so checking that each one is actually attainable is part of solving the equation rather than an afterthought.
True or False?
A solution to $\tan x = k$ exists for every value of $k$ .
True.
Tangent is unbounded, so no value of $k$ is out of reach and a root of that form is never discarded.
It is only $\sin x = k$ and $\cos x = k$ that need $- 1 \leq k \leq 1$ .
Why can a quadratic trigonometric equation have four solutions in one interval?
Because each of the two roots of the quadratic is itself an equation with several solutions in that interval.
$\sin x = - \frac{1}{3}$ and $\sin x = \frac{1}{2}$ each contribute two angles between $0 °$ and $360 °$ , giving four in all.
Facing a trigonometric equation, what is the first thing to check?
Whether it involves a function of the angle, such as $2 x$ or $θ - 30 °$ , rather than the angle on its own.
That has to be settled before anything else, because everything which follows depends on which angle you are actually solving for.
What are you trying to reduce every trigonometric equation to?
A simple equation of the form $\sin x = k$ , $\cos x = k$ or $\tan x = k$ , in one trigonometric function.
Rearranging, factorising and substituting an identity are all just different ways of getting there.
How does the degree of an equation help you choose an identity?
A linear equation mixing sine and cosine usually wants $\tan x \equiv \frac{\sin x}{\cos x}$ .
A quadratic one, carrying a squared function, usually wants $\sin^{2} x + \cos^{2} x \equiv 1$ .
Do you check that a solution exists before or after looking for more solutions?
Check that it exists first.
A root that no angle can produce contributes nothing at all to the solution set, so there is nothing there to go looking for.
True or False?
Each trigonometric equation has one correct method that has to be used to solve it.
False.
A sketch, the CAST diagram and a substituted identity can all reach the same answers.
The shape of the equation is what makes one route shorter than another, not any rule about which one is allowed.

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