True or False? There is one correct method for solving a trigonometric equation.

False. Sketching a graph, using the CAST diagram, applying an identity and factorising a quadratic are all legitimate, and most equations can be done more than one way. Marks are for correct solutions, not for a particular route.

In solving a trigonometric equation, when is a sketch of the graph more useful than the CAST diagram?

When the interval covers more than one revolution, or when you want to see the pattern the solutions follow. CAST is quicker for a single revolution, but it only ever gives one revolution's worth, so you still have to extend to the interval.

What are the two things that most often go wrong at the end of a trigonometric equation?

Missing solutions. There is usually more than one in the interval, and stopping at the first is the commonest error. Solutions outside the interval. Check that every value you write down actually lies inside the range you were given.

International A Level (IAL)MathsEdexcelPure 2FlashcardsTrigonometryTrigonometric Equations

Trigonometric Equations (Edexcel International A Level (IAL) Maths: Pure 2): Flashcards

Exam code: YMA01

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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 say that the two sides are identical.
Complete the two identities you have to know:
$\tan θ \equiv______$
$\sin^{2} θ + \cos^{2} θ \equiv______$
The completed identities are:
$\tan θ \equiv \frac{\sin θ}{\cos θ}$
$\sin^{2} θ + \cos^{2} θ \equiv 1$
Note that $\sin^{2} θ$ means ${(\sin θ)}^{2}$ , the whole function squared.
Where does the identity $\tan θ \equiv \frac{\sin θ}{\cos θ}$ come from?
From dividing $\sin θ$ by $\cos θ$ in a right-angled triangle.
That gives $\frac{opposite \div hypotenuse}{adjacent \div hypotenuse}$ , and the hypotenuse cancels to leave $\frac{opposite}{adjacent}$ , which is $\tan θ$ .
True or False?
$\tan θ$ has no value at all when $\cos θ = 0$ .
True.
The quotient form of $\tan θ$ carries $\cos θ$ in its denominator, and dividing by zero has no meaning.
Which theorem is $\sin^{2} θ + \cos^{2} θ \equiv 1$ built from?
It comes from Pythagoras' theorem.
Dividing $a^{2} + b^{2} = c^{2}$ through by $c^{2}$ gives ${(\frac{a}{c})}^{2} + {(\frac{b}{c})}^{2} = 1$ , and in a right-angled triangle those two fractions are $\sin θ$ and $\cos θ$ .
You have to show that one trigonometric expression is identical to another. How do you decide what to substitute?
Compare the two sides and look at what has disappeared.
If $\tan x$ appears on one side and not on the other, it must have been replaced using an identity, so that is the substitution to make.
Define the CAST diagram.
A diagram of the four quadrants showing which trigonometric functions are positive in each.
Going anticlockwise from $0 °$ : All in the first quadrant, Sine in the second, Tangent in the third, Cosine in the fourth.
Once you have the principal value of a trigonometric equation, how does the CAST diagram give you the other solutions?
Draw your angle from the horizontal in its own quadrant, then draw the same angle from the horizontal in all four quadrants.
Read off the angles in the quadrants where your function has the right sign. That gives every solution between $0 °$ and $360 °$ .
How do you find the extra solutions to a trigonometric equation when the interval you are given is wider than one revolution?
Add and subtract $360 °$ from each solution you already have, keeping any that land inside the interval.
For $\cos x = \frac{1}{2}$ with $- 360 ° \leq x \leq 720 °$ , the solutions $60 °$ and $300 °$ give $- 300 °$ , $- 60 °$ , $60 °$ , $300 °$ , $420 °$ and $660 °$ .
Give the radian equivalents of the following common angles:
$180 ° =______$
$270 ° =______$
The four common angles are:
$90 ° = \frac{π}{2}, 180 ° = π$
$270 ° = \frac{3 π}{2}, 360 ° = 2 π$
Check your calculator is in the right mode before you start.
How do you solve an equation like $\cos (θ - 30 °) = 0.5$ ?
Transform the interval first. Put $Z = θ - 30 °$ and subtract $30 °$ from each end, so $0 ° \leq θ \leq 360 °$ becomes $- 30 ° \leq Z \leq 330 °$ .
Solve $\cos Z = 0.5$ in that interval, then add $30 °$ back to every solution.
How is solving $\sin 2 x = \frac{1}{\sqrt{2}}$ different from solving an equation like $\cos (θ - 30 °) = 0.5$ ?
The interval is multiplied rather than shifted: $0 ° \leq x \leq 360 °$ becomes $0 ° \leq 2 x \leq 720 °$ .
Solve for $2 x$ across the wider interval, then divide every solution by 2 instead of subtracting a shift.
True or False?
If you solve $\sin 2 x = \frac{1}{\sqrt{2}}$ by finding the values of $2 x$ , those values are the solutions of the equation.
False.
They are solutions for the transformed variable, not for $x$ or $θ$ .
Every one has to be converted back: add or subtract to undo the shift, or divide by the multiplier.
What makes a trigonometric equation quadratic, and what do you do first?
It contains $\sin^{2} θ$ , $\cos^{2} θ$ or $\tan^{2} x$ .
If a plain trigonometric function appears as well, use an identity so that everything is in terms of one function, then rearrange so the equation equals zero.
Complete the substitution that turns this into a quadratic in $\cos θ$ :
$2 \sin^{2} θ = - 3 \cos θ$
becomes
$2 (1 -______) = - 3 \cos θ$
The substitution gives:
$2 (1 - \cos^{2} θ) = - 3 \cos θ$
which rearranges to $2 \cos^{2} θ - 3 \cos θ - 2 = 0$ .
When solving a quadratic trigonometric equation, how does replacing the trigonometric function with a single letter help?
It turns the equation into an ordinary quadratic you can factorise on sight.
$2 \cos^{2} θ - 3 \cos θ - 2 = 0$ becomes $2 C^{2} - 3 C - 2 = 0$ , which factorises as $(2 C + 1) (C - 2) = 0$ . Put $\cos θ$ back afterwards.
True or False?
A quadratic trigonometric equation always gives two sets of solutions.
False.
It gives two values for the trigonometric function, but one of them may be impossible.
$(2 \cos θ + 1) (\cos θ - 2) = 0$ gives $\cos θ = - \frac{1}{2}$ or $\cos θ = 2$ , and $\cos θ = 2$ has no solutions at all.
Which values of $k$ make $\sin x = k$ and $\cos x = k$ solvable, and how is $\tan x = k$ different?
$\sin x = k$ and $\cos x = k$ only have solutions when $- 1 \leq k \leq 1$ .
$\tan x = k$ has solutions for every value of $k$ .
Why should you factorise $\tan^{2} x = 2 \tan x$ rather than divide through by $\tan x$ ?
Dividing by a trigonometric function loses solutions.
Dividing gives only $\tan x = 2$ . Factorising to $\tan x (\tan x - 2) = 0$ keeps $\tan x = 0$ as well, and both are needed.
What do you do when a trigonometric equation gives a negative principal value but the interval is $0 ° \leq x \leq 360 °$ ?
Do not discard it: use it to find the solutions that are in the interval.
For $\sin x = - \frac{1}{4}$ the calculator gives $- 14.5 °$ , and the solutions in range are $180 ° + 14.5 °$ and $360 ° - 14.5 °$ .
In what order should you deal with an unfamiliar trigonometric equation?
Handle any function of the angle first, transforming the interval. Then use an identity to get everything into one trigonometric function; finally decide whether what remains is linear or quadratic.
Only once the equation reads $\sin x = k$ , $\cos x = k$ or $\tan x = k$ is the principal value any use.
True or False?
There is one correct method for solving a trigonometric equation.
False.
Sketching a graph, using the CAST diagram, applying an identity and factorising a quadratic are all legitimate, and most equations can be done more than one way.
Marks are for correct solutions, not for a particular route.
In solving a trigonometric equation, when is a sketch of the graph more useful than the CAST diagram?
When the interval covers more than one revolution, or when you want to see the pattern the solutions follow.
CAST is quicker for a single revolution, but it only ever gives one revolution's worth, so you still have to extend to the interval.
What are the two things that most often go wrong at the end of a trigonometric equation?
Missing solutions. There is usually more than one in the interval, and stopping at the first is the commonest error.
Solutions outside the interval. Check that every value you write down actually lies inside the range you were given.

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