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Trigonometric Equations & Identities (SQA National 5 Maths): Flashcards

Exam code: X847 75

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FrontPeriod & Related Angles
What does knowing the period of $\sin x$ let you do?

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What does knowing the period of $\sin x$ let you do?
Adding or subtracting $360 °$ to any angle gives another angle with exactly the same sine.
So since $\sin 30 ° = 0.5$ , it follows that $\sin 390 ° = 0.5$ as well.
How far apart are two angles with the same tangent?
Two such angles differ by a multiple of $180 °$ , since that is the period of $\tan x$ .
So $\tan 45 ° = 1$ means that $\tan 225 ° = 1$ as well.
Define related angles.
Related angles are the other values of $x$ that give the same value of $\sin x$ or $\cos x$ .
They exist because both graphs are symmetric, so each height is reached more than once in a full turn.
Complete the two related-angle rules for the interval from $0 °$ to $360 °$ .
$for \sin x the related angle is______- x$
$for \cos x the related angle is______- x$
The completed rules are:
$for \sin x the related angle is 180 - x$
$for \cos x the related angle is 360 - x$
The two differ because the sine graph is symmetric about $90 °$ and the cosine graph about $180 °$ .
True or False?
Two different angles between $0 °$ and $360 °$ can have exactly the same sine.
True.
The sine graph rises and falls through every height between $- 1$ and $1$ more than once in a full turn.
Both $30 °$ and $150 °$ have a sine of $0.5$ .
Given that $\cos 60 ° = 0.5$ what other angle has the same cosine?
Use $360 - x$ , which gives $360 - 60 = 300 °$ .
So $\cos 300 °$ is also $0.5$ , and adding $360 °$ to either would give further angles.
How can symmetry give an angle whose cosine is the negative of a known one?
An angle as far past $90 °$ as another is before it has the opposite cosine.
Since $60 °$ is $30 °$ before $90 °$ , $120 °$ is $30 °$ after it, and $\cos 120 ° = - 0.5$ .
What must a trigonometric equation be rearranged into first?
Rearrange it into the form $\sin x = \dots$ , $\cos x = \dots$ or $\tan x = \dots$ before doing anything else.
So $2 \sin x - 1 = 0$ becomes $\sin x = 0.5$ .
True or False?
The inverse sine key on a calculator gives all the solutions of $\sin x = 0.5$ .
False.
The calculator gives only the first solution, here $30 °$ , and the rest have to be found from related angles and the period.
A trigonometric equation usually has several solutions inside the interval a question specifies.
How do you find every solution of a trigonometric equation in an interval?
Take the inverse function for the first solution, then use related angles and the period to generate the others.
Keep only those that lie inside the interval the question gives, such as $0 ° \leq x \leq 360 °$ .
What do you do when the calculator gives a negative first solution?
Add the period to it to get a positive solution inside the interval.
Since $\sin^{- 1} (- 0.5) = - 30 °$ , adding $360 °$ gives $330 °$ , and related angles then give any others.
Complete the rearrangement of this equation into the standard form.
$16 \sin x ° + 7 = 11 \Rightarrow \sin x ° =______$
The completed rearrangement is:
$16 \sin x ° + 7 = 11 \Rightarrow \sin x ° = 0.25$
Subtracting $7$ leaves $16 \sin x ° = 4$ , and dividing by $16$ gives $0.25$ .
Solve $\sin x ° = 0.25$ for $x$ between $0$ and $360$ to one decimal place.
The inverse sine gives $x = 14.477 \dots$ , and the related angle is $180 - 14.477 \dots$ .
Rounding both to one decimal place gives $x = 14.5$ or $x = 165.5$ .
How is the second solution of $\tan x = 1$ found?
Add $180 °$ to the first, because the tangent graph has no symmetry rule of its own and simply repeats.
So $\tan^{- 1} (1) = 45 °$ gives a second solution of $45 + 180 = 225 °$ .
Define an identity.
An identity is a statement that is true for every value of the variable, not merely for particular ones.
An equation like $3 x + 4 = 13$ holds only for one value of $x$ , whereas $\cos^{2} x ° + \sin^{2} x ° = 1$ holds for all of them.
Complete the two trigonometric identities you need to know.
$\cos^{2} x ° + \sin^{2} x ° =______$
$\tan x ° =______$
The completed identities are:
$\cos^{2} x ° + \sin^{2} x ° = 1$
$\tan x ° = \frac{\sin x °}{\cos x °}$
These two are the only ones required, and neither is given on the Formulae List.
What two rearrangements of $\cos^{2} x ° + \sin^{2} x ° = 1$ are useful?
The two forms are $\cos^{2} x ° = 1 - \sin^{2} x °$ and $\sin^{2} x ° = 1 - \cos^{2} x °$ .
Each lets you swap an expression in one squared function for one in the other, which is usually what a question wants.
True or False?
$3 \sin^{2} x ° + 3 \cos^{2} x °$ simplifies to a number with no $x$ in it.
True.
Factorising gives $3 (\sin^{2} x ° + \cos^{2} x °)$ , and the bracket is always equal to $1$ .
The whole expression is therefore $3$ , whatever $x$ happens to be.
How do you simplify an expression that mixes all three trigonometric functions?
Replace $\tan x °$ with $\frac{\sin x °}{\cos x °}$ , which leaves only sines and cosines to cancel.
So $\sin x ° \cos x ° \tan x °$ becomes $\sin^{2} x °$ once the cosines cancel.
How is $\frac{\sin x ° \cos x °}{\tan x °}$ written in its simplest form?
Replace the $\tan x °$ and turn the division into a multiplication by the reciprocal.
That gives $\sin x ° \cos x ° \times \frac{\cos x °}{\sin x °}$ , and cancelling the sines leaves $\cos^{2} x °$ .
How is $5 - 2 \sin^{2} x °$ rewritten so that it uses $\cos^{2} x °$ instead?
Substitute $\sin^{2} x ° = 1 - \cos^{2} x °$ , giving $5 - 2 (1 - \cos^{2} x °)$ .
Expanding carefully leaves $5 - 2 + 2 \cos^{2} x °$ , which is $3 + 2 \cos^{2} x °$ .

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