How do you prove a trigonometric identity?

Start with one side only , usually the more complicated one, and use the identities to transform it step by step until it becomes the other side. Do not move terms from one side to the other, because that assumes the result you are trying to prove.

GCSEFurther MathsAQAFlashcards6. GeometryTrigonometric Graphs & Equations

Trigonometric Graphs & Equations (AQA GCSE Further Maths): Flashcards

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What are the key points of $y = \sin x$ from $0 °$ to $360 °$ ?

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What are the key points of $y = \sin x$ from $0 °$ to $360 °$ ?
It starts at $(0, 0)$ , rises to a maximum at $(90 °, 1)$ , returns to zero at $(180 °, 0)$ , falls to a minimum at $(270 °, - 1)$ and comes back to $(360 °, 0)$ .
Sketching those five points and joining them with a smooth curve is enough.
What are the key points of $y = \cos x$ from $0 °$ to $360 °$ ?
It starts at its maximum $(0, 1)$ , crosses zero at $(90 °, 0)$ , reaches its minimum at $(180 °, - 1)$ , crosses zero again at $(270 °, 0)$ and returns to $(360 °, 1)$ .
The distinguishing feature is that it starts at a maximum rather than at zero.
How are the graphs of $y = \sin x$ and $y = \cos x$ related?
They are the same wave, shifted by $90 °$ along the $x$ -axis, since $\cos x = \sin (x + 90 °)$ .
So remembering one shape gives you the other.
Complete these periods, meaning the interval after which each graph repeats:
$y = \sin x$ and $y = \cos x$ repeat every $______$ .
$y = \tan x$ repeats every $______$ .
The completed statements are:
$y = \sin x$ and $y = \cos x$ repeat every $360 °$ .
$y = \tan x$ repeats every $180 °$ .
The shorter period of $\tan x$ is the reason it behaves differently from the other two when you look for solutions.
What does the graph of $y = \tan x$ look like?
It passes through $0 °$ , $180 °$ and $360 °$ , and rises steeply towards vertical asymptotes at $90 °$ and $270 °$ .
It has no maximum or minimum, and its values are not restricted to between $- 1$ and $1$ .
True or False?
$\sin x$ and $\cos x$ are never greater than 1 or less than $- 1$ .
True.
Both graphs oscillate between $- 1$ and $1$ , so those are the only values they take.
This is worth knowing as a check: an equation such as $\cos x = 2$ has no solutions at all.
What symmetry does the sine graph have between $0 °$ and $180 °$ ?
It is symmetrical about the vertical line $x = 90 °$ , so $\sin (180 ° - x) = \sin x$ .
For example $\sin 20 °$ and $\sin 160 °$ are equal.
Why does $y = \tan x$ have an asymptote at $90 °$ ?
Because $\tan x$ is a fraction with $\cos x$ on the bottom, and $\cos 90 ° = 0$ .
Dividing by zero is undefined, so the curve shoots off rather than taking a value there.
What happens to these graphs beyond $360 °$ , or below $0 °$ ?
The same pattern continues forever in both directions, because the graphs are periodic.
So a trigonometric equation has infinitely many solutions, which is why questions always state a range of angles to work within.
Complete the two trigonometric identities:
$\tan θ = \frac{\sin θ}{______}$
$\sin^{2} θ +______= 1$
The completed identities are:
$\tan θ = \frac{\sin θ}{\cos θ}$
$\sin^{2} θ + \cos^{2} θ = 1$
Both have to be memorised, and both hold for every value of $θ$ , which is what makes them identities rather than equations.
What are the two rearrangements of $\sin^{2} θ + \cos^{2} θ = 1$ ?
$\sin^{2} θ = 1 - \cos^{2} θ$
$\cos^{2} θ = 1 - \sin^{2} θ$
These are the forms you actually use, because they let you replace one squared ratio with the other.
True or False?
$\sin^{2} θ$ means $\sin (θ^{2})$ .
False.
$\sin^{2} θ$ means ${(\sin θ)}^{2}$ , so you take the sine first and then square the answer.
The two are completely different: for $θ = 30 °$ the first gives $0.25$ and the second gives $\sin 900 °$ .
Given that $\sin θ = \frac{3}{5}$ and $\cos^{2} θ = 1 - \sin^{2} θ$ , why are there two possible values of $\cos θ$ ?
Because $\cos^{2} θ = 1 - \frac{9}{25} = \frac{16}{25}$ , and square-rooting gives $\cos θ = \pm \frac{4}{5}$ .
The identity fixes $\cos^{2} θ$ but not the sign, so extra information about the angle is needed to choose between them.
Simplify $\tan θ \cos θ$ .
Replacing $\tan θ$ gives $\frac{\sin θ}{\cos θ} \times \cos θ$ , and the $\cos θ$ terms cancel:
$\tan θ \cos θ = \sin θ$
Rewriting $\tan$ in terms of $\sin$ and $\cos$ is usually the move that starts a simplification.
How do you prove a trigonometric identity?
Start with one side only, usually the more complicated one, and use the identities to transform it step by step until it becomes the other side.
Do not move terms from one side to the other, because that assumes the result you are trying to prove.
Why is the calculator's answer usually not the whole answer?
The inverse functions $\sin^{- 1}$ , $\cos^{- 1}$ and $\tan^{- 1}$ return a single angle, but the graphs repeat, so the given range normally contains more than one solution.
Sketching the graph and drawing a horizontal line across it shows you how many to look for.
If $x$ is a solution, complete each rule for finding another one:
For $\sin$ , another solution is $______- x$ .
For $\cos$ , another solution is $______- x$ .
For $\tan$ , another solution is $x +______$ .
The completed rules are:
For $\sin$ , another solution is $180 ° - x$ .
For $\cos$ , another solution is $360 ° - x$ .
For $\tan$ , another solution is $x + 180 °$ .
Each one comes from the symmetry of that graph, so you can rebuild them from a sketch if you forget.
Solve $\sin x = 0.5$ for $0 ° \leq x \leq 360 °$ .
The calculator gives $x = \sin^{- 1} 0.5 = 30 °$ , and the sine symmetry gives $180 ° - 30 °$ :
$x = 30 ° and x = 150 °$
There are exactly two, because the line $y = 0.5$ crosses the sine curve twice in this range.
Solve $3 \sin x + 1 = 0$ for $0 ° \leq x \leq 360 °$ .
Rearrange first to $\sin x = - \frac{1}{3}$ . The calculator returns $- 19.47 °$ , which is outside the range, so use it as a reference angle instead.
Sine is negative in the second half of the range, giving $180 ° + 19.47 °$ and $360 ° - 19.47 °$ :
$x = 199.5 ° and x = 340.5 °$
True or False?
$\tan x = 2$ has two solutions between $0 °$ and $360 °$ .
True.
The calculator gives $63.4 °$ , and adding $180 °$ gives $243.4 °$ .
Note that the second one does not come from $180 ° - x$ , which is the sine rule for extra solutions and would give a wrong answer here.
Solve $4 \sin^{2} x - 1 = 0$ for $0 ° \leq x \leq 360 °$ .
Rearranging gives $\sin^{2} x = \frac{1}{4}$ , so $\sin x = \pm \frac{1}{2}$ , and each sign gives two angles:
$x = 30 °, 150 °, 210 °, 330 °$
Forgetting the negative square root is what turns four solutions into two.
How do you solve $2 \cos^{2} x + \cos x - 1 = 0$ ?
Treat $\cos x$ as the unknown and factorise, exactly as you would a quadratic in $x$ :
$(2 \cos x - 1) (\cos x + 1) = 0$
That gives $\cos x = \frac{1}{2}$ or $\cos x = - 1$ , and each is then solved separately over the given range.
Why might one bracket of a factorised trig equation give no solutions?
Because it can lead to a value outside the possible range, such as $\sin x = 2$ or $\cos x = - 1.28$ .
Since $\sin$ and $\cos$ only take values between $- 1$ and $1$ , that bracket is rejected and only the other one is solved.
How do you decide which identity to substitute into an equation?
Choose the substitution that leaves the equation in one trigonometric ratio only.
So an equation containing $\sin^{2} x$ and $\cos x$ should have the $\sin^{2} x$ replaced, because replacing the $\cos x$ instead would introduce a square root.
Using $\sin^{2} x = 1 - \cos^{2} x$ , show that $2 \sin^{2} x - \cos x = 0$ becomes a quadratic in $\cos x$ .
Substituting gives $2 (1 - \cos^{2} x) - \cos x = 0$ , which expands to $2 - 2 \cos^{2} x - \cos x = 0$ .
Multiplying through by $- 1$ and reordering gives:
$2 \cos^{2} x + \cos x - 2 = 0$
Arranging it with a positive squared term makes the next step, factorising or using the quadratic formula, easier to see.

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