A calculator gives one angle. The domain decides how many there really are, and the quadrant signs decide where they sit.
$\sin^{-1}$ hands you one angle and stops. Every other solution has to be found by you, from the sign of the ratio and the size of the domain.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
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Solve $\sin\theta = 0.5$ on your calculator. It gives one answer. Sketch a sine curve from $0^\circ$ to $360^\circ$ and count how many times it reaches $0.5$. Where did the others go?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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One from the calculator, the rest from the quadrants
Work through the core explanation before applying it.
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One from the calculator, the rest from the quadrants
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Take the related acute angle from $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$ of the positive value. Then place a solution in every quadrant where the ratio has the required sign, and keep only those inside the given domain.
ASTC: All, Sine, Tangent, Cosine positive related acute angle $\alpha = $ inverse of $|$value$|$
Solve for the whole bracket first
For $\sin(2\theta) = 0.5$ on $0 \leq \theta \leq 360^\circ$, solve for $2\theta$ on $0 \leq 2\theta \leq 720^\circ$. Widen the domain before you solve, then halve at the end.
The sign chooses the quadrants
A negative ratio never means a negative angle here. It means the solutions sit in the two quadrants where that ratio is negative.
Match the units to the question
A domain written with $\pi$ wants radians and a calculator in radian mode; a domain in degrees wants degrees. Answers in the wrong unit earn nothing.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
The inverse trigonometric functions return only one angle, the related acute angle.
ASTC records which ratios are positive in each quadrant: All, Sine, Tangent, Cosine.
A solution exists in every quadrant where the ratio has the required sign.
The domain decides how many of those solutions are kept.
Understand
Concepts
Why a calculator returns one answer when the equation has several.
Why the sign of the ratio, not the sign of the angle, selects the quadrants.
Why a multiple-angle equation needs its domain widened before solving.
Can do
Skills
Solve a trigonometric equation on a restricted domain in degrees or radians.
Find all solutions of an equation with a negative ratio.
Solve equations of the form $\sin(n\theta) = k$ by widening the domain.
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Key terms
Restricted domainThe interval the question limits solutions to. Like this: $0^\circ \leq \theta \leq 360^\circ$ keeps one full revolution.
Related acute angleThe acute angle from the inverse function applied to the positive value. Like this: for $\sin\theta = -0.5$ the related acute angle is $\sin^{-1}(0.5) = 30^\circ$.
ASTCWhich ratios are positive by quadrant: All, Sine, Tangent, Cosine. Like this: in the third quadrant only tangent is positive.
QuadrantOne of the four regions of the unit circle, each $90^\circ$ wide. Like this: the second quadrant runs from $90^\circ$ to $180^\circ$.
Widening the domainMultiplying the domain by the same factor as the angle before solving. Like this: for $2\theta$ on $0^\circ$ to $360^\circ$, solve on $0^\circ$ to $720^\circ$.
General solutionEvery solution, written with a term that adds full revolutions. Like this: $\theta = 30^\circ + 360^\circ n$ covers one family of solutions.
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Why one answer is never the answer
Work through the core explanation before applying it.
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Why one answer is never the answer
core concept
$\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$ are functions, so each returns exactly one angle. But $\sin\theta = 0.5$ is satisfied at $30^\circ$ and at $150^\circ$, and again every $360^\circ$ after each of those.
The calculator gives the related acute angle. Everything else you place yourself, using the sign of the ratio and the domain you were given.
So the method has three parts: find the related acute angle from the positive value, decide which quadrants the sign allows, then keep only the solutions inside the domain.
Always take the inverse of the positive value. For $\cos\theta = -0.5$, work with $\cos^{-1}(0.5) = 60^\circ$ and then place the solutions in the quadrants where cosine is negative. Feeding the negative straight in gives one angle and hides the other.
Quick check: how many solutions does $\sin\theta = 0.5$ have on $0^\circ \leq \theta \leq 360^\circ$?
An inverse trigonometric function returns only the related acute angle. Find it from the positive value, place a solution in each quadrant the sign allows, then keep those inside the domain.
Pause, copy the three-part method and the note about taking the inverse of the positive value, into your book.
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Using ASTC to place the solutions
core concept
We just saw that the calculator gives one angle. That raises a question: which quadrants do the others live in? This card answers it → the ones where the ratio has the sign the equation demands.
ASTC records where each ratio is positive, reading anticlockwise from the first quadrant: A for all of them, then S for sine, then T for tangent, then C for cosine.
For $\sin\theta = 0.5$, sine is positive, so solutions sit in quadrants 1 and 2: $\theta = 30^\circ$ and $\theta = 180^\circ - 30^\circ = 150^\circ$.
For $\cos\theta = -0.5$, cosine is negative, so quadrants 2 and 3: $\theta = 180^\circ - 60^\circ = 120^\circ$ and $\theta = 180^\circ + 60^\circ = 240^\circ$.
The quadrant formulas, in terms of the related acute angle $\alpha$. Quadrant 1: $\alpha$. Quadrant 2: $180^\circ - \alpha$. Quadrant 3: $180^\circ + \alpha$. Quadrant 4: $360^\circ - \alpha$. In radians, replace $180^\circ$ with $\pi$ and $360^\circ$ with $2\pi$.
For which equation do the solutions lie in quadrants 2 and 3?
ASTC gives the sign of each ratio by quadrant. Place the related acute angle $\alpha$ using $\alpha$, $180^\circ - \alpha$, $180^\circ + \alpha$ and $360^\circ - \alpha$, keeping the quadrants where the ratio has the required sign.
Pause, copy ASTC, the four quadrant formulas, and both worked examples, into your book.
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Multiple angles, and radians
core concept
We just saw how to handle $\sin\theta = k$. That raises a question: what if the angle is $2\theta$? This card answers it → widen the domain to match, solve, then divide back.
For $\sin(2\theta) = 0.5$ on $0^\circ \leq \theta \leq 360^\circ$, the bracket $2\theta$ runs over $0^\circ \leq 2\theta \leq 720^\circ$. Solve on the wider domain first.
On $0^\circ$ to $720^\circ$: $2\theta = 30^\circ, 150^\circ, 390^\circ, 510^\circ$. Halving gives $\theta = 15^\circ, 75^\circ, 195^\circ, 255^\circ$, four solutions rather than two.
In radians the method is identical, with $\pi$ for $180^\circ$. For $\cos\theta = \dfrac{1}{2}$ on $0 \leq \theta \leq 2\pi$: $\theta = \dfrac{\pi}{3}$ and $\theta = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$.
Widening the domain is what stops you losing solutions. Solving $\sin(2\theta) = 0.5$ on the original domain finds two answers when there are four. The number of solutions multiplies by the same factor as the angle.
Fill the blank: to solve $\sin(3\theta) = k$ for $0^\circ \leq \theta \leq 360^\circ$, first solve for $3\theta$ on $0^\circ$ to degrees.
For $\sin(n\theta) = k$, multiply the domain by $n$, solve for the whole bracket on that wider domain, then divide every solution by $n$. The number of solutions multiplies by $n$. In radians the method is the same with $\pi$ in place of $180^\circ$.
Pause, copy the widen-solve-divide order, the four solutions of $\sin(2\theta) = 0.5$, and the radian example, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · A POSITIVE RATIO
Solve $\sin\theta = 0.5$ for $0^\circ \leq \theta \leq 360^\circ$.
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
Solve $\sin\theta = \dfrac{\sqrt{3}}{2}$ for $0^\circ \leq \theta \leq 360^\circ$.
Solve $\cos\theta = -\dfrac{\sqrt{2}}{2}$ for $0^\circ \leq \theta \leq 360^\circ$.
Solve $\tan\theta = 1$ for $0^\circ \leq \theta \leq 360^\circ$.
How many solutions does $\cos(3\theta) = 0.4$ have on $0^\circ \leq \theta \leq 360^\circ$?
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Revisit the sine curve you sketched
Run the quick drill and copy the summary into your book.
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Revisit the sine curve you sketched
At the start your calculator gave one solution of $\sin\theta = 0.5$ while the curve met $0.5$ twice. Name both solutions, and explain in one sentence why the inverse function could only ever return one of them.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 33 marks
Q1. Solve $\cos\theta = -\dfrac{1}{2}$ for $0^\circ \leq \theta \leq 360^\circ$, showing the related acute angle and the quadrants you used. (3 marks)
Q3. Explain why solving $\sin(2\theta) = k$ on the original domain rather than the widened one loses solutions. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $\alpha = 60^\circ$, sine positive in quadrants 1 and 2, so $\theta = 60^\circ, 120^\circ$. Practice 2: $\alpha = 45^\circ$, cosine negative in quadrants 2 and 3, so $\theta = 135^\circ, 225^\circ$. Practice 3: $\alpha = 45^\circ$, tangent positive in quadrants 1 and 3, so $\theta = 45^\circ, 225^\circ$. Practice 4: six.
Q1 (3 marks): Related acute angle $\cos^{-1}\left(\dfrac{1}{2}\right) = 60^\circ$ [1]. Cosine is negative in quadrants 2 and 3 [1]. $\theta = 120^\circ$ and $\theta = 240^\circ$ [1].
Q2 (4 marks): Widen the domain: $0^\circ \leq 2\theta \leq 720^\circ$ [1]. Related acute angle $\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = 60^\circ$, and sine is negative in quadrants 3 and 4 [1]. $2\theta = 240^\circ, 300^\circ, 600^\circ, 660^\circ$ [1]. $\theta = 120^\circ, 150^\circ, 300^\circ, 330^\circ$ [1].
Q3 (2 marks): As $\theta$ runs across the given domain, $2\theta$ runs across twice that interval, so the equation has twice as many solutions [1]. Solving on the original domain finds only the solutions of the first revolution of the bracket and discards the rest, so half the answers are lost [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Quadrant Quest
earn bronze · silver · gold
Find every solution of a trigonometric equation on a restricted domain, in degrees and radians. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.