Lesson 6 ~45 min Trigonometry D · Path +95 XP

Solving Trigonometric Equations

A trigonometric equation almost never has one solution. The calculator returns a single value, the circle supplies the others, and the interval stated in the question decides which of them belong in the answer.

Today's hook: Type $\sin^{-1}(0.5)$ into a calculator and it returns $30°$. That is one solution out of infinitely many, and in most questions it is not even the one being asked for. Knowing where the others are is the difference between a third of the marks and all of them.

0/5QUESTS

Think First

warm-up

The equation $\sin\theta = 0.5$ has solutions at $30°$ and $150°$. Write down two more, using what you know about turning all the way round the circle. Then say how many solutions the equation has in total if no interval is stated.

Record your answer in your workbook.

The Big Idea

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Solving takes three steps. Find the related acute angle from the size of the value, ignoring its sign. Use the sign to decide which quadrants are allowed. Then build every solution in the stated interval.

$$\text{size} \to \text{related angle}, \qquad \text{sign} \to \text{quadrants}$$

The picture is always a line crossing the circle. For sine it is a horizontal line at the given height; for cosine a vertical line at the given position. Either crosses a circle at most twice, which is why these equations have two solutions per turn, or one, or none.

$\text{related angle} + \text{quadrants}$

Ignore the sign at first

Find the related angle from the size of the value, then use the sign to place it.

Read the interval

It decides how many answers to give. Missing solutions is the commonest loss of marks.

Substitute back

Check each answer in the original equation, signs included.

What You'll Master

objectives

Know

That a trigonometric equation generally has more than one solution
That the related angle comes from the size and the quadrants from the sign
That tangent equations have one solution per $180°$, unlike sine and cosine

Understand

Why sine and cosine equations have at most two solutions per turn
Why a calculator returns only one of them

Can Do

Solve $\sin\theta = c$, $\cos\theta = c$ and $\tan\theta = c$ over a stated interval
Use exact values to solve without a calculator where possible
Find both the acute and the obtuse angle with a given sine

Words You Need

vocabulary

SolutionA value of the angle satisfying the equation.

Domain of solutionThe interval of angles the question restricts the answer to.

Inverse function keyThe calculator key returning one angle with a given ratio.

General solutionA formula giving every solution, including those beyond one turn.

Principal valueThe single value a calculator returns.

Why There Is Never Just One

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Solving $\sin\theta = c$ means finding every angle whose point on the unit circle sits at height $c$. Draw the horizontal line at that height and look at where it meets the circle.

A straight line meets a circle at most twice, so within one full turn there are at most two solutions. There are exactly two when $-1 < c < 1$, exactly one when $c = 1$ or $c = -1$, where the line touches at a single point, and none when $|c| > 1$.

For cosine the picture is the same with a vertical line, at the given horizontal position.

And because the point goes round again, adding any whole number of full turns to a solution gives another. So without a stated interval there are infinitely many solutions.

A calculator returns exactly one. The inverse key is built to give a single value, called the principal value, and it has no way of knowing which of the many you want. Supplying the rest is your job, and the interval in the question tells you how far to go.

The Method

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Three steps, in this order.

1. Find the related acute angle from the size of the value, ignoring its sign. Use exact values if you recognise the number, or the calculator's inverse key otherwise.

2. Decide which quadrants from the sign of the value, using the quadrant rules from Lesson 3.

3. Build the solutions in each allowed quadrant, using the related angle, and keep those inside the stated interval.

Take $\sin\theta = -\tfrac{1}{2}$ for $0° \leq \theta \leq 360°$.

Step 1. The size is $\tfrac{1}{2}$, and $\sin 30° = \tfrac{1}{2}$, so the related angle is $30°$.
Step 2. The value is negative, and sine is negative in the third and fourth quadrants.
Step 3. Third quadrant: $180° + 30° = 210°$. Fourth quadrant: $360° - 30° = 330°$.

So $\theta = 210°$ or $330°$, which is exactly what the diagram shows.

Building the angle in each quadrant

Given a related angle $\alpha$: the first-quadrant solution is $\alpha$; the second is $180° - \alpha$; the third is $180° + \alpha$; the fourth is $360° - \alpha$. These are the Lesson 3 formulas run backwards.

Tangent Behaves Differently

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Tangent equations follow the same three steps but produce a different count, because the tangent graph takes every value once per branch.

$\tan\theta = c$ has exactly one solution in each $180°$ interval, whatever $c$ is, so exactly two in a full turn. And the two are always $180°$ apart, since the period is $180°$.

Solve $\tan\theta = -1$ for $0° \leq \theta \leq 360°$.

The size is $1$, and $\tan 45° = 1$, so the related angle is $45°$. Tangent is negative in the second and fourth quadrants. Second: $180° - 45° = 135°$. Fourth: $360° - 45° = 315°$.

So $\theta = 135°$ or $315°$, and note that $315° - 135° = 180°$, as the period requires. That difference is a free check on any tangent answer.

There is no value of $c$ for which a tangent equation has no solution, unlike sine and cosine, because the tangent graph is unbounded and passes through every real value in every branch.

Reading the Interval

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The interval is not decoration. It decides the answer, and different intervals give genuinely different answer sets for the same equation.

For $\cos\theta = \tfrac{1}{2}$, the related angle is $60°$ and cosine is positive in the first and fourth quadrants.

On $0° \leq \theta \leq 360°$: $\theta = 60°$ or $300°$.
On $0° \leq \theta \leq 180°$: only $\theta = 60°$, since $300°$ is outside.
On $0° \leq \theta \leq 720°$: $60°$, $300°$, $420°$ and $660°$, adding a full turn to each.
On $-180° \leq \theta \leq 180°$: $60°$ and $-60°$, since $300°$ is outside but $300° - 360° = -60°$ is inside and is the same point.

Two habits prevent most errors here.

Write the interval down before you start, and check each candidate against it.
Check the endpoints, since $\leq$ includes them and $<$ does not.

Losing a solution loses a mark just as surely as getting one wrong, and a solution outside the interval loses one too. Both are avoided by the same discipline.

Common Pitfalls

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Giving only the calculator's answer.

Fix: the calculator gives one value. Use the quadrant rules to find the others and the interval to decide which to keep.

Using the negative value in the inverse key and then adding a sign again.

Fix: find the related angle from the size only, then place it with the sign. Feeding a negative into the inverse key gives a negative angle, which then needs converting anyway.

Reporting solutions that lie outside the stated interval.

Fix: check each candidate against the interval before writing it down, and watch whether the endpoints are included.

Expecting a tangent equation to have no solutions when the value is large.

Fix: tangent is unbounded, so $\tan\theta = 1000$ has solutions like any other value. It is sine and cosine that are limited to $[-1,1]$.

Watch Me Solve It · 3 examples

Watch Me Solve It · An exact-value sine equation

+15 XP per step

PROBLEM

Solve $\sin\theta = \dfrac{\sqrt{3}}{2}$ for $0° \leq \theta \leq 360°$.

1

Find the related angle from the size

$\sin 60° = \tfrac{\sqrt{3}}{2}$

Recognised from the special triangle, so no calculator is needed. The related angle is $60°$.
2

Use the sign to choose quadrants

The value is positive, and sine is positive in the first and second quadrants.
3

Build a solution in each

$\text{first: } 60°$

$\text{second: } 180° - 60° = 120°$
4

Check the interval and substitute back

Both $60°$ and $120°$ lie in $0°$ to $360°$, so both are kept. Checking: $\sin 120° = \sin 60° = \tfrac{\sqrt{3}}{2}$ by the supplementary relationship, as required.

Answer$\theta = 60°$ or $120°$

Watch Me Solve It · A negative cosine equation

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PROBLEM

Solve $\cos\theta = -\dfrac{1}{\sqrt{2}}$ for $0° \leq \theta \leq 360°$.

1

Find the related angle from the size only

$\cos 45° = \tfrac{1}{\sqrt{2}}$

Ignore the minus sign at this stage. The related angle is $45°$.
2

Use the sign to choose quadrants

The value is negative, and cosine is negative in the second and third quadrants.
3

Build a solution in each

$\text{second: } 180° - 45° = 135°$

$\text{third: } 180° + 45° = 225°$
4

Verify both

$\cos 135° = -\cos 45° = -\tfrac{1}{\sqrt{2}} \ \checkmark$

$\cos 225° = -\cos 45° = -\tfrac{1}{\sqrt{2}} \ \checkmark$

Both check out, and both lie in the interval.

Answer$\theta = 135°$ or $225°$

Watch Me Solve It · A calculator value over an unusual interval

+15 XP per step

PROBLEM

Solve $\tan\theta = 2.5$ for $-180° \leq \theta \leq 180°$, giving answers to the nearest degree.

1

Find the related angle

$\tan^{-1}(2.5) \approx 68.2°$

Not an exact value, so the calculator is used. The related angle is about $68°$.
2

Choose quadrants from the sign

The value is positive, and tangent is positive in the first and third quadrants.
3

Build both solutions in a standard turn

$\text{first: } 68°$

$\text{third: } 180° + 68° = 248°$
4

Adjust to the stated interval

$248° - 360° = -112°$

The value $248°$ lies outside $-180°$ to $180°$, so subtract a full turn to reach the same point at $-112°$, which is inside. So the answers are $68°$ and $-112°$. Checking the period: $68° - (-112°) = 180°$, as a tangent equation requires.

Answer$\theta \approx 68°$ or $-112°$

Brain Trainer · 5 problems

Brain Trainer · Solve and count

5 problems

Five items on trigonometric equations. Work each one, then reveal the answer.

1 Solve $\sin\theta = \tfrac{1}{2}$ for $0° \le \theta \le 360°$.

Related angle $30°$; sine positive in quadrants one and two.$30°$ and $150°$
2 Solve $\cos\theta = -1$ for $0° \le \theta \le 360°$.

The point is at the far left of the circle.$180°$
3 How many solutions does $\sin\theta = 2$ have?

Sine never exceeds $1$.None
4 Solve $\tan\theta = 1$ for $0° \le \theta \le 360°$.

Related angle $45°$; tangent positive in quadrants one and three.$45°$ and $225°$
5 $\sin\theta = 0.9$ and one solution is $64°$. Find the other for $0° \le \theta \le 360°$.

Supplementary angles share a sine.$116°$

Complete in your workbook.

Multiple Choice · 5 questions

MC1

How many

+10 XP

For $0° \leq \theta \leq 360°$, the equation $\cos\theta = 0.4$ has:

MC2

Choosing quadrants

+10 XP

To solve $\sin\theta = -0.6$, the quadrants to use are:

MC3

Building the angle

+10 XP

If the related angle is $\alpha$, the third-quadrant solution is:

MC4

Tangent

+10 XP

For $0° \leq \theta \leq 360°$, the equation $\tan\theta = 8$ has:

MC5

The interval

+10 XP

Solving $\cos\theta = \tfrac{1}{2}$ for $0° \leq \theta \leq 180°$ gives:

Short Answer · 3 questions

Solve with exact values

+15 XP

SHORT ANSWER

Solve each equation for $0° \leq \theta \leq 360°$, giving exact answers and showing the related angle and quadrants used.
(a) $\cos\theta = \tfrac{\sqrt{3}}{2}$
(b) $\sin\theta = -\tfrac{\sqrt{2}}{2}$
(c) $\tan\theta = -\sqrt{3}$
(d) $\sin\theta = 1$

Write your working in your book.

Intervals matter

+15 XP

SHORT ANSWER

Consider the equation $\sin\theta = \tfrac{1}{2}$.
(a) Find all solutions for $0° \leq \theta \leq 360°$.
(b) Find all solutions for $0° \leq \theta \leq 720°$.
(c) Find all solutions for $-360° \leq \theta \leq 0°$.
(d) Explain why the equation has infinitely many solutions if no interval is stated, and describe them all in one sentence.

Write your working in your book.

Acute and obtuse

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SHORT ANSWER

(a) Given $\sin\theta = 0.62$ with $0° \leq \theta \leq 180°$, find both possible values of $\theta$ to the nearest degree.
(b) Given $\cos\theta = 0.62$ over the same interval, explain why there is only one value, and find it.
(c) Explain what feature of the two graphs accounts for the difference between (a) and (b).
(d) A problem requires an obtuse angle whose sine is $0.62$. State which of your answers applies and explain why a calculator alone would not have produced it.

Write your working in your book.

Stretch Challenge · Every solution, and equations that need rearranging

+25 XP

CHALLENGE

(a) Write a formula giving every solution of $\sin\theta = \tfrac{1}{2}$, for any integer number of turns.
(b) Solve $2\sin\theta + 1 = 0$ for $0° \leq \theta \leq 360°$, and explain why the rearrangement must come first.
(c) Solve $\sin^2\theta = \tfrac{1}{4}$ for $0° \leq \theta \leq 360°$, and explain why this equation has twice as many solutions as $\sin\theta = \tfrac{1}{2}$.

(a) The two solutions in one turn are $30°$ and $150°$, and adding any whole number of full turns to either gives another solution. So every solution is $\theta = 30° + 360°n$ or $\theta = 150° + 360°n$, where $n$ is any integer, positive, negative or zero. Taking $n = 0$ recovers the two basic solutions, $n = 1$ gives $390°$ and $510°$, and $n = -1$ gives $-330°$ and $-210°$, matching the answers found by hand in the previous question.

(b) Rearranging: $2\sin\theta = -1$, so $\sin\theta = -\tfrac{1}{2}$. The related angle is $30°$ and sine is negative in the third and fourth quadrants, giving $\theta = 180° + 30° = 210°$ and $\theta = 360° - 30° = 330°$. The rearrangement must come first because the three-step method applies to an equation of the form $\sin\theta = c$, with the ratio alone on one side; until that form is reached there is no value $c$ whose size gives a related angle and whose sign chooses quadrants. Attempting to read quadrants off the unrearranged equation would mean reading the sign of $2\sin\theta + 1$, which is not the sign of the sine.

(c) Taking the square root of both sides gives $\sin\theta = \pm\tfrac{1}{2}$, and both signs must be kept, because squaring destroys sign information and undoing it admits both. The positive case gives $30°$ and $150°$, as before. The negative case gives $210°$ and $330°$, from part (b). So there are four solutions: $30°$, $150°$, $210°$ and $330°$. It has twice as many because $\sin^2\theta = \tfrac{1}{4}$ asks only that the sine have size $\tfrac{1}{2}$, placing no condition on its sign, whereas $\sin\theta = \tfrac{1}{2}$ demands a particular sign as well. Geometrically, the horizontal lines at height $\tfrac{1}{2}$ and at height $-\tfrac{1}{2}$ each cross the circle twice, and the squared equation accepts all four crossings while the original accepts only two. Forgetting the $\pm$ and reporting two solutions is the standard error here, and it is the same one met when solving $x^2 = 4$.

Quick Review

recap

Three steps

Related angle from the size, quadrants from the sign, then the interval

Building angles

$\alpha$, $180°-\alpha$, $180°+\alpha$, $360°-\alpha$

Tangent

One solution per $180°$, always two per turn

The calculator

Returns one value; the rest are yours to find

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