Standard Trigonometric Equations

What you will get from this section. We will start with the circle definition of the trigonometric functions, establish three basic properties of sine, and use them to find every solution of a sine equation. The examples then show how the same reasoning extends to equations with trigonometric expressions on both sides. Cosine and tangent have analogous results, which we will state at the end.

In this article, I am deliberately revisiting a familiar A-level topic, developing it from the basic definitions rather than simply quoting results without fully explaining them. My aim is to show what a thorough understanding of even a basic topic can achieve, and how this depth of understanding can greatly simplify some TMUA questions.

TMUA Relevance Score: 8/10

Defining the trigonometric functions

Consider a circle with centre $O=(0,0)$ and radius $r>0$. Starting at the point $(r,0)$ on the positive $x$-axis, rotate a point $P$ through an angle $\theta$ around the circle. Positive angles represent anticlockwise rotation, while negative angles represent clockwise rotation. We may also make any number of complete turns.

Suppose the coordinates of $P$ after this rotation are $(x,y)$, as shown below.

We define

$$\sin\theta=\frac{y}{r},\qquad \cos\theta=\frac{x}{r},\qquad \tan\theta=\frac{y}{x}.$$

These definitions give sine and cosine for every real angle $\theta$, and tangent whenever $x\neq0$. For acute angles, they agree with the familiar trigonometric ratios in a right-angled triangle.

Throughout this article, angles are measured in degrees. The results are, of course, equally valid in radians, with the angles converted accordingly.

Three basic properties of sine

Let the point corresponding to $\theta$ now be labelled $P(a,b)$, so that $\sin\theta=b/r$. The diagram below shows this position together with the positions corresponding to $180-\theta$ and $-\theta$. Each is a possible position of the same moving point $P$.

Property 1: Periodicity

Starting from $P(a,b)$, any number of complete turns in either direction returns us to exactly the same point. Each additional rotation through $360^\circ$ or $-360^\circ$ is a complete turn, so it leaves $P$ in the position reached after the original rotation through $\theta$. Therefore,

Property 2: Reflection in the $y$-axis does not change the sine

The point corresponding to $180-\theta$ is the reflection of $P(a,b)$ in the $y$-axis, so its coordinates are $(-a,b)$. Its vertical coordinate is unchanged. Hence

Property 3: Reversing the angle changes the sign of the sine

The point corresponding to $-\theta$ is the reflection of $P(a,b)$ in the $x$-axis, so its coordinates are $(a,-b)$. Therefore,

or, equivalently, $\sin\theta=-\sin(-\theta)$. This is the statement that sine is an odd function.

Although the diagram shows an acute $\theta$, none of these arguments requires it to be acute. The same rotations and reflections apply wherever $P$ lies on the circle, so all three properties hold for every real $\theta$.

Basic properties. For every real $\theta$ and every integer $n$,

$$\sin\theta=\sin(\theta+360n),$$ $$\sin\theta=\sin(180-\theta),$$ $$\sin\theta=-\sin(-\theta).$$

A motivating example: solving $\sin x=\frac12$ graphically

Consider the equation $\sin x=\frac12$. We can solve it graphically by finding the intersections of the curve $y=\sin x$ and the horizontal line $y=\frac12$.

The graph below shows the sine curve in red and the horizontal line in blue. The horizontal axis is measured in degrees.

In the interval $0\leq x<360$, there are two intersections, at $x=30$ and $x=150$. These exact values agree with the familiar results $\sin30=\sin150=\frac12$.

However, the sine curve continues in both directions, repeating every $360$. Each of these two intersections therefore repeats after adding or subtracting any multiple of $360$.

The filled circles show the family

while the hollow circles show the family

We can describe these two infinite sets compactly:

Every intersection belongs to one of these families, so together they give all solutions of $\sin x=\frac12$. This example illustrates how a seemingly simple equation can have two infinite, periodic families of solutions.

It turns out that we can use the basic properties of sine to find these general solutions purely algebraically. Combined with some familiar trigonometric identities, this approach can greatly simplify some tricky trigonometric equations.

Standard sine equations and their general solutions

By a standard sine equation, we mean an equation of the form $\sin(ax+b)=c$, where $a$, $b$ and $c$ are real constants and $a\neq0$. Here $a$ and $b$ are coefficients, not the coordinates used in the circle diagram. The expression $ax+b$ is the whole angle being passed into sine.

We will now use the basic properties to establish the following equivalence entirely algebraically.

For all real $x$, $a$, $b$ and $c$, with $a\neq0$, and every integer $n$,

$$\begin{aligned} \sin(ax+b)=c \quad\Longleftrightarrow\quad &\sin(ax+b\pm360n)=c\\ &\text{or}\quad \sin(180-(ax+b)\pm360n)=c. \end{aligned}$$

Proof

Property 1 gives

Combining Properties 1 and 2 gives

All three expressions therefore have exactly the same value. Consequently, the original expression equals $c$ if and only if either transformed expression equals $c$, proving the equivalence.

Notice that we made no assumption about $c$. The argument applies unchanged when $c=1$ or $c=-1$. It also applies when $c>1$ or $c<-1$: in those cases, all three equations have no real solutions for $x$ because sine always lies between $-1$ and $1$.

A practical general solution for $\sin(ax+b)=c$

Suppose now that $-1\leq c\leq1$, so that $\sin^{-1}c$ exists. This gives the unique angle in $[-90,90]$ whose sine is $c$, called the principal value. It does not mean a reciprocal.

Starting from this angle and applying Properties 1 and 2 gives two families of solutions:

Set 1: $\quad ax+b=\sin^{-1}c\pm360n$.

Set 2: $\quad ax+b=180-\sin^{-1}c\pm360n$.

In each family, $n$ may be any integer. Since integers can be positive or negative, writing $+360n$ alone describes the same family as $\pm360n$.

We have found two families of solutions. The graph in our motivating example suggests that these should account for every solution, and this is indeed the case. A rigorous proof that there are no other solutions is possible, but we omit it to keep the article brief.

When $c=1$ or $c=-1$, the two families describe the same set of solutions, as expected. If $c>1$ or $c<-1$, the original equation has no real solutions, and the real inverse sine is undefined.

These formulas provide a simple practical algebraic method: subtract $b$, divide by $a$, and keep only the values of $x$ in any specified interval. Remember to divide the period term by $a$ as well.

You might wonder: how useful is all this? After all, these standard equations are... standard, and they look rather simple. 🤔 Well, firstly, they can be made to look far more menacing than they really are. Secondly, an otherwise very tricky trigonometric equation can sometimes be transformed into a standard one—and then your hope for a simple equation may come true after all. 😎 As the examples below will demonstrate.

Example 1: A standard sine equation

Solve $\sin(2x+20)=-0.5$ for $0\leq x\leq360$.

Example 1 solution

Since $\sin^{-1}(-0.5)=-30$, direct application of the practical general solution gives

Subtracting $20$ and dividing by $2$ gives

Therefore, the solutions in the specified interval are $x=95,155,275,335$.

The algebra is direct — the abstraction means you barely have to think about anything geometric or graphical.

Example 2: Sine expressions on both sides

Solve $\sin(3x+50)=\sin(x+10)$ for $0\leq x\leq360$.

Example 2 solution

This time, the right-hand side is already a sine expression. In one respect, this is actually simpler: we do not even need to evaluate an inverse sine! We can use $x+10$ as our starting angle and apply the same two families directly:

Simplifying and making $x$ the subject gives

Therefore, the solutions are $x=30,120,160,210,300,340$.

And that's it — unceremoniously solved! No hard graphical or geometrical thinking required.

Example 3: Transforming cosine into sine

Solve $-\cos(2x)=\sin(4x)$ for $0\leq x\leq360$.

Example 3 solution

This equation looks rather more menacing because it contains both sine and cosine, together with a negative sign. However, recall the familiar GCSE identity

Combining this with Property 3 allows us to rewrite the left-hand side:

The original equation is therefore equivalent to

We are now in exactly the same situation as Example 2. Applying the same general result gives

Simplifying and making $x$ the subject gives

Therefore, the solutions are

And there it is: after applying two basic identities, this rather menacing trigonometric equation has become two ordinary linear equations. Our hope of a simple equation has come true! 😎

Example 4: A modulus, twenty solutions, and one short sum

Find the sum of all distinct solutions of

in the interval $-720<x<720$.

Example 4 solution

A modulus inside sine, cosine on the other side, and an interval covering four complete turns. Drawing the graphs accurately would be quite a task! Instead, we will use our general solutions to turn the equation into very ordinary algebra.

First, write $\cos(3x)=\sin(90-3x)$. The expression inside the modulus changes sign at $x=-15$, so we consider two cases.

Case 1: $x\geq-15$

Here $|2x+30|=2x+30$, giving

Our general result gives

Rearranging, and reindexing where convenient, gives

But the second family is already contained in the first, because

It contributes no additional solutions.

We therefore need only the members of

that satisfy $-15\leq x<720$.

There are no solutions between $-15$ and $0$, and the first positive solution is $12$. Since $0<12<72$ and $720=10\cdot72$, the allowed indices are $n=0,1,\ldots,9$. There are exactly ten solutions: an arithmetic sequence starting at $12$, with common difference $72$.

We may or may not need to enumerate them. Let's check out Case 2 first!

Case 2: $x<-15$

Here $|2x+30|=-2x-30$, giving

Our general result gives

Rearranging gives

This time, the first family is already contained in the second: $120=-24+2\cdot72$, and $360$ is a multiple of $72$. Again, it contributes no additional solutions.

We therefore need only the members of

that satisfy $-720<x<-15$.

The solution closest to zero is $-24$. Counting downwards in steps of $72$, the allowed indices are again $n=0,1,\ldots,9$. There are exactly ten solutions: an arithmetic sequence starting at $-24$, with common difference $-72$.

Sum without enumeration

We now have two arithmetic sequences, each containing ten terms. Their common differences are opposite, so pairing corresponding terms gives

There are ten pairs. Therefore, the required sum is simply

There are other ways to tackle this question, but this method is particularly efficient: our practical general solutions turn a potentially messy graphical problem into manageable algebra, and a simple pairing observation finishes the job!

Summary

Cosine and tangent have analogous results, proved in similar ways. To keep this article brief, we omit those proofs. Equations involving cosine, tangent or angles in radians can be approached in the same way. The results are collected below; for radians, replace $180$ with $\pi$ and $360$ with $2\pi$.

Basic properties

Here $n\in\mathbb Z$, and the identities apply wherever the functions are defined.

Function	Periodicity	Supplementary angles	Negative angles
Sine	$\sin(\theta\pm360n)=\sin\theta$	$\sin(180-\theta)=\sin\theta$	$\sin(-\theta)=-\sin\theta$
Cosine	$\cos(\theta\pm360n)=\cos\theta$	$\cos(180-\theta)=-\cos\theta$	$\cos(-\theta)=\cos\theta$
Tangent	$\tan(\theta\pm180n)=\tan\theta$	$\tan(180-\theta)=-\tan\theta$	$\tan(-\theta)=-\tan\theta$

General solutions

Here $a\neq0$ and $n\in\mathbb Z$. The inverse functions use their principal values: $\sin^{-1}c\in[-90,90]$, $\cos^{-1}c\in[0,180]$ and $\tan^{-1}c\in(-90,90)$. If $c$ is outside the stated range, there are no real solutions.

Equation	Condition	General solutions
$\sin(ax+b)=c$	$-1\leq c\leq1$	$ax+b=\sin^{-1}c\pm360n$ or $ax+b=180-\sin^{-1}c\pm360n$
$\cos(ax+b)=c$	$-1\leq c\leq1$	$ax+b=\cos^{-1}c\pm360n$ or $ax+b=-\cos^{-1}c\pm360n$
$\tan(ax+b)=c$	$c\in\mathbb R$	$ax+b=\tan^{-1}c\pm180n$