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1.1 Equation structure and trigonometric identities

Use radian measure, unit-circle definitions, and identity families to solve trig equations carefully.

Course contents

MATH1025: Preparatory mathematics

Preparatory notes for algebraic technique, trigonometry, and proof-aware worked examples.

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Chapter 0-1Foundations and early methods2 sections
0.10.1 Course foundations and notation1.11.1 Equation structure and trigonometric identitiesCurrent section

Why this chapter matters

This chapter is about structure, not memorization. In trigonometry, the same expression can be rewritten in several useful ways, but those rewrites are only valuable if you know what they preserve:

  • some steps preserve the solution set exactly;
  • some steps only produce candidate solutions;
  • some steps are valid only when the domain is checked first.

The goal is to read trig expressions as objects with geometry, symmetry, and algebraic behavior.

Radian measure and standard position

Definition

Radian measure

If an angle subtends an arc of length s on a circle of radius r, then its radian measure is

θ=sr.\theta = \frac{s}{r}.

When r=1r=1, the radian measure is literally the arc length on the unit circle. That is why radians are the natural unit for trigonometry and calculus.

Since one full revolution is 2π2\pi radians or 360∘360^\circ, we have

π radians=180∘,\pi \text{ radians} = 180^\circ,

so

1 radian=180π∘,1∘=π180 radians.1 \text{ radian} = \frac{180}{\pi}^\circ, \qquad 1^\circ = \frac{\pi}{180} \text{ radians}.

Definition

Angles in standard position

An angle in the xy-plane is in standard position if its vertex is at the origin and its initial ray lies on the positive x-axis.

  • Counterclockwise rotation is positive.
  • Clockwise rotation is negative.

From now on, every angle in this chapter is measured in radians unless the problem explicitly says otherwise.

Worked example

Convert between degrees and radians

The angle 150∘150^\circ equals

150∘=150⋅π180=5π6.150^\circ = 150 \cdot \frac{\pi}{180} = \frac{5\pi}{6}.

Likewise, -45^\circ = -\pi/4.

Area and arc length

Two useful geometric formulas follow immediately from the definition of radian measure:

s=rθ,A=12r2θ.s = r\theta, \qquad A = \frac{1}{2}r^2\theta.

These formulas are not separate facts to memorize. They are the same angle measure translated into length and area.

Trigonometric functions on the unit circle

Definition

Trigonometric functions

Place an angle θ\theta in standard position on a circle of radius r. Let the terminal ray meet the circle at P(x,y). Then

sin⁡θ=yr,cos⁡θ=xr,tan⁡θ=yx,\sin\theta = \frac{y}{r}, \qquad \cos\theta = \frac{x}{r}, \qquad \tan\theta = \frac{y}{x},

and

csc⁡θ=ry,sec⁡θ=rx,cot⁡θ=xy.\csc\theta = \frac{r}{y}, \qquad \sec\theta = \frac{r}{x}, \qquad \cot\theta = \frac{x}{y}.

On the unit circle, r=1r=1, so sine and cosine become the y- and x- coordinates directly. That is the geometric reason the unit circle is so useful: it turns trigonometric values into coordinates.

Some values are undefined because their denominators vanish. For example, tan(\pi/2) is undefined because \cos(\pi/2)=0.

Worked example

Exact values from the unit circle

The standard values at 0, \pi/6, \pi/4, \pi/3, and \pi/2 are:

| θ\theta | sin⁡θ\sin\theta | cos⁡θ\cos\theta | tan⁡θ\tan\theta | | --- | --- | --- | --- | | 0 | 0 | 1 | 0 | | \pi/6 | 1/2 | \sqrt3/2 | 1/\sqrt3 | | \pi/4 | \sqrt2/2 | \sqrt2/2 | 1 | | \pi/3 | \sqrt3/2 | 1/2 | 3\sqrt3 | | \pi/2 | 1 | 0 | undefined |

These values are the building blocks for the rest of the chapter.

Common mistake

Do not treat degrees as the default

If a problem says \pi/3, it means radians, not degrees. In this course, 60∘60^\circ and \pi/3 are the same angle written in different units.

Symmetry and periodicity

The unit circle immediately gives the basic symmetry rules:

sin⁡(−θ)=−sin⁡θ,cos⁡(−θ)=cos⁡θ,tan⁡(−θ)=−tan⁡θ.\sin(-\theta) = -\sin\theta, \qquad \cos(-\theta) = \cos\theta, \qquad \tan(-\theta) = -\tan\theta.

Reflection across the y-axis and the origin also gives the familiar angle shifts such as π−θ\pi-\theta, π+θ\pi+\theta, and 2π+θ2\pi+\theta.

Theorem

Periodicity of trigonometric functions

sin⁡(θ+2π)=sin⁡θ,cos⁡(θ+2π)=cos⁡θ,\sin(\theta+2\pi)=\sin\theta, \qquad \cos(\theta+2\pi)=\cos\theta,sec⁡(θ+2π)=sec⁡θ,csc⁡(θ+2π)=csc⁡θ,\sec(\theta+2\pi)=\sec\theta, \qquad \csc(\theta+2\pi)=\csc\theta,tan⁡(θ+π)=tan⁡θ,cot⁡(θ+π)=cot⁡θ.\tan(\theta+\pi)=\tan\theta, \qquad \cot(\theta+\pi)=\cot\theta.

Periodicity is not merely a trick for reducing angles. It tells you what part of the circle controls each function.

Worked example

Use symmetry to simplify a value

Because sin⁡(−θ)=−sin⁡θ\sin(-\theta)=-\sin\theta, we get

sin⁡(−π/6)=−sin⁡(π/6)=−12.\sin(-\pi/6) = -\sin(\pi/6) = -\frac12.

Because cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta, we get

cos⁡(−π/6)=cos⁡(π/6)=32.\cos(-\pi/6) = \cos(\pi/6) = \frac{\sqrt3}{2}.

The basic identities

Theorem

Pythagorean identities

sin⁡2θ+cos⁡2θ=1,1+tan⁡2θ=sec⁡2θ,1+cot⁡2θ=csc⁡2θ.\sin^2\theta + \cos^2\theta = 1, \qquad 1+\tan^2\theta = \sec^2\theta, \qquad 1+\cot^2\theta = \csc^2\theta.

The first identity is the coordinate statement x2+y2=1x^2+y^2=1 on the unit circle. The other two follow by dividing through by cos⁡2θ\cos^2\theta or sin⁡2θ\sin^2\theta where those quantities are nonzero.

Worked example

Turn one identity into another

Starting from

sin⁡2θ+cos⁡2θ=1,\sin^2\theta + \cos^2\theta = 1,

divide by cos⁡2θ\cos^2\theta to obtain

tan⁡2θ+1=sec⁡2θ.\tan^2\theta + 1 = \sec^2\theta.

This step is valid only where cos⁡θ≠0\cos\theta \neq 0.

Common mistake

Do not divide without checking the denominator

If a step divides by sin⁡θ\sin\theta or cos⁡θ\cos\theta, the identity may become invalid at the angles where that quantity is zero. Always keep track of the restricted domain.

Compound angle formulas

The compound angle formulas are the workhorse identities of the chapter. They let you turn sums and differences of angles into algebraic expressions.

Proof

Why the cosine difference formula is the key step

Theorem

Compound angle formulas

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\betasin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\betacos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\betacos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\betatan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}tan⁡(α−β)=tan⁡α−tan⁡β1+tan⁡αtan⁡β\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}

These formulas explain most of the angle-algebra manipulations you will see in later work.

Worked example

Derive a sine addition formula from the cosine one

If you know

cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β,\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta,

then replacing α\alpha by \pi/2-\alpha and β\beta by −β-\beta gives

sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β.\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.

The other addition and subtraction formulas follow in the same way.

Double angle, product-to-sum, and sum-to-product

Theorem

Double angle formulas

sin⁡(2α)=2sin⁡αcos⁡α\sin(2\alpha)=2\sin\alpha\cos\alphacos⁡(2α)=cos⁡2α−sin⁡2α=2cos⁡2α−1=1−2sin⁡2α\cos(2\alpha)=\cos^2\alpha-\sin^2\alpha =2\cos^2\alpha-1 =1-2\sin^2\alphatan⁡(2α)=2tan⁡α1−tan⁡2α\tan(2\alpha)=\frac{2\tan\alpha}{1-\tan^2\alpha}

These are just the compound-angle formulas with the same angle used twice.

Theorem

Product-to-sum formulas

sin⁡αcos⁡β=12(sin⁡(α+β)+sin⁡(α−β))\sin\alpha\cos\beta=\frac12\bigl(\sin(\alpha+\beta)+\sin(\alpha-\beta)\bigr)cos⁡αsin⁡β=12(sin⁡(α+β)−sin⁡(α−β))\cos\alpha\sin\beta=\frac12\bigl(\sin(\alpha+\beta)-\sin(\alpha-\beta)\bigr)sin⁡αsin⁡β=−12(cos⁡(α+β)−cos⁡(α−β))\sin\alpha\sin\beta=-\frac12\bigl(\cos(\alpha+\beta)-\cos(\alpha-\beta)\bigr)cos⁡αcos⁡β=12(cos⁡(α+β)+cos⁡(α−β))\cos\alpha\cos\beta=\frac12\bigl(\cos(\alpha+\beta)+\cos(\alpha-\beta)\bigr)

Theorem

Sum-to-product formulas

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A-\sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A-\cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}

Worked examples

Worked example

Rewrite a linear combination as a single sine

Express 3cos⁡x−sin⁡x\sqrt3\cos x - \sin x in the form rsin⁡(x+α)r\sin(x+\alpha).

We compare

rsin⁡(x+α)=rsin⁡xcos⁡α+rcos⁡xsin⁡α.r\sin(x+\alpha)=r\sin x\cos\alpha+r\cos x\sin\alpha.

Matching coefficients gives

rsin⁡α=3,rcos⁡α=−1.r\sin\alpha=\sqrt3, \qquad r\cos\alpha=-1.

So r=3+1=2r=\sqrt{3+1}=2, and we may choose \alpha=2\pi/3. Therefore

3cos⁡x−sin⁡x=2sin⁡ ⁣(x+2π3).\sqrt3\cos x-\sin x = 2\sin\!\left(x+\frac{2\pi}{3}\right).

If we solve 3cos⁡x−sin⁡x=1\sqrt3\cos x-\sin x=1, then

2sin⁡ ⁣(x+2π3)=1,2\sin\!\left(x+\frac{2\pi}{3}\right)=1,

so

sin⁡ ⁣(x+2π3)=12.\sin\!\left(x+\frac{2\pi}{3}\right)=\frac12.

Hence

x+2π3=π6+2kπorx+2π3=5π6+2kπ,x+\frac{2\pi}{3}=\frac{\pi}{6}+2k\pi \quad\text{or}\quad x+\frac{2\pi}{3}=\frac{5\pi}{6}+2k\pi,

which gives

x=−π2+2kπorx=−π6+2kπ,k∈Z.x=-\frac{\pi}{2}+2k\pi \quad\text{or}\quad x=-\frac{\pi}{6}+2k\pi, \qquad k\in Z.

Worked example

A compact algebraic identity

Show that

cos⁡4θ+sin⁡4θ=1+cos⁡2(2θ)2=3+cos⁡(4θ)4.\cos^4\theta+\sin^4\theta=\frac{1+\cos^2(2\theta)}{2} =\frac{3+\cos(4\theta)}{4}.

Start with

(sin⁡2θ+cos⁡2θ)2=1.(\sin^2\theta+\cos^2\theta)^2=1.

Expanding gives

sin⁡4θ+cos⁡4θ+2sin⁡2θcos⁡2θ=1.\sin^4\theta+\cos^4\theta+2\sin^2\theta\cos^2\theta=1.

So

sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ.\sin^4\theta+\cos^4\theta =1-2\sin^2\theta\cos^2\theta.

Use sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta)=2\sin\theta\cos\theta to get

sin⁡2θcos⁡2θ=14sin⁡2(2θ),\sin^2\theta\cos^2\theta=\frac14\sin^2(2\theta),

and then use sin⁡2(2θ)=1−cos⁡2(2θ)\sin^2(2\theta)=1-\cos^2(2\theta) or cos⁡(4θ)=1−2sin⁡2(2θ)\cos(4\theta)=1-2\sin^2(2\theta) to reach the stated forms.

Common mistake

Back-substitution is not optional

After using identities to solve a trigonometric equation, always check the candidate solutions in the original equation. A transformation may introduce extra roots or hide domain restrictions.

Quick checks

Quick check

Convert 210∘210^\circ to radians.

Use the degree-radian conversion formula.

Solution

Answer

Quick check

What are sin⁡(−θ)\sin(-\theta), cos⁡(−θ)\cos(-\theta), and tan⁡(−θ)\tan(-\theta)?

Use symmetry of the unit circle.

Solution

Answer

Quick check

Why is \tan(\pi/2) undefined?

Look at the denominator in the definition.

Solution

Answer

Quick check

Show the route from sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 to 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta.

State the division step carefully.

Solution

Answer

Exercises

  1. Convert 330∘330^\circ to radians and compute sin⁡(330∘)\sin(330^\circ) and cos⁡(330∘)\cos(330^\circ).
  2. Use a periodicity identity to simplify tan⁡(θ+π)\tan(\theta+\pi).
  3. Prove 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta from the Pythagorean identity.
  4. Rewrite 3cos⁡x−sin⁡x\sqrt3\cos x-\sin x as rsin⁡(x+α)r\sin(x+\alpha) and solve 3cos⁡x−sin⁡x=1\sqrt3\cos x-\sin x = 1.
  5. Show that \cos^4\theta+\sin^4\theta = (3+\cos 4\theta)/4.

The point of the exercises is not speed. It is to make the identities feel like tools that you can derive, explain, and reuse.

Prerequisites

0.1

0.1 Course foundations and notation

Key terms in this unit

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