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Module 3 · L4 of 15 ~35 min ⚡ +95 XP available

Pythagorean Identities

One simple equation — $\sin^2\theta + \cos^2\theta = 1$ — sits at the heart of trigonometry. Every Pythagorean identity is born from this single fact about the unit circle. In this lesson you'll derive the two companion identities involving $\tan^2\theta$, $\sec^2\theta$, $\cot^2\theta$, and $\csc^2\theta$, then learn to use all three to simplify expressions and find missing trig values.

Today's hook — Starting from $\sin^2\theta + \cos^2\theta = 1$, divide every term by $\cos^2\theta$. What do you get? The second Pythagorean identity appears instantly — and a third one comes from dividing by $\sin^2\theta$ instead. One identity, three forms. This lesson shows you why they matter and how to wield them in exam questions.

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Recall — your gut answer first

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Starting from $\sin^2\theta + \cos^2\theta = 1$, how could you derive an identity involving $\tan^2\theta$ and $\sec^2\theta$? Write your gut answer before reading on.

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The three identities at a glance

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There are exactly three Pythagorean identities, all derived from the unit circle equation $x^2 + y^2 = 1$. Commit all three to memory — they appear in every trig exam:

Starting point: $\sin^2\theta + \cos^2\theta = 1$.

Divide by $\cos^2\theta$ to get $\tan^2\theta + 1 = \sec^2\theta$.

Divide by $\sin^2\theta$ to get $1 + \cot^2\theta = \csc^2\theta$.

$\sin^2\theta + \cos^2\theta = 1$

Identity 1 — the foundation

$\sin^2\theta + \cos^2\theta = 1$. Valid for all $\theta$. Rearranges to $\sin^2\theta = 1 - \cos^2\theta$ or $\cos^2\theta = 1 - \sin^2\theta$.

Identity 2 — the tan/sec form

$1 + \tan^2\theta = \sec^2\theta$. From dividing by $\cos^2\theta$. Note: $\tan^2\theta = \sec^2\theta - 1$.

Identity 3 — the cot/csc form

$1 + \cot^2\theta = \csc^2\theta$. From dividing by $\sin^2\theta$. Note: $\cot^2\theta = \csc^2\theta - 1$.

What you'll master

Know

Key facts

The three Pythagorean identities and all their rearrangements
How to derive each identity from $\sin^2\theta + \cos^2\theta = 1$
The domain conditions for each identity

Understand

Concepts

How each identity is derived from the unit circle equation $x^2 + y^2 = 1$
Why the identities hold for all real $\theta$ (not just acute angles)
The logical connection between all three identities

Can do

Skills

Use identities to simplify trigonometric expressions
Find the exact value of one trig function given another and a quadrant
Prove simple trig identities using substitution

Key terms

Pythagorean identityAn identity derived from the unit circle equation $x^2 + y^2 = 1$, where $x = \cos\theta$ and $y = \sin\theta$.

Trigonometric identityAn equation that is true for all values of the variable in its domain — not just for specific angles.

Unit circleThe circle $x^2 + y^2 = 1$ centred at the origin. Defines $\sin\theta$ and $\cos\theta$ as the $y$- and $x$-coordinates of a point on the circle.

Quadrant determinationThe quadrant in which $\theta$ lies determines the signs of all trig functions. Used to choose the correct sign when taking a square root.

Deriving the three Pythagorean identities

core concept

On the unit circle, the point corresponding to angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$. Since this point lies on $x^2 + y^2 = 1$:

$$\sin^2\theta + \cos^2\theta = 1 \tag{1}$$

Derive identity (2): Divide every term in (1) by $\cos^2\theta$ (valid when $\cos\theta \ne 0$):

$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} \;\Longrightarrow\; \tan^2\theta + 1 = \sec^2\theta \tag{2}$$

Derive identity (3): Divide every term in (1) by $\sin^2\theta$ (valid when $\sin\theta \ne 0$):

$$\frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta} \;\Longrightarrow\; 1 + \cot^2\theta = \csc^2\theta \tag{3}$$

Important: these hold for ALL real $\theta$. A common misconception is that $\sin^2\theta + \cos^2\theta = 1$ only works for acute angles. It follows directly from the unit circle equation, which is valid for every angle. There is no restriction on $\theta$ for identity (1); identities (2) and (3) only require $\cos\theta \ne 0$ and $\sin\theta \ne 0$ respectively (i.e. the functions must be defined).

Identity 1: ^2 + ^2 = 1 (all ); Identity 2: 1 + ^2 = ^2 (where 0) — derived by dividing (1) by ^2

Pause — copy all three Pythagorean identities into your book: $\sin^2\theta+\cos^2\theta=1$; $1+\tan^2\theta=\sec^2\theta$ (divide by $\cos^2\theta$); $1+\cot^2\theta=\csc^2\theta$ (divide by $\sin^2\theta$).

Quick check: Which of the following is the correct form of the second Pythagorean identity?

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · FIND A MISSING TRIG VALUE

If $\sin\theta = \dfrac{3}{5}$ and $\dfrac{\pi}{2} < \theta < \pi$, find $\cos\theta$ using a Pythagorean identity.

Apply identity (1): $\sin^2\theta + \cos^2\theta = 1$
$\left(\dfrac{3}{5}\right)^2 + \cos^2\theta = 1$
$\dfrac{9}{25} + \cos^2\theta = 1$
$\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$

Substitute the known value and solve for $\cos^2\theta$.

PROBLEM 2 · SIMPLIFY A TRIG EXPRESSION

Simplify $\dfrac{1}{1-\sin^2\theta}$.

Recognise $1 - \sin^2\theta$ from identity (1): $\sin^2\theta + \cos^2\theta = 1 \;\Rightarrow\; \cos^2\theta = 1 - \sin^2\theta$

Look for a sub-expression that matches a rearrangement of a known identity.

PROBLEM 3 · PROVE A TRIG IDENTITY

Show that $\dfrac{\sec^2\theta - 1}{\sec^2\theta} = \sin^2\theta$.

Work on the left-hand side (LHS). From identity (2): $\sec^2\theta - 1 = \tan^2\theta$
So $\dfrac{\sec^2\theta - 1}{\sec^2\theta} = \dfrac{\tan^2\theta}{\sec^2\theta}$

Always work from one side only (usually the more complex side). Identify and substitute identity (2).

Did you get this? True or false: $\sin^2\theta + \cos^2\theta = 1$ is only valid for $0 \le \theta \le \dfrac{\pi}{2}$.

Simplifying and factorising with identities

core technique

We just saw that $\sin^2\theta+\cos^2\theta=1$ yields $1+\tan^2\theta=\sec^2\theta$ (divide by $\cos^2\theta$) and $1+\cot^2\theta=\csc^2\theta$ (divide by $\sin^2\theta$). That raises a question: given a messy expression like $\sin^4\theta-\cos^4\theta$ or $\frac{1}{1-\sin^2\theta}$, which identity do you reach for, and how do you spot the pattern? This card answers it → difference of two squares unlocks $\sin^4\theta-\cos^4\theta = \sin^2\theta-\cos^2\theta$; direct substitution $1-\sin^2\theta = \cos^2\theta$ simplifies the fraction to $\sec^2\theta$.

The Pythagorean identities are tools for rewriting expressions. The key skill is recognising a pattern and making the right substitution:

When you see $1 - \sin^2\theta$ or $1 - \cos^2\theta$, replace with the other squared trig function
When you see $\sec^2\theta - 1$, replace with $\tan^2\theta$
When you see $\csc^2\theta - 1$, replace with $\cot^2\theta$
When you need a single trig function in the answer, convert everything to $\sin$ and $\cos$ first, then simplify

Example — factorise: $\cos^2\theta - 1 + \cos^2\theta = 2\cos^2\theta - 1$. This does not simplify via identities directly, but $\sin^4\theta - \cos^4\theta = (\sin^2\theta + \cos^2\theta)(\sin^2\theta - \cos^2\theta) = 1 \cdot (\sin^2\theta - \cos^2\theta) = \sin^2\theta - \cos^2\theta$.

Exam technique. When simplifying, always state which identity you are using as a brief comment in your working: "using $\sin^2\theta + \cos^2\theta = 1$". This earns method marks even if you make an arithmetic error later.

^4 - ^4 = (^2 - ^2)(^2 + ^2) = ^2 - ^2 — always look for difference of two squares; ^2^2 + ^4 = ^2(^2 + ^2) = ^2 — factorise first, then apply identity

Pause — copy both factorisation patterns into your book: $\sin^4\theta-\cos^4\theta = \sin^2\theta-\cos^2\theta$ (difference of squares with $\sin^2+\cos^2=1$), and $\frac{1}{1-\sin^2\theta} = \sec^2\theta$.

Fill the gap: $\sin^2\theta\cos^2\theta + \cos^4\theta = \cos^2\theta(\sin^2\theta + \cos^2\theta) = $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Thinking the identity only holds for acute angles

$\sin^2\theta + \cos^2\theta = 1$ is valid for ALL real $\theta$, not just $0 \le \theta \le \dfrac{\pi}{2}$. This follows directly from the unit circle definition, where the point $(\cos\theta, \sin\theta)$ lies on $x^2 + y^2 = 1$ for every angle $\theta$.

Trap 02

Forgetting to use the quadrant when taking square roots

After using an identity to find $\cos^2\theta = \dfrac{16}{25}$, you get $\cos\theta = \pm\dfrac{4}{5}$. Choosing the wrong sign loses the mark. Always check the quadrant: if $\dfrac{\pi}{2} < \theta < \pi$ (Q2), then $\cos\theta < 0$. If you do not state the quadrant reasoning, you will lose marks even with the right numerical answer.

Trap 03

Confusing $\tan^2\theta + 1$ with $1 + \tan^2\theta$

The identity is $1 + \tan^2\theta = \sec^2\theta$, which is the same as $\tan^2\theta + 1 = \sec^2\theta$ (addition commutes). What students sometimes write incorrectly is $\sec^2\theta + 1 = \tan^2\theta$ (wrong — this would require $\tan^2\theta > \sec^2\theta$, which contradicts the identity). Always rearrange carefully.

Did you get this? True or false: $\sec^2\theta - 1 = \tan^2\theta$ is a valid rearrangement of a Pythagorean identity.

Work mode · how are you completing this lesson?

Activities · practice with the identities

If $\cos\theta = -\dfrac{5}{13}$ and $\pi < \theta < \dfrac{3\pi}{2}$, find $\sin\theta$ exactly.

Simplify $\sin^2\theta\cos^2\theta + \cos^4\theta$.

Show that $\dfrac{\sec^2\theta - 1}{\sec^2\theta} = \sin^2\theta$. State each identity you use.

If $\tan\theta = 2$ and $\theta$ is in Q3, find $\sec\theta$ using identity 2. Then find $\cos\theta$.

Explain in your own words why there are three Pythagorean identities rather than just one.

Odd one out: Three of these are valid Pythagorean identities or rearrangements. Which one is NOT?

Revisit your thinking

Earlier you were asked how to derive an identity involving $\tan^2\theta$ and $\sec^2\theta$ from $\sin^2\theta + \cos^2\theta = 1$.

The answer: divide every term by $\cos^2\theta$. This gives $\dfrac{\sin^2\theta}{\cos^2\theta} + 1 = \dfrac{1}{\cos^2\theta}$, which simplifies to $\tan^2\theta + 1 = \sec^2\theta$. Then dividing by $\sin^2\theta$ instead gives the third identity $1 + \cot^2\theta = \csc^2\theta$. So all three identities share the same source — Pythagoras applied to the unit circle.

Why are there three rather than just one? Because there are three natural "scales" for the equation: unchanged (giving sin and cos), divided by $\cos^2\theta$ (giving tan and sec), and divided by $\sin^2\theta$ (giving cot and csc). Each scale exposes a different pair of trig functions.

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Multiple choice

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Short answer

ApplyBand 42 marks

Q1. If $\cos\theta = -\dfrac{5}{13}$ and $\pi < \theta < \dfrac{3\pi}{2}$, find $\sin\theta$. (2 marks)

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ApplyBand 42 marks

Q2. Simplify $\sin^2\theta\cos^2\theta + \cos^4\theta$. (2 marks)

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AnalyseBand 52 marks

Q3. Show that $\dfrac{\sec^2\theta - 1}{\sec^2\theta} = \sin^2\theta$. (2 marks)

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Comprehensive answers (click to reveal)

Activity 1: 1. $\sin^2\theta = 1 - 25/169 = 144/169$; Q3 so $\sin\theta = -12/13$ · 2. $\cos^2\theta(\sin^2\theta + \cos^2\theta) = \cos^2\theta$ · 3. LHS $= \tan^2\theta/\sec^2\theta = \frac{\sin^2\theta}{\cos^2\theta} \cdot \cos^2\theta = \sin^2\theta$ = RHS · 4. $\sec^2\theta = 5$; $\sec\theta = -\sqrt{5}$ (Q3, $\cos\theta < 0$); $\cos\theta = -1/\sqrt{5}$ · 5. Three identities arise because we can divide by $\cos^2\theta$, $\sin^2\theta$, or leave unchanged — each choice exposes a different pair of reciprocal trig functions

Q1 (2 marks): $\sin^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$ [1]. Since $\pi < \theta < \frac{3\pi}{2}$ (Q3), $\sin\theta < 0$, so $\sin\theta = -\dfrac{12}{13}$ [1].

Q2 (2 marks): $\sin^2\theta\cos^2\theta + \cos^4\theta = \cos^2\theta(\sin^2\theta + \cos^2\theta)$ [1] $= \cos^2\theta \cdot 1 = \cos^2\theta$ [1].

Q3 (2 marks): LHS $= \dfrac{\sec^2\theta - 1}{\sec^2\theta} = \dfrac{\tan^2\theta}{\sec^2\theta}$ [using identity 2, 1 mark] $= \dfrac{\sin^2\theta/\cos^2\theta}{1/\cos^2\theta} = \sin^2\theta$ = RHS [1].

Boss battle · The Identity Master

earn bronze · silver · gold

Five timed questions on Pythagorean identities. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering identity questions. Lighter alternative to the boss.

Key takeaways

Identity 1$\sin^2\theta + \cos^2\theta = 1$ — from $x^2+y^2=1$ on the unit circle. Valid for all $\theta$.

Identity 2$1 + \tan^2\theta = \sec^2\theta$ — obtained by dividing identity 1 by $\cos^2\theta$.

Identity 3$1 + \cot^2\theta = \csc^2\theta$ — obtained by dividing identity 1 by $\sin^2\theta$.

Finding missing valuesSubstitute into the appropriate identity, solve for the unknown, then use the quadrant to determine the sign of the square root.

Simplifying expressionsRecognise sub-expressions ($1-\sin^2\theta$, $\sec^2\theta-1$, etc.) and replace using rearranged identities.

Proving identitiesWork from one side only. Never rearrange both sides. State which identity you use in each step.

Next lesson: Compound Angle Formulas.

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← Lesson 3 · Graphs of Reciprocal Trig Functions Lesson 5 · Compound Angle Formulas →

Module overview · Extension 1 · Checkpoint 1