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Module 2 · L7 of 15 ~35 min ⚡ +50 XP in Learn · +25 to complete

Pythagorean Identities

The Pythagorean theorem is one of the most famous results in mathematics. But did you know it hides inside every trigonometric function? In this lesson you will discover the three Pythagorean identities that connect sine, cosine, tangent, and their reciprocals — identities so powerful they appear in almost every trigonometry problem you will ever solve.

Today's hook — You already know that $\sin^2 \theta + \cos^2 \theta = 1$. What do you think happens if you divide every term in this equation by $\sin^2 \theta$? And what happens if you divide every term by $\cos^2 \theta$? Try to predict the results before reading on.
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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

01
Recall — your gut answer first
+5 XP warm-up

You already know that $\sin^2 \theta + \cos^2 \theta = 1$. What do you think happens if you divide every term in this equation by $\sin^2 \theta$? And what happens if you divide every term by $\cos^2 \theta$? Try to predict the results before reading on.

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02
The three Pythagorean identities
+5 XP to read

One simple identity, two divisions, three powerful results. The second and third identities are derived by dividing the first identity by $\cos^2 \theta$ and $\sin^2 \theta$ respectively. All three are true for every value of $\theta$ where the functions are defined.

The fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$ comes directly from the unit circle equation $x^2 + y^2 = 1$. Dividing by $\cos^2 \theta$ gives the tangent-secant identity. Dividing by $\sin^2 \theta$ gives the cotangent-cosecant identity.

Three Pythagorean Identities sin²θ + cos²θ = 1 1 + tan²θ = sec²θ 1 + cot²θ = csc²θ
sin² + cos² = 1
Identity 1
$\sin^2 \theta + \cos^2 \theta = 1$. Use when you know $\sin \theta$ and want $\cos \theta$, or vice versa.
Identity 2
$1 + \tan^2 \theta = \sec^2 \theta$. Derived by dividing Identity 1 by $\cos^2 \theta$ (provided $\cos \theta \neq 0$).
Identity 3
$1 + \cot^2 \theta = \csc^2 \theta$. Derived by dividing Identity 1 by $\sin^2 \theta$ (provided $\sin \theta \neq 0$).

Fundamental identity: $\sin^2 \theta + \cos^2 \theta = 1$ (from the unit circle $x^2 + y^2 = 1$); Divide by $\cos^2 \theta$: $1 + \tan^2 \theta = \sec^2 \theta$ (provided $\cos \theta \neq 0$)

Pause — copy the fundamental identity $\sin^2\theta + \cos^2\theta = 1$ and the derived identity $1 + \tan^2\theta = \sec^2\theta$ (obtained by dividing through by $\cos^2\theta$) into your book.

True or false: The identity $1 + \tan^2 \theta = \sec^2 \theta$ is derived by dividing $\sin^2 \theta + \cos^2 \theta = 1$ by $\sin^2 \theta$.

03
What you'll master
Know

Key facts

  • The three Pythagorean identities
  • How to derive the tangent-secant and cotangent-cosecant identities
  • Common rearrangements of each identity
Understand

Concepts

  • Why all three identities come from the unit circle
  • How division transforms one identity into another
  • When each identity is most useful
Can do

Skills

  • Prove the three Pythagorean identities from first principles
  • Use the identities to find missing trig values
  • Simplify trigonometric expressions using identities
04
Key terms
Pythagorean identityAn equation connecting trig functions derived from $x^2 + y^2 = 1$ on the unit circle.
Fundamental identity$\sin^2 \theta + \cos^2 \theta = 1$ — the base from which all others are derived.
Derived identityAn identity obtained by algebraic manipulation of the fundamental identity.
RestrictionA condition on $\theta$ required for an algebraic step to be valid (e.g. $\cos \theta \neq 0$).
RearrangementRewriting an identity to isolate a different term (e.g. $\sin^2 \theta = 1 - \cos^2 \theta$).
VerificationChecking that an identity holds for a specific value of $\theta$.
05
Deriving and using the identities
core concept · +3 XP at end

We just saw that dividing $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$ gives $\tan^2\theta + 1 = \sec^2\theta$. That raises a question: where do these identities actually come from — and how do we use them to rearrange into a form we need? This card answers it → the derivation starts from the unit circle equation $x^2 + y^2 = 1$ and the rearrangements are shown step by step.

The fundamental identity comes directly from the unit circle equation $x^2 + y^2 = 1$, where $x = \cos \theta$ and $y = \sin \theta$.

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

Deriving Identity 2: Divide every term by $\cos^2 \theta$ (provided $\cos \theta \neq 0$):

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

Deriving Identity 3: Divide every term by $\sin^2 \theta$ (provided $\sin \theta \neq 0$):

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

Each identity is useful in different situations. Memorise the rearrangements too:

  • $\sin^2 \theta = 1 - \cos^2 \theta$  ·  $\cos^2 \theta = 1 - \sin^2 \theta$
  • $\tan^2 \theta = \sec^2 \theta - 1$  ·  $\sec^2 \theta - \tan^2 \theta = 1$
  • $\cot^2 \theta = \csc^2 \theta - 1$  ·  $\csc^2 \theta - \cot^2 \theta = 1$
Why physicists love these identities. In the study of waves and oscillations, energy is often split into "kinetic" and "potential" components. In simple harmonic motion, the total energy is constant — just like $\sin^2 \theta + \cos^2 \theta = 1$. The Pythagorean identity appears in the equations of pendulums, springs, and even quantum mechanical wave functions.

Start the derivation with the unit circle: at angle $\theta$, the point is $(\cos \theta, \sin \theta)$, and $x^2 + y^2 = 1$; Dividing by $\cos^2 \theta$ gives $\tan^2 \theta + 1 = \sec^2 \theta$ — tangent always pairs with...

Pause — copy the unit-circle derivation pathway: $(x^2 + y^2 = 1) \to (\sin^2\theta + \cos^2\theta = 1)$, then divide by $\cos^2\theta$ to get $\tan^2\theta + 1 = \sec^2\theta$ into your book.

Quick check: Which identity do you obtain by dividing $\sin^2 \theta + \cos^2 \theta = 1$ by $\cos^2 \theta$?

06
Worked Example — Proving the tangent-secant identity
+5 XP for trying first

We just saw the derivation: dividing $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$ yields $\tan^2\theta + 1 = \sec^2\theta$. That raises a question: how do you write this as a formal proof — starting from one side and reaching the other? This card answers it → begin with the fundamental identity, divide by $\cos^2\theta$, simplify, and always state the restriction $\cos\theta \neq 0$.

Prove that $1 + \tan^2 \theta = \sec^2 \theta$.

Your turn first. Try the proof yourself before viewing the solution.

Proof structure: start from $\sin^2 \theta + \cos^2 \theta = 1$, divide through, simplify; Always state the restriction: "provided $\cos \theta \neq 0$"

Pause — copy the proof structure (start from the fundamental identity, divide, simplify) and the requirement to state "provided $\cos\theta \neq 0$" into your book.

Fill the blanks: drag each token into the matching blank to complete the proof step.

cos²θ tan²θ sec²θ sin²θ

Dividing $\sin^2 \theta + \cos^2 \theta = 1$ by ___ gives ___ $+ 1 =$ ___.

07
Worked Example — Finding a missing value
+5 XP for trying first

We just saw how to prove an identity by manipulating one side to match the other. That raises a question: identities are also useful backwards — if I know $\sin\theta$, can I find $\cos\theta$ without a calculator? This card answers it → yes: rearrange $\sin^2\theta + \cos^2\theta = 1$ and substitute the known value.

If $\sec \theta = \frac{5}{4}$ and $\theta$ is acute, find $\tan \theta$.

Your turn first. Try it yourself before viewing the solution.

Identify which identity links the known and unknown trig functions; Substitute the given value, isolate the unknown squared term, then take the square root

Pause — copy the three-step procedure: (1) identify which identity links known to unknown, (2) substitute, (3) isolate the squared term then take the square root into your book.

Quick check: If $\tan \theta = 2$ and $\theta$ is acute, which identity gives $\sec \theta$?

08
Worked Example — Simplifying an expression
+5 XP for trying first

We just saw how to use a Pythagorean identity to find a missing trig value from a known one. That raises a question: what if the expression contains $\sec^2\theta - 1$ — does that simplify to something neater? This card answers it → yes: $\sec^2\theta - 1 = \tan^2\theta$ is a rearrangement of Identity 2; spotting this pattern is the key skill.

Simplify $\frac{\sec^2 \theta - 1}{\tan \theta}$.

Your turn first. Try it yourself before viewing the solution.

Simplification strategy: identify a Pythagorean identity "lurking" in the numerator or denominator; $\sec^2 \theta - 1 = \tan^2 \theta$ is the rearrangement of Identity 2 — recognise it instantly

Pause — copy the simplification strategy (look for a Pythagorean identity lurking in the expression) and the rearrangement $\sec^2\theta - 1 = \tan^2\theta$ into your book.

Odd one out: Three of these expressions simplify to $1$ using Pythagorean identities. Which one does NOT?

09
Common traps
Trap 1 — Mixing up the second and third identities

Students sometimes write $1 + \tan^2 \theta = \csc^2 \theta$. The identity with tangent always pairs with secant, not cosecant. Remember the pairs: sine-cosine, tangent-secant, cotangent-cosecant.

Trap 2 — Forgetting the "1" in the second and third identities

Some students write $\tan^2 \theta = \sec^2 \theta$ without the $+1$. Memorise the complete equation: $1 + \tan^2 \theta = \sec^2 \theta$.

Trap 3 — Not stating the restriction when dividing

When you divide by $\cos^2 \theta$, you are assuming $\cos \theta \neq 0$. In formal proofs, this restriction should be mentioned. Always note: "provided $\cos \theta \neq 0$" when dividing by $\cos^2 \theta$.

10
Drill — build fluency
+2 XP per reveal

Use the appropriate identity to find each exact missing value.

1

If $\tan \theta = 2$ and $\theta$ is acute, find $\sec \theta$.

2

If $\csc \theta = 3$ and $\frac{\pi}{2} < \theta < \pi$, find $\cot \theta$.

3

If $\cos \theta = -\frac{2}{3}$ and $\pi < \theta < \frac{3\pi}{2}$, find $\sin \theta$.

4

If $\sec \theta = -\frac{5}{3}$ and $\frac{\pi}{2} < \theta < \pi$, find $\tan \theta$.

5

Simplify $\sin^2 \theta(1 + \cot^2 \theta)$.

11
Revisit — one identity, two divisions
+5 XP for checking

Return to your original answer from Section 01. Dividing $\sin^2 \theta + \cos^2 \theta = 1$ by $\cos^2 \theta$:

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

Dividing by $\sin^2 \theta$:

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

One simple identity, two divisions, three powerful results. This is the elegance of the Pythagorean identities.

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next.

02
Short answer
Apply Band 4–5 3 marks

Prove the cotangent-cosecant identity

Prove that $1 + \cot^2 \theta = \csc^2 \theta$ starting from $\sin^2 \theta + \cos^2 \theta = 1$. State any restrictions. (3 marks)

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View comprehensive answer

Proof:

Start with $\sin^2 \theta + \cos^2 \theta = 1$.

Divide every term by $\sin^2 \theta$ (provided $\sin \theta \neq 0$):

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

Simplify:

$$1 + \cot^2 \theta = \csc^2 \theta \quad \text{QED}$$

Restriction: $\sin \theta \neq 0$, i.e. $\theta \neq n\pi$ for $n \in \mathbb{Z}$.

Apply Band 5–6 3 marks

Finding sin and cos from tangent

If $\tan \theta = -\frac{4}{3}$ and $\frac{3\pi}{2} < \theta < 2\pi$, find the exact values of $\sin \theta$ and $\cos \theta$. Show all working. (3 marks)

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View comprehensive answer

Working:

Using $1 + \tan^2 \theta = \sec^2 \theta$:

$$\sec^2 \theta = 1 + \frac{16}{9} = \frac{25}{9}$$

In QIV, $\sec \theta > 0$, so $\sec \theta = \frac{5}{3}$, giving $\cos \theta = \frac{3}{5}$.

Using $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$$\sin \theta = -\frac{4}{3} \times \frac{3}{5} = -\frac{4}{5}$$

Check: $\frac{16}{25} + \frac{9}{25} = 1$ ✓

Answer: $\sin \theta = \mathbf{-\frac{4}{5}}$, $\cos \theta = \mathbf{\frac{3}{5}}$

Analyse Band 6 3 marks

Simplify using identities

Simplify $\frac{\sin^2 \theta}{1 - \cos \theta}$ to a single trigonometric expression. Justify each step. (3 marks)

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View comprehensive answer

Working:

Replace $\sin^2 \theta$ with $1 - \cos^2 \theta$:

$$\frac{1 - \cos^2 \theta}{1 - \cos \theta} = \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 - \cos \theta}$$

Cancel (where $\cos \theta \neq 1$):

$$= 1 + \cos \theta$$

Answer: $\mathbf{1 + \cos \theta}$ (where $\cos \theta \neq 1$)

Key insight: recognising $1 - \cos^2 \theta$ as a difference of squares is the critical step.

01
Boss battle
earn bronze · silver · gold

Five timed questions on Pythagorean identities. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).

Enter the arena
02
Science Jump · Pythagorean identities
arcade practice

Climb platforms, hit checkpoints, and answer trig identity questions. Quick recall, lighter than the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

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