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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Orient and recall
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Worksheets
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
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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The three Pythagorean identities
Work through the core explanation before applying it.
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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.
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 $\sin^2 \theta$.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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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
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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 $\theta$.
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Deriving and using the identities
Work through the core explanation before applying it.
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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 $\cos \theta \neq 0$):
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 $\cos^2 \theta$?
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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.
SOLUTION
2 marks · HSC band 4+
Step 1: Start with the fundamental identity.
$$\sin^2 \theta + \cos^2 \theta = 1$$
Step 2: Divide every term by $\cos^2 \theta$ (provided $\cos \theta \neq 0$).
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.
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.
Step 4: Take the square root. Since $\theta$ is acute, $\tan \theta > 0$.
$$\tan \theta = \frac{3}{4}$$
Answer: $\mathbf{\frac{3}{4}}$
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 $\sec \theta$?
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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?
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Common traps
Meet the mistakes that cost marks, then do it yourself.
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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 $\cos^2 \theta$.
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Drill, build fluency
Work through the core explanation before applying it.
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Drill, build fluency
+2 XP per reveal
Use the appropriate identity to find each exact missing value.
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If $\tan \theta = 2$ and $\theta$ is acute, find $\sec \theta$.
Restriction: $\sin \theta \neq 0$, i.e. $\theta \neq n\pi$ for $n \in \mathbb{Z}$.
ApplyBand 5–63 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)
