Pythagorean identities — sin² + cos² = 1, derivations of the other two: Worked examples

The scenario matrix

Before working anything, let us list every kind of situation these identities create. Each worked example below is tagged with the cell it covers.

Cell	Situation	Twist it tests	Example
A	Given a ratio, angle in Quadrant I	all functions positive — the "easy" sign	Ex 1
B	Given a ratio, angle in Quadrant II	$sin > 0$ , $cos < 0$ — pick the right sign	Ex 2
C	Given a ratio, angle in Quadrant III	both $sin < 0$ and $cos < 0$	Ex 3
D	Given a ratio, angle in Quadrant IV	$sin < 0$ , $cos > 0$	Ex 4
E	Degenerate / undefined input	$cos θ = 0$ : where $tan, sec$ die	Ex 5
F	Limiting behaviour	as $θ \to 9 0^{\circ}$ , what blows up?	Ex 6
G	Identity proof (no numbers)	algebra only, using Trig identity proofs — strategy	Ex 7
H	Real-world word problem	a physical distance forces the identity	Ex 8
I	Exam-style twist	mix with Double angle formulas / Solving trig equations	Ex 9

The mnemonic is "All Students Take Calculus" — A, S, T, C going anticlockwise from Quadrant I. The figure below is the map we will point back to in every example.

Figure — Pythagorean identities — sin² + cos² = 1, derivations of the other two

Cell E — Degenerate / undefined input

Worked example Ex 5 · What are

$tan θ$ and $sec θ$ at $θ = 9 0^{\circ}$ ? Does $1 + tan^{2} θ = sec^{2} θ$ hold there? Forecast: at $9 0^{\circ}$ the unit-circle point is $(0, 1)$ , so $cos 9 0^{\circ} = 0$ . Anything with $cos$ in the denominator is in trouble. Predict "undefined."

$tan 9 0^{\circ} = \frac{sin 9 0 ^{\circ}}{cos 9 0 ^{\circ}} = \frac{1}{0}$ — undefined (division by zero). Why this step? $tan = sin / cos$ ; the denominator vanishes.
$sec 9 0^{\circ} = \frac{1}{cos 9 0 ^{\circ}} = \frac{1}{0}$ — also undefined. Why this step? $sec$ is the reciprocal of $cos$ ; same zero denominator.
Because both sides are undefined, the identity $1 + tan^{2} θ = sec^{2} θ$ does not apply at $θ = 9 0^{\circ}$ . Why this step? An equation between two non-existent numbers is meaningless — you cannot plug in a value that isn't defined.

Verify: the master identity is fine though — $sin^{2} 9 0^{\circ} + cos^{2} 9 0^{\circ} = 1 + 0 = 1.$ ✓ Only the divided children fail, exactly where we divided by $0$ .

Cell F — Limiting behaviour

Worked example Ex 6 · As

$θ \to 9 0^{\circ}$ from below, what happens to $tan θ$ and $sec θ$ , and does $sec^{2} θ - tan^{2} θ$ still equal $1$ ? Forecast: near $9 0^{\circ}$ , $cos θ$ is a tiny positive number, so $1/ cos θ$ is huge. Both $tan$ and $sec$ should shoot to $+ \infty$ . But their difference of squares — guess it stays put.

Take a sequence of angles marching toward $9 0^{\circ}$ :

At $θ = 8 9^{\circ}$ : $cos = 0.01745$ , so $sec = 57.30$ , $tan = 57.29$ . Why this step? Watch the raw numbers grow as $cos$ shrinks.
At $θ = 89. 9^{\circ}$ : $sec = 572.96$ , $tan = 572.96$ (to 2 dp). Both explode together. Why this step? $cos$ is now $\approx 0.001745$ ; reciprocals blow up.
Yet $sec^{2} θ - tan^{2} θ = (sec - tan) (sec + tan)$ . Each factor: $sec + tan \to \infty$ but $sec - tan \to 0$ at just the right rate. Their product stays exactly $1$ — because $sec^{2} - tan^{2} = 1$ is the rearranged identity, true for every angle where it is defined. Why this step? The identity is algebraic, not a limit — it holds at each $8 9^{\circ}$ , $89. 9^{\circ}$ , $89.9 9^{\circ}$ exactly, even as the pieces run away to infinity.

Verify: at $θ = 8 9^{\circ}$ , $sec^{2} - tan^{2} = 57.298 7^{2} - 57.290 0^{2} = 1.0000.$ ✓ The blow-ups cancel perfectly.

Cell G — Pure identity proof (no numbers)

Worked example Ex 7 · Prove

$\frac{cos θ}{1 - sin θ} = \frac{1 + sin θ}{cos θ}$ . Forecast: cross-multiplying gives $cos^{2} θ = (1 - sin θ) (1 + sin θ)$ . The right side is a difference of squares — smells like the master identity. Predict "it collapses to $cos^{2} = cos^{2}$ ."

Cross-multiply (allowed when denominators $\neq = 0$ ): claim becomes $cos^{2} θ = (1 - sin θ) (1 + sin θ)$ . Why this step? Clearing fractions turns a "prove equal" into a "prove products equal" — cleaner. This is the standard move in Trig identity proofs — strategy.
Expand the right side by difference of squares: $(1 - sin θ) (1 + sin θ) = 1 - sin^{2} θ$ . Why this step? $(a - b) (a + b) = a^{2} - b^{2}$ with $a = 1, b = sin θ$ .
By the master identity, $1 - sin^{2} θ = cos^{2} θ$ . Why this step? This is exactly $sin^{2} + cos^{2} = 1$ rearranged — the whole point of the problem.
So both sides equal $cos^{2} θ$ . Identity proved (for $cos θ \neq = 0$ and $sin θ \neq = 1$ ).

Verify numerically at $θ = 3 0^{\circ}$ : LHS $= \frac{cos 3 0 ^{\circ}}{1 - sin 3 0 ^{\circ}} = \frac{0.86603}{0.5} = 1.73205$ ; RHS $= \frac{1 + sin 3 0 ^{\circ}}{cos 3 0 ^{\circ}} = \frac{1.5}{0.86603} = 1.73205.$ ✓

Cell H — Real-world word problem

Worked example Ex 8 · A ladder of length

$1 m$ leans against a wall, making angle $θ$ with the ground. Its foot is $cos θ$ metres from the wall and its top is $sin θ$ metres up. Show that "horizontal reach² + height² = ladder length²", and if the foot is $0.6 m$ out, how high is the top? Forecast: this is Pythagoras on a right triangle with hypotenuse $1$ — so the two squared distances must add to $1^{2}$ . Guess the height for a $0.6$ foot before computing.

The ladder, the wall, and the ground form a right triangle: legs $cos θ$ (along ground) and $sin θ$ (up the wall), hypotenuse $1$ (the ladder). Why this step? The wall meets the ground at a right angle — that's what makes Pythagoras apply. See the figure.
Pythagoras: $cos^{2} θ + sin^{2} θ = 1^{2} = 1$ — the master identity, wearing a hard hat. Why this step? The physical distances are the trig coordinates, so the geometric fact and the identity are the same statement.
Foot distance $= cos θ = 0.6$ . Height $= sin θ = 1 - cos^{2} θ = 1 - 0.36 = 0.64 = 0.8 m$ . Why this step? Solve the identity for the unknown leg; take $+$ because a height is a positive distance (physically Q1).

Verify (units and identity): $0. 6^{2} + 0. 8^{2} = 0.36 + 0.64 = 1.00 (m^{2}) = (1 m)^{2} .$ ✓ The top is $0.8 m$ up — a real "3-4-5" ladder.

Cell I — Exam-style twist (mixes double angle)

Worked example Ex 9 · Given

$sin θ = \frac{3}{5}$ with $θ$ in Quadrant II, find $cos 2 θ$ two different ways and confirm they agree. Forecast: the Double angle formulas give $cos 2 θ$ from $sin θ$ and $cos θ$ . Since the identity feeds double-angle constantly, expect a clean fraction. Guess whether $cos 2 θ$ is positive or negative.

First get $cos θ$ . $cos^{2} θ = 1 - \frac{9}{25} = \frac{16}{25}$ , and Q2 ⇒ $cos θ = - \frac{4}{5}$ . Why this step? We need $cos θ$ ; the master identity plus the Q2 sign rule (Cell B logic) delivers it.
Way 1: $cos 2 θ = cos^{2} θ - sin^{2} θ = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$ . Why this step? One standard double-angle form. It uses both squares — which is why the Pythagorean identity underlies it.
Way 2: $cos 2 θ = 1 - 2 sin^{2} θ = 1 - 2 \cdot \frac{9}{25} = 1 - \frac{18}{25} = \frac{7}{25}$ . Why this step? The " $1 - 2 sin^{2}$ " form is just Way 1 after replacing $cos^{2}$ by $1 - sin^{2}$ — the master identity in action. Getting the same answer confirms both.

Verify: both routes give $\frac{7}{25} = 0.28$ (positive). ✓ Note the sign of $cos θ$ did not matter here because $cos 2 θ$ only uses $cos^{2} θ$ — a subtle exam trap avoided.

Recall Case-coverage self-test

Which cell did we NOT need a $\pm$ decision for? ::: Cells E and F (undefined / limiting) and G (pure algebra) — the four quadrant cells A–D each needed a sign choice. In Ex 9, why didn't the sign of $cos θ$ change the answer? ::: $cos 2 θ$ depends only on $cos^{2} θ$ and $sin^{2} θ$ , which erase the sign. Where exactly do the child identities fail? ::: Where the divisor is zero — $cos θ = 0$ kills $tan, sec$ ; $sin θ = 0$ kills $cot, csc$ .

Connections

Yeh note Hinglish mein →
Unit Circle definition of sin and cos — the source of $(cos θ, sin θ)$
Pythagoras Theorem — the engine behind every cell
Reciprocal and Quotient trig identities — needed for $tan, sec, cot, csc$
Trig identity proofs — strategy — Cell G's toolkit
Double angle formulas — Cell I
Solving trig equations — where sign-choice really bites

Worked examples — Pythagorean identities — sin² + cos² = 1, derivations of the other two

The scenario matrix

Cell A — Quadrant I

Cell B — Quadrant II

Cell C — Quadrant III

Cell D — Quadrant IV

Cell E — Degenerate / undefined input

Cell F — Limiting behaviour

Cell G — Pure identity proof (no numbers)

Cell H — Real-world word problem

Cell I — Exam-style twist (mixes double angle)

Connections