Pythagorean identities — sin² + cos² = 1, derivations of the other two: Question bank

Before we start, one reminder of notation so nothing is used before it is defined:

$sin^{2} θ$ means $(sin θ)^{2}$ — square the output, not the angle.
$tan θ = \frac{sin θ}{cos θ}$ — the quotient of the two unit-circle coordinates: how steep the ray is (up-amount divided by right-amount). It is undefined wherever $cos θ = 0$ .
$sec θ = \frac{1}{cos θ}$ , $csc θ = \frac{1}{sin θ}$ , $cot θ = \frac{cos θ}{sin θ}$ — these are the reciprocal and quotient functions. Each is undefined wherever its denominator is $0$ .

The picture below is the object every claim on this page secretly refers to — keep glancing back at it.

True or false — justify

Read each claim, decide, then reveal the reasoning.

sin^{2} θ + cos^{2} θ = 1

holds for negative angles like

θ = - 40°

True. The identity came from squaring the unit-circle coordinates (see figure s02), and squaring erases the sign — so it holds for every real

θ

, positive, negative, or beyond

360°

1 + tan^{2} θ = sec^{2} θ

holds for every angle

θ

==False — but for a subtle reason: it does not become a wrong number, it stops being a statement at all.== Wherever

cos θ = 0

(like

θ = 90°

tan θ

and

sec θ

are undefined, so the equation has no meaning there. Everywhere the functions are defined, it is perfectly true.

sin^{2} θ + cos^{2} θ = 1

then

sin θ + cos θ = 1

False. You cannot "square-root each term" of a sum:

a^{2} + b^{2} \neq = a + b

. At

θ = 45°

sin θ + cos θ = 2 \approx 1.41

, not

1

sec^{2} θ - tan^{2} θ

and

csc^{2} θ - cot^{2} θ

are the same constant.

True. Both equal

1

: the first is

1 + tan^{2} = sec^{2}

rearranged, the second is

1 + cot^{2} = csc^{2}

rearranged.

cos^{2} θ = 1 - sin^{2} θ

lets you write

cos θ = 1 - sin^{2} θ

for all

θ

False. Only the square is fixed; the root needs a sign.

cos θ = \pm 1 - sin^{2} θ

, and the quadrant of

θ

chooses

+

-

(see the sign chart, figure s03).

sin^{2} θ

can be negative for some

θ

False. It's a real number squared, so

sin^{2} θ \geq 0

always — likewise

0 \leq sin^{2} θ \leq 1

since

∣ sin θ ∣ \leq 1

The identity

sin^{2} θ + cos^{2} θ = 1

only works in the first quadrant where the triangle is "real".

False. Squaring the legs

∣ cos θ ∣

and

∣ sin θ ∣

removes the absolute values, so the same equation survives in all four quadrants and on the axes.

sec^{2} θ \geq 1

for every

θ

where it is defined.

True.

sec^{2} θ = 1 + tan^{2} θ

and

tan^{2} θ \geq 0

, so

sec^{2} θ \geq 1

always (equality only when

tan θ = 0

Spot the error

Each line contains ONE flawed step. Name it.

cos^{2} θ = \frac{1}{4} \Rightarrow cos θ = \frac{1}{2}

The lost

\pm

. Correctly

cos θ = \pm \frac{1}{2}

; you must use the quadrant (figure s03) to pick the sign, because both

+ \frac{1}{2}

and

- \frac{1}{2}

square to

\frac{1}{4}

1 + tan^{2} 90° = sec^{2} 90°

, so

\infty = \infty

, fine."

You can't plug in

90°

at all.

tan 90°

and

sec 90°

are undefined (division by

cos 90° = 0

). Note the nuance: as

θ \to 90°

both

tan^{2} θ

and

sec^{2} θ

grow without bound, and their difference stays

1

the whole way — but a limit that diverges is not a value, so "

\infty = \infty

" is not an equation you can write. The identity is a statement about numbers, and at

90°

there simply are no numbers to compare.

sin^{2} θ = sin (θ^{2})

, so

sin^{2} 30° = sin 900°

Confusing the exponent's target.

sin^{2} θ \equiv (sin θ)^{2}

(square the output), which is nothing like

sin (θ^{2})

(square the angle first).

"To simplify

1 - cos^{2} θ

, write it as

1 - cos θ

Same square-root-of-a-difference trap.

1 - cos^{2} θ = ∣ sin θ ∣

(from the master identity), not

1 - cos θ

"Divide

sin^{2} + cos^{2} = 1

cos^{2}

to get

tan^{2} + 1 = csc^{2}

Wrong reciprocal on the right.

\frac{1}{cos ^{2} θ} = sec^{2} θ

, so the result is

tan^{2} θ + 1 = sec^{2} θ

;

csc

comes from dividing by

sin^{2}

sin^{2} θ + cos^{2} θ = 1

, so

2 sin θ cos θ = 1

as well."

Invented a false consequence.

2 sin θ cos θ = sin 2 θ

(a double-angle formula), which varies between

- 1

and

1

— it is not fixed at

1

"Since

sec^{2} θ - tan^{2} θ = 1

, then

sec θ - tan θ = 1

too."

Un-squaring a difference term by term.

sec θ - tan θ

is generally not

1

; only the squared combination collapses to

1

Why questions

Short "why does the maths behave this way" prompts.

Why does dividing the master identity by

cos^{2} θ

specifically produce the tan–sec family?

Because

\frac{sin θ}{cos θ} = tan θ

and

\frac{1}{cos θ} = sec θ

— dividing by

cos^{2}

turns each term into exactly those functions squared.

Why must we attach

cos θ \neq = 0

to the identity

1 + tan^{2} θ = sec^{2} θ

Because both

tan = sin / cos

and

sec = 1/ cos

divide by

cos θ

; where

cos θ = 0

that division is illegal, so the identity's terms don't exist.

Why does squaring in Pythagoras let the identity survive in every quadrant?

The legs of the triangle are the lengths

∣ cos θ ∣

and

∣ sin θ ∣

(figure s01); squaring removes the absolute-value signs, so negative coordinates give the same squared result.

Why is

sec^{2} θ \geq 1

but

sin^{2} θ \leq 1

sec θ = 1/ cos θ

and

∣ cos θ ∣ \leq 1

makes

∣ sec θ ∣ \geq 1

; meanwhile

∣ sin θ ∣ \leq 1

caps

sin^{2} θ

1

Why do we say the two "child" identities carry no new information?

They are the same equation

sin^{2} + cos^{2} = 1

merely divided by

cos^{2}

sin^{2}

; no independent fact is added — the geometry is identical.

Why does the quadrant, not the identity, decide the sign of

cos θ

The identity only fixes

cos^{2} θ

; two angles with the same

sin

can sit in different quadrants and have opposite-signed cosines (figure s03), so extra info (the quadrant) is required. See Solving trig equations.

Why is "

sin θ = tan θ cos θ

" always safe but "

tan θ = sin θ / cos θ

" is not?

The first is a multiplication and never divides; the second, which is the definition of

tan θ

, divides by

cos θ

and so fails where

cos θ = 0

Edge cases

The degenerate and boundary inputs the topic quietly hides.

θ = 0°

, what happens to

1 + cot^{2} θ = csc^{2} θ

It breaks:

sin 0° = 0

makes

cot 0°

and

csc 0°

undefined (division by

0

), so the identity does not apply there.

θ = 90°

, is

sin^{2} θ + cos^{2} θ = 1

still valid?

Yes.

sin 90° = 1, cos 90° = 0

give

1 + 0 = 1

. The master identity never divides, so it survives every angle including the axes.

For

θ = 180°

: does

1 + tan^{2} θ = sec^{2} θ

hold?

Yes. Here

cos 180° = - 1 \neq = 0

, so

tan 180° = 0

sec 180° = - 1

, and

1 + 0 = (- 1)^{2} = 1

. ✓

What is the smallest value

sec^{2} θ

can take, and where?

Its minimum is

1

, reached whenever

tan θ = 0

(i.e.

θ = 0°, 180°, \dots

), because

sec^{2} = 1 + tan^{2}

and

tan^{2} \geq 0

Can

sin^{2} θ

and

cos^{2} θ

ever be equal, and what is each then?

Yes, when

θ = 45°, 135°, \dots

; then each equals

\frac{1}{2}

, since they must still sum to

1

For a huge angle like

θ = 1000°

, does the master identity still hold?

Yes. Sine and cosine repeat every

360°

, and the identity holds for all real

θ

, so periodicity guarantees it stays true.

Connections

Unit Circle definition of sin and cos
Pythagoras Theorem
Reciprocal and Quotient trig identities
Trig identity proofs — strategy
Double angle formulas
Solving trig equations