Unit circle definition of trig functions — all 6 trig functions for any angle: Question bank

First, three pictures we lean on

Before the traps, let us pin down the three visual ideas the questions keep referencing, so nothing is used before it is drawn.

Figure — Unit circle definition of trig functions — all 6 trig functions for any angle

True or false — justify

Every answer below must include the reason, not just T or F.

cos θ

can be equal to

2

False.

cos θ = x

is a coordinate on a circle of radius

1

, so

- 1 \leq x \leq 1

. A point with

x = 2

is off the circle entirely.

tan θ

can be equal to

2

True.

tan θ = y / x

is a ratio of two coordinates, not a coordinate itself, so it is not capped at

1

— near the vertical axis

x

is tiny and the ratio grows without bound.

sec θ = 0.5

is possible for some real

θ

False.

sec θ = 1/ x

and

∣ x ∣ \leq 1

, so

∣ sec θ ∣ \geq 1

always. A secant strictly between

- 1

and

1

never happens; same for

csc

sin θ > 0

then

θ

must be in Q1 or Q2.

True.

sin θ = y

, and

y > 0

means the landing point is in the upper half of the circle, which is exactly quadrants I and II.

sin θ + cos θ

can exceed

1

True. Although each is at most

1

, they need not peak together. At

θ = 45° = \frac{π}{4}

both equal

\frac{2}{2} \approx 0.707

, and their sum is

2 \approx 1.414 > 1

sin^{2} θ + cos^{2} θ = 1

only holds for acute angles.

False. It holds for every angle, because it is just the circle equation

x^{2} + y^{2} = 1

rewritten — the point always sits on the unit circle no matter how far you rotate.

cos (- θ) = cos θ

for all

θ

True. Rotating clockwise instead of counterclockwise mirrors the point across the x-axis, which keeps

x

unchanged. Cosine is even.

sin (- θ) = sin θ

for all

θ

False. Mirroring across the x-axis flips the y-coordinate, so

sin (- θ) = - sin θ

. Sine is odd, not even.

tan θ

is undefined at the same angles as

csc θ

False.

tan

and

sec

break when

x = cos θ = 0

(

90°, 270°, \dots

, i.e.

\frac{π}{2}, \frac{3 π}{2}, \dots

);

csc

and

cot

break when

y = sin θ = 0

(

0°, 180°, \dots

, i.e.

0, π, \dots

). They fail at different places.

Adding

720°

(

= 4 π

) to an angle changes none of its six trig values.

True.

720° = 2 \times 360°

is two full loops back to the identical point, and every trig value depends only on that point (periodicity).

Spot the error

Each line contains a flawed statement. Reveal the fix.

cos 150° > 0

because the triangle I drew has positive side lengths."

Wrong: side lengths are positive but coordinates carry signs. In Q2 the point sits left of the axis, so

x = cos 150° < 0

; the answer is

- \frac{3}{2}

sin θ

is the x-coordinate because 'sine' sounds vertical... wait, horizontal."

Wrong association.

sin θ = y

(vertical),

cos θ = x

(horizontal). Anchor it: at

θ = 0

you are at

(1, 0)

, and

cos 0° = 1

— cosine grabbed the

x

tan 90° = \frac{1}{0}

, which is a huge number, so

tan 90° \approx 1 0^{9}

Wrong:

\frac{1}{0}

is undefined, not "very large". As

θ \to 90°

the value blows up (its limit is infinite), but at exactly

90° = \frac{π}{2}

it has no value at all.

sin 210° = sin 30° = \frac{1}{2}

since

210° = 180° + 30°

Wrong: the reference angle

30°

only fixes the magnitude.

210°

is in Q3 where

y < 0

, so

sin 210° = - \frac{1}{2}

cos 400°

is undefined because

400°

is more than a full turn."

Wrong: angles past

360°

(

2 π

) are fine — just loop back.

cos 400° = cos (400° - 360°) = cos 40°

, a perfectly ordinary value.

sec θ = \frac{1}{s i n θ}

because sec pairs with sine alphabetically."

Wrong:

sec θ = \frac{1}{c o s θ} = \frac{1}{x}

. It is

csc θ

that equals

\frac{1}{s i n θ}

. Memorize by reciprocal, not by name.

"Since

tan

repeats every

360°

tan (θ + 180°) \neq = tan θ

Wrong:

tan

actually repeats every

180°

(

π

). Adding

180°

sends

(x, y) \to (- x, - y)

, so

\frac{- y}{- x} = \frac{y}{x}

— the ratio is unchanged.

Why questions

Why is

sin^{2} θ + cos^{2} θ = 1

never "

= 1

approximately" but exactly?

Because it is not a numerical coincidence — it is the definition of the unit circle

x^{2} + y^{2} = 1

with

x = cos θ, y = sin θ

substituted in. Every landing point is on the circle by construction.

Why can

tan θ

be any real number while

sin θ

cannot?

sin θ = y

is a bounded coordinate (

∣ y ∣ \leq 1

), but

tan θ = y / x

is a ratio whose denominator shrinks toward

0

near the vertical axis, letting the value run off to

\pm \infty

Why does the naive right triangle fail for

θ = 150° = \frac{5 π}{6}

A right triangle only makes acute angles

0°

–

90°

, and its sides are lengths (always positive). The unit circle keeps rotating past

90°

and lets the coordinates go negative, which the triangle alone cannot express.

Why is cosine even but sine odd, geometrically?

Negating the angle reflects the landing point across the x-axis. Reflection across the x-axis leaves

x

alone (so

cos

even) but flips

y

(so

sin

odd).

Why do

sec

and

csc

never take values strictly between

- 1

and

1

They are reciprocals of

cos

and

sin

, whose magnitudes are at most

1

. The reciprocal of a number with magnitude

\leq 1

has magnitude

\geq 1

, so

∣ sec ∣, ∣ csc ∣ \geq 1

Why is

tan θ = \frac{sin θ}{cos θ}

rather than

\frac{cos θ}{sin θ}

On the right triangle inside the circle,

tan = \frac{opposite}{adjacent} = \frac{y}{x} = \frac{s i n θ}{c o s θ}

. The reversed ratio

x / y

is a different function, cotangent.

Why does knowing the quadrant plus the reference angle pin down a value completely?

The reference angle gives the magnitude of

x

and

y

; the quadrant gives the sign of each. Magnitude plus sign uniquely determines the coordinate, hence the trig value.

Edge cases

θ = 0°

(

0

rad), which of the six functions are undefined and why?

The point is

(1, 0)

, so

y = 0

cot θ = x / y

and

csc θ = 1/ y

divide by zero, so both are undefined; the other four are fine.

θ = 90°

(

\frac{π}{2}

), which functions blow up (limit runs to infinity)?

The point is

(0, 1)

, so

x = 0

. Both

tan θ = y / x

and

sec θ = 1/ x

are undefined there, and their one-sided limits diverge to

\pm \infty

— that is the precise sense of "blow up". Meanwhile

sin 90° = 1, cos 90° = 0, cot 90° = 0, csc 90° = 1

What is

tan 180°

(

tan π

), and is it undefined?

The point is

(- 1, 0)

, so

y = 0, x = - 1

tan 180° = y / x = 0/ (- 1) = 0

— perfectly defined and equal to zero, not undefined and not blowing up.

cos θ = - 1

achievable, and where?

Yes, at

θ = 180° = π

, where the landing point is

(- 1, 0)

. This is the minimum possible value of cosine since

x \geq - 1

on the circle.

Can

sin θ

and

cos θ

both be

0

at the same angle?

No. That would require the point

(0, 0)

, the origin, which is not on the unit circle (

0^{2} + 0^{2} = 0 \neq = 1

). At least one coordinate is always nonzero.

What happens to

sec θ

θ

approaches

90°

(

\frac{π}{2}

) from below versus above?

From below (

θ \to 90 °^{-}

x \to 0^{+}

sec θ \to + \infty

; from above (

θ \to 90 °^{+}

x \to 0^{-}

sec θ \to - \infty

. The two sides blow up to opposite infinities — a genuine vertical asymptote, not a mere gap.

Is there an angle where

tan θ = cot θ

Yes: whenever

y / x = x / y

, i.e.

x^{2} = y^{2}

, at

θ = 45°, 135°, \dots

(

\frac{π}{4}, \frac{3 π}{4}, \dots

). There

tan θ = \pm 1 = cot θ

For a full loop, how many distinct angles in

[0°, 360°)

(i.e.

[0, 2 π)

) give

sin θ = \frac{1}{2}

Two:

30° = \frac{π}{6}

(Q1) and

150° = \frac{5 π}{6}

(Q2), the two places where

y = \frac{1}{2}

meets the circle. Sine hitting a value in

(0, 1)

always meets the upper circle twice.

Connections

Use these to see where each trap really lives in the wider theory:

Right-triangle trigonometry (SOH-CAH-TOA) — the acute-angle special case; the "positive side lengths" error comes from mistaking a triangle side for a signed coordinate.
Reference angles — the tool that gives magnitude; almost every "spot the error" item confuses magnitude with the final signed value.
Pythagorean identities — the "exactly $1$ " true/false item is really the circle equation, and all three identities descend from it.
Even and odd functions — the even-cosine / odd-sine reflection is why the negative-angle traps work as they do.
Periodicity and $2\pi$ — behind the $400°$ and $720°$ ( $4 π$ ) items: full loops return to the same point.
Radian measure — the second language for every angle here; converting fluently kills "big angle" panic.
Graphs of trig functions — where "blow up" becomes visible as a vertical asymptote in the $tan$ and $sec$ graphs.