Complementary angle relationships — sin(90−θ) = cos θ etc.: Question bank

This page is a rapid-fire self-test. Every line is a Question ::: Answer reveal. Read the question, say your answer out loud with a reason, then reveal. The answers here always explain the why — bare "yes/no" teaches nothing.

Three figures do the heavy lifting on this page — glance at them before each section.

Figure 1 — the swap in the triangle. This is the single picture behind every question below.

Recall The one idea behind every question here

In a right triangle the two acute angles are $θ$ and $90° - θ$ . What is opposite one angle is adjacent to the other (Figure 1: the orange side is opposite $θ$ but adjacent to $90° - θ$ ). That single swap of roles is the whole family of cofunction identities: $sin (90° - θ) = cos θ$ , $tan (90° - θ) = cot θ$ , $sec (90° - θ) = csc θ$ . See Right Triangle Trigonometry and Unit Circle for the two viewpoints.

Figure 2 — the unit-circle reflection. The identity is really a mirror across the line $y = x$ : reflecting the point at angle $θ$ gives the point at angle $90° - θ$ , swapping the $x$ and $y$ coordinates — i.e. swapping cos and sin.

Figure 3 — the two graphs are one shift apart. Cosine is sine slid left by $90°$ ; wherever the curves cross is where a function equals its cofunction.

Two words you must never mix up:

Complementary = sum to 90° (a corner).
Supplementary = sum to 180° (a straight line).

True or false — justify

True or false:

sin (90° - θ) = cos θ

holds for every real

θ

, not just acute ones.

True — on the unit circle (Figure 2) reflecting the angle-

θ

point across

y = x

always sends

(cos θ, sin θ)

(sin θ, cos θ)

, whatever quadrant

θ

sits in; the triangle only motivates the acute case.

True or false: complementary angles add to 180°.

False — that is supplementary. Complementary angles add to 90°, the "C for Corner" case where the cofunction swaps actually work.

True or false:

sin (180° - θ) = cos θ

False — that swap only happens at 90°. For supplementary angles

sin (180° - θ) = sin θ

(same function, not the co-function).

True or false:

cos (90° - θ) = sin θ

is just the previous identity read backwards.

True — replacing

θ

90° - θ

sin (90° - θ) = cos θ

turns it into

cos (90° - θ) = sin θ

, so the two are the same statement viewed from the other acute angle.

True or false: if

sin A = cos B

then

A

and

B

must be complementary.

False in general —

A + B = 90°

is one family, but because sine repeats every

360°

and has a mirror at

180°

, whole extra families like

A = 180° + B^{'}

also solve it; complementary is guaranteed only when both angles are restricted to acute values.

True or false:

tan (90° - θ) = cot θ

can be derived purely from the sine and cosine cofunction identities.

True —

tan (90° - θ) = \frac{sin ( 90° - θ )}{cos ( 90° - θ )} = \frac{cos θ}{sin θ} = cot θ

, no new geometry needed.

True or false:

sec (90° - θ) = csc θ

is undefined exactly where

csc θ

is undefined.

True — both blow up when

sin θ = 0

(i.e.

θ = 0°, 180°, \dots

), which is the same set of bad points, so the two sides fail together and remain equal wherever they exist.

True or false: in radians the identity becomes

sin (\frac{π}{2} - θ) = cos θ

True —

90°

and

\frac{π}{2}

are the same angle, so you just substitute; the geometry is unchanged.

True or false:

sin (90° - θ) = cos θ

means

sin

and

cos

are the same function.

False — they are shifts of each other (Figure 3), not equal; the cosine graph is the sine graph slid left by

90°

, and they agree only where the two curves cross.

Spot the error

Find the error: "

sin (90° - \frac{π}{6}) = cos \frac{π}{6} = \frac{3}{2}

Units are mixed — you cannot subtract a radian measure

\frac{π}{6}

from a degree measure

90°

. Write either

sin (90° - 30°)

sin (\frac{π}{2} - \frac{π}{6})

, never one of each.

Find the error: "

sin (90° - 50°) = sin (40°)

, so the cofunction identity gives

sin 40°

You did arithmetic, not an identity.

90° - 50°

40°

, but the point of the identity is

sin (90° - 50°) = cos 50°

— you're meant to swap to the co-function, giving a form that often matches other terms.

Find the error: "

cos (90° - θ) = \frac{1}{s i n ( 90° - θ )}

That confuses the cofunction rule with the reciprocal rule.

cos (90° - θ) = sin θ

; the reciprocal of

sin (90° - θ)

csc (90° - θ) = sec θ

, a different quantity.

Find the error: "

tan (90° - θ) = \frac{1}{t a n ( 90° - θ )}

because tan and cot are reciprocals."

Nonsense — that equation says a number equals its own reciprocal (only true at

\pm 1

). The correct statement is

tan (90° - θ) = cot θ = \frac{1}{t a n θ}

, with

θ

not

90° - θ

under the fraction.

Find the error: "Since

sin (3 x) = cos (2 x)

, set

3 x = 2 x

x = 0

You matched functions without converting first. Rewrite

cos (2 x) = sin (90° - 2 x)

, then

3 x = 90° - 2 x

, giving

x = 18°

— the two arguments are equal only after both sides are sines.

Find the error: "

sec (90° - A) \cdot sin A = cos A

sec (90° - A) = csc A = \frac{1}{s i n A}

, so the product is

\frac{1}{s i n A} \cdot sin A = 1

, not

cos A

. The result is the constant

1

for all valid

A

Find the error: "

cot (90° - θ) = cot θ

by symmetry."

The cofunction of cotangent is tangent, not itself:

cot (90° - θ) = tan θ

. It equals itself only where

tan θ = cot θ

(at

θ = 45°

Why questions

Why does subtracting from

90°

specifically cause the swap, and not

80°

100°

Because the two acute angles of a right triangle always sum to exactly

90°

; only at

90°

does "the other acute angle" exist, making one angle's opposite side become the other's adjacent side (Figure 1).

Why is

sin (90° - θ) = cos θ

called a cofunction identity?

"Co-" is short for complement: cosine literally means "sine of the complementary angle," and each pair (sin/cos, tan/cot, sec/csc) is a function and its complement-partner.

Why do we bother converting

sin (90° - θ)

cos θ

instead of just evaluating?

In equation-solving and proofs you need the form to match other terms;

cos 50°

can cancel or combine with another

cos

, whereas the number

0.643

hides that structure. See Solving Trigonometric Equations.

Why does

sin (3 x) = sin (90° - 2 x)

let us "cancel the sines" — and is that the whole story?

Only on the acute interval

0° < x < 90°

, where sine is one-to-one (Figure 3, rising portion), do equal outputs force equal inputs, giving

3 x = 90° - 2 x

, i.e.

x = 18°

. Over the full range sine also repeats, so the general solution is

3 x = 90° - 2 x + 360° k

3 x = 180° - (90° - 2 x) + 360° k

for integer

k

; the acute restriction just keeps the single answer.

Why is

tan (90° - θ)

related to

cot θ

rather than

tan θ

Tangent is opposite-over-adjacent; from the complementary angle those two sides swap places (Figure 1), which flips the ratio upside-down — and an upside-down tangent is exactly the cotangent.

Why can we get the sec/csc and cot identities "for free" once we have sin/cos/tan?

Because sec, csc, cot are just reciprocals of cos, sin, tan; taking

1/ (\cdot)

of both sides of a known identity preserves equality, turning each cofunction pair into its reciprocal cofunction pair.

Why does the cosine graph look like the sine graph shifted, and how is that the same identity?

Start from

cos θ = sin (90° - θ)

; sine is odd, so

sin (90° - θ) = sin (- (θ - 90°)) = - sin (θ - 90°)

, and one more step gives

cos θ = sin (θ + 90°)

— an explicit left shift of $90°$ (Figure 3). The algebraic identity and the graph shift are two faces of one fact (see Even and Odd Functions).

Why do the identities never sprout a minus sign, even though sine and cosine go negative in quadrants II–IV?

Because

θ

and

90° - θ

always land in matching sign situations: as

θ

sweeps a full turn, the reflection across

y = x

(Figure 2) carries the sign along with the coordinate swap, so

sin (90° - θ) = cos θ

holds with a plus sign in every quadrant — no sign chart needed for the

90° - θ

form (contrast this with

sin (90° + θ) = + cos θ

but

cos (90° + θ) = - sin θ

, where signs do appear).

Edge cases

What is

sin (90° - 0°)

, and does the identity survive at

θ = 0

sin 90° = 1 = cos 0°

, so it holds cleanly;

θ = 0

is a degenerate triangle (one angle vanishes) yet the unit-circle identity still gives the right value.

What happens to

tan (90° - θ) = cot θ

when

θ = 0

cot 0°

is undefined (division by

sin 0° = 0

), and indeed

tan (90°)

is also undefined — both sides fail simultaneously, so the identity is respected as a "both-undefined" match.

Does

sin (90° - θ) = cos θ

still work for a negative angle like

θ = - 30°

Yes —

sin (120°) = \frac{3}{2} = cos (- 30°)

; the unit-circle proof never assumed

θ

was positive, so signs are handled automatically.

Does the identity behave the same for an obtuse input like

θ = 120°

(a quadrant-II angle)?

Yes —

sin (90° - 120°) = sin (- 30°) = - \frac{1}{2}

and

cos 120° = - \frac{1}{2}

agree; the reflection across

y = x

(Figure 2) works in every quadrant, so no extra sign correction is ever needed for the

90° - θ

pattern.

θ = 45°

, what is special about the cofunction identities?

45°

is its own complement (

90° - 45° = 45°

), so every identity collapses to a self-statement:

sin 45° = cos 45°

tan 45° = cot 45° = 1

— the symmetry point where function and co-function meet (the crossing in Figure 3).

What does

sec (90° - 90°)

evaluate to, and is that consistent with

csc 90°

sec 0° = 1

and

csc 90° = 1

, so both give

1

— consistent; the edge input

θ = 90°

(complement

0°

) is a valid, well-defined case here.

Is there any acute angle where

sin θ = cos θ

but

θ

is not its own complement?

No — for acute angles

sin θ = cos θ

forces

θ = 90° - θ

, i.e.

θ = 45°

; the equality of a function and its cofunction pins you to the self-complementary angle.