This deep-dive drills the parent note Co-function identities (index 3.1.11) into every case a problem can throw at you. We build each symbol from scratch, forecast before solving, and verify every number.
Before anything, the one rule this whole page rests on:
If any of those words feel shaky, revisit Right Triangle Trigonometry (SOH-CAH-TOA) and Complementary and Supplementary Angles first.
Here is every class of problem this topic can produce. The worked examples below each carry a cell tag so you can see the whole space is covered.
Cell
What makes it different
Example
A. Plain convert
Angle already looks like 90° − θ ; just fire the rule
Ex 1
B. Collapse a sum
Two different angles that are complements; combine with Pythagorean Identities
Ex 2
C. Ratio → 1
Numerator/denominator become identical after a swap
Ex 3
D. Radian form
2 π − θ instead of degrees
Ex 4
E. Sign / wrong-quadrant trap
Angle is > 90° , so it is not a complement — must reroute via Reference Angles and Supplementary Identities
Ex 5
F. Degenerate / boundary
θ = 0° or θ = 90° : check the rule still holds and nothing blows up
Ex 6
G. Chained / nested
Apply the identity twice, or inside another function
Ex 7
H. Real-world word problem
Sun-angle / ramp geometry, needs a figure
Ex 8
I. Exam twist (prove-it)
Prove an expression is constant using the identity + algebra
Ex 9
Everything below fills these nine cells.
sec 58° into a co-function.
Forecast: which co-function partners with sec , and what complement do you need? Guess before reading.
Step 1. Write 58° = 90° − 32° .
Why this step? The rule only fires when the angle is written as "90° minus something." That "something" is the complement 32° .
Step 2. Apply sec ( 90° − θ ) = csc θ with θ = 32° .
Why this step? sec 's co-function partner is csc (seC ↔ Csc). The complement's angle carries over.
Answer: sec 58° = csc 32° .
Verify: numerically sec 58° = 1/ cos 58° ≈ 1.8871 and csc 32° = 1/ sin 32° ≈ 1.8871 . Same value ✔.
cos 2 22° + cos 2 68° .
Forecast: 22° and 68° — what do they add to? Predict the final number.
Step 1. Notice 68° = 90° − 22° , so cos 68° = cos ( 90° − 22° ) = sin 22° .
Why this step? The two angles are complements, so the co-function rule turns two different angles into one angle — the trick that makes the sum collapse.
Step 2. Substitute: cos 2 22° + sin 2 22° .
Why this step? Now both terms share the single angle 22° , which is exactly the shape Pythagorean Identities wants.
Step 3. cos 2 22° + sin 2 22° = 1 .
Answer: 1 .
Verify: cos 2 22° + cos 2 68° ≈ 0.8597 + 0.1403 = 1.0000 ✔.
cot 75° tan 15° .
Forecast: is this bigger than, less than, or exactly 1 ? Guess first.
Step 1. 75° = 90° − 15° , so cot 75° = cot ( 90° − 15° ) = tan 15° .
Why this step? cot ( 90° − θ ) = tan θ (coT ↔ Tan). This makes the denominator identical to the numerator.
Step 2. tan 15° tan 15° = 1 .
Why this step? Any nonzero quantity over itself is 1 ; and tan 15° = 0 , so we're safe to divide.
Answer: 1 .
Verify: tan 15° ≈ 0.2679 , cot 75° = 1/ tan 75° ≈ 0.2679 ; ratio ≈ 1 ✔.
sin ( 2 π − 7 2 π ) without the 2 π .
Forecast: in radians, 2 π is the complement. Which function does sin turn into?
Step 1. Match the pattern sin ( 2 π − θ ) = cos θ with θ = 7 2 π .
Why this step? Radians are just another way to measure angles (see Radian Measure ); 2 π rad = 90° , so the same co-function rule applies — only the units of the angle changed.
Answer: cos 7 2 π .
Verify: 2 π − 7 2 π = 14 7 π − 4 π = 14 3 π . Then sin 14 3 π ≈ 0.6235 and cos 7 2 π ≈ 0.6235 ✔.
Worked example A student writes
sin 130° = cos ( 90° − 130° ) = cos ( − 40° ) . Is this a valid co-function use? Simplify sin 130° correctly.
Forecast: does the co-function rule even apply when the angle is bigger than 90° ? Guess yes/no.
Step 1. Check the requirement: co-function pairing needs two angles that sum to exactly 90° (both acute). Here 130° > 90° , so it has no acute complement — the identity is the wrong tool.
Why this step? The whole geometric picture came from a right triangle whose two other angles are acute. Feed in 130° and there is no such triangle; blindly writing 90° − 130° = − 40° is meaningless as a triangle angle.
Step 2. Use the correct tool instead: 130° = 180° − 50° , a supplementary relationship (see Reference Angles and Supplementary Identities ). For sine, sin ( 180° − θ ) = sin θ .
Why this step? 130° lives in Quadrant II where sine is positive; the reference angle is 50° , and the supplementary rule (not the co-function rule) governs this case.
Step 3. sin 130° = sin 50° . If you want a co-function form of that, sin 50° = cos ( 90° − 50° ) = cos 40° .
Why this step? Now 50° is acute, so the co-function rule is legal and turns it into cos 40° .
Answer: sin 130° = sin 50° = cos 40° (and it is positive , ≈ 0.766 ).
Verify: sin 130° ≈ 0.7660 , cos 40° ≈ 0.7660 ✔. The naive "cos ( − 40° ) " happens to also equal 0.766 here only because cos is even and cos ( − 40° ) = cos 40° — a coincidence of sign, not a licence to skip the check.
Common mistake Why the shortcut is dangerous
Applying 90° − θ to a non-acute angle sometimes gives the right number (as above, thanks to cos being even) and sometimes gives nonsense. Never rely on the co-function rule outside acute angles — route obtuse angles through supplementary/reference identities first.
Worked example Test the co-function rule at the edges
θ = 0° and θ = 90° for tan ( 90° − θ ) = cot θ . Where (if anywhere) does something blow up?
Forecast: cot 0° — is that a finite number? Predict what breaks.
Step 1. At θ = 0° : left side tan ( 90° − 0° ) = tan 90° , which is undefined (vertical, division by cos 90° = 0 ). Right side cot 0° = sin 0° cos 0° = 0 1 , also undefined .
Why this step? The identity is an equation ; for it to be honest, both sides must be undefined at the same inputs. They are — the identity is consistent even at its breakdown point.
Step 2. At θ = 90° : left side tan ( 90° − 90° ) = tan 0° = 0 . Right side cot 90° = sin 90° cos 90° = 1 0 = 0 . Both 0 ✔.
Why this step? This is the opposite extreme; confirming both sides equal 0 shows the rule holds right up to the boundary.
Step 3. Sanity on the safe interior: at θ = 30° , tan 60° = 3 ≈ 1.732 and cot 30° = 3 ✔.
Answer: The identity holds for all θ where both functions are defined; it is jointly undefined at θ = 0° (where cot 0° and tan 90° both explode). No contradiction.
Verify: tan 60° = 3 and cot 30° = 3 , so their difference is 0 ✔.
cos ( 90° − ( 90° − θ ) ) .
Forecast: apply the rule once, then again. Where do you land — back at θ ?
Step 1. Innermost first: let u = 90° − θ . The expression is cos ( 90° − u ) .
Why this step? Naming the inner complement u keeps the pattern cos ( 90° − u ) visible so we don't misapply.
Step 2. cos ( 90° − u ) = sin u = sin ( 90° − θ ) .
Why this step? One co-function swap: cos ( 90° − u ) = sin u . Then put u back.
Step 3. sin ( 90° − θ ) = cos θ .
Why this step? A second co-function swap collapses sin ( 90° − θ ) straight to cos θ .
Answer: cos θ . (Two complements undo each other, but each swap flips the function — cos → sin → cos lands you back on cosine, as it must, since 90° − ( 90° − θ ) = θ .)
Verify: at θ = 25° : cos ( 90 − ( 90 − 25 )) = cos 25° ≈ 0.9063 ✔.
Worked example A straight ramp leans against a wall. The angle the ramp makes with the
ground is θ = 35° . What angle does it make with the wall , and how does the ramp's "sine seen from the ground" relate to its "cosine seen from the wall"?
Forecast: the ground and the wall meet at a right angle. What must the two ramp-angles add to?
Step 1. Ground ⟂ wall means the corner is 90° . The ramp, ground, and wall form a right triangle, so the angle at the wall is ϕ = 90° − 35° = 55° (look at the figure: the two coloured angles fill the square corner).
Why this step? The two non-right angles of any right triangle are complementary — the geometric backbone of every co-function identity.
Step 2. From the ground's viewpoint, the ramp's height = opposite, so sin 35° = ramp height . From the wall's viewpoint, that same height is adjacent , so cos 55° = ramp height .
Why this step? One physical side is "opposite" to one observer and "adjacent" to the other — this is why sin 35° = cos 55° .
Step 3. Therefore sin 35° = cos ( 90° − 35° ) = cos 55° .
Answer: wall angle = 55° , and sin 35° = cos 55° .
Verify: sin 35° ≈ 0.5736 and cos 55° ≈ 0.5736 ✔; angles 35° + 55° = 90° ✔.
Worked example Prove that
sin θ ⋅ sec ( 90° − θ ) = 1 for every θ where both are defined.
Forecast: which co-function does sec ( 90° − θ ) become? Then what cancels?
Step 1. sec ( 90° − θ ) = csc θ .
Why this step? seC ↔ Csc under complement; this converts the unknown-looking term into something built from sin θ .
Step 2. csc θ = sin θ 1 .
Why this step? csc is by definition the reciprocal of sin — now the product will telescope.
Step 3. sin θ ⋅ sin θ 1 = 1 (valid whenever sin θ = 0 ).
Why this step? A quantity times its reciprocal is 1 ; we flag sin θ = 0 as the domain condition.
Answer: sin θ ⋅ sec ( 90° − θ ) = 1 . ∎
Verify: at θ = 50° : sin 50° ⋅ sec 40° = 0.7660 × 1.3054 ≈ 1.0000 ✔.
Recall Self-test: match each problem to its matrix cell
sin 80° cos 10° is which cell? ::: Cell C (ratio → 1).
sin 145° is which cell, and which tool? ::: Cell E — supplementary/reference, NOT co-function.
tan 90° vs cot 0° is which cell? ::: Cell F (degenerate boundary; both undefined).
cos 2 18° + cos 2 72° is which cell? ::: Cell B (collapse a sum) → equals 1 .