Noesis
3.1.11 · D5Advanced Trigonometry

Question bank — Co-function identities

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Toolbox refresher (read before the traps)

Two things trip students on this page more than the identities themselves: the reciprocal functions and what the ratios even mean. The figure below is the one triangle everything on this page is built on.

Figure — Co-function identities

WHY — the geometric derivation

Look at the figure. The right angle sits at ; the two acute angles are at and at . Since a triangle's angles sum to and one is ,

Now watch which side is which for each angle (the figure prints both labels on each leg):

  • The pink leg is opposite — but it touches , so it is adjacent to .
  • The blue leg is adjacent to — but it faces , so it is opposite .

So the same physical side is "opposite" from 's view and "adjacent" from 's view. Feed that into SOH-CAH-TOA: Both equal , and , therefore No formula was invented — the swap of the word oppositeadjacent, forced by the two viewpoints, is the whole reason the function name flips to its co-partner.

True or false — justify

Every line is Statement ::: Verdict + one-sentence reason. Decide true or false and why before revealing.

holds for every real , not just acute ones.
True — the triangle proof covers acute , and the angle-difference derivation ( from ) extends it to all real .
.
False — the complement rule swaps the function, so , not .
even when .
True — at both sides equal , which is exactly why ; the identity does not break at the "symmetric" angle.
because sine distributes over subtraction.
False — sine is a function, not a multiplier, so it does not distribute; the correct value is .
For to work, the two angles must add to .
True — makes them complementary, which is the only condition the co-function rule needs.
is a valid co-function move.
True but sneaky — and , so it does give the right number; still, this is safer handled as a supplementary/reference-angle problem since .
.
True — cosecant's co-partner is secant, and .
The co-function identities require to be measured in degrees.
False — they work in radians too; just replace with .
can be derived without any new triangle.
True — it comes from taking the reciprocal of the already-proved , so no fresh geometry is needed.
In radians, .
True — and , which equals ; the identity works identically once becomes .

Spot the error

Each line states a flawed line of work; the reveal names the mistake and fixes it.

"."
Tangent does not distribute over subtraction, and is undefined anyway; the correct identity is .
" since the angle is nearly the same."
The function must change to its co-partner; the correct result is . "Looks close" is not a rule.
" because both are co-functions."
They are equal only if the angles are complements; here both angles are , and , so the equality fails — actually .
" so I'll write and stop."
Backwards setup: the identity is , and here the argument is , not . Do it properly: , and (supplementary rule, since sits in Quadrant II where sine is positive), so .
"."
The swap is being ignored: cotangent's co-partner is tangent, so .
" by the reciprocal rule."
The reciprocal step is applied to the wrong function: , not .
" because both sines are near ."
Estimating instead of using the identity: , so the sum is by the Pythagorean identity.
"Since , then too."
is the supplementary case, not the complementary one; the correct rule is .

Why questions

does jumping to the complement change the function name at all?
Because from the complement's viewpoint the opposite and adjacent sides are swapped — the pink leg is opposite but adjacent to — so the ratio that was "sine" (opp/hyp) for one angle becomes "cosine" (adj/hyp) for the other, exactly as the double-labelled legs show in the figure.
must the two acute angles of a right triangle be complementary?
All three angles sum to and one is , leaving exactly shared between the other two — this is the geometric backbone from Complementary and Supplementary Angles.
can we get all six co-function identities from just the sine and cosine ones?
Because tan, cot, sec, csc are all built from sine and cosine — e.g. derive , substituting the proven into the definition , with no new geometry.
is and not ?
We substitute and into , giving with no sign flip — both replacements are positive swaps.
does dividing Pythagoras by give exactly ?
Because and , so becomes after dividing every term by .
do co-function identities count as a special case of the angle-difference formulas?
Setting the first angle to in (and its sine twin) makes , , collapsing the formula straight into , so they are one specific slice of Angle Sum and Difference Formulas.
does "co-" appear in cosine, cotangent, and cosecant but not in sine, tangent, secant?
Because each "co-" function is literally the plain function of the complement: cosine complement's sine, cotangent complement's tangent, cosecant complement's secant.

Edge cases

, what does say, and does it hold?
It gives , i.e. — true, and it corresponds to a fully degenerate triangle where one acute angle has shrunk to zero.
, is still meaningful?
Yes — the left side is and the right side is , which is perfectly defined (cot is only undefined where , i.e. at /), so the identity holds cleanly here.
\cot\dfrac{\pi}{5}\theta=\dfrac{\pi}{5}\dfrac{\pi}{2}90°\text{What happens to }\sec(90°-\theta)=\csc\theta \text{ at }\theta=0°?
The right side is undefined (division by zero), and the left side is also undefined — both blow up together, so the identity is consistent everywhere it exists.
\sin(90°-210°)=\sin(-120°)=-\tfrac{\sqrt3}{2}\cos210°=-\tfrac{\sqrt3}{2}\text{Does the identity survive negative angles, e.g. }\theta=-30°?
Yes — as a true identity it holds for all reals: , matching .
\theta=45°90°-\theta=45°\sin45°=\cos45°\text{Can a co-function identity apply when one "angle" exceeds }90°?
Algebraically yes (the angle-difference proof needs no triangle), but geometrically no — if then is negative, so no genuine right triangle exists; read the sign of the result from its quadrant instead.

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