Chapter 11.4 — Exercise 11.3 — Complementary Angles
Trigonometric ratios of complementary angles. This is Lesson 4 of 5 in Chapter 11: Trigonometry.
Two Angles That Always Add to 90°
Any acute angle in a right triangle has a built-in partner — the triangle's other acute angle, whatever's left over from 90° once the first one is subtracted. Exercise 11.3 connects every ratio of one to the matching ratio of the other, without ever needing to know either angle's actual numeric value.
Where the Six Relationships Come From
In right triangle ABC (right angle at B), if ∠A=θ then ∠C=90°−θ automatically, since the three angles must sum to 180°. Applying the ratio definitions to ∠C directly — using the exact same triangle, just measured from the other acute corner — gives every relationship at once.
sin(90°−θ) = cosθ cos(90°−θ) = sinθ tan(90°−θ) = cotθ
cosec(90°−θ) = secθ sec(90°−θ) = cosecθ cot(90°−θ) = tanθsinC=AB/AC by definition — but AB/AC is also exactly what cosA already meant, since AB is adjacent to A. That single overlap, one side playing "opposite" for one angle and "adjacent" for the other, is the entire reason every complementary-angle relationship holds. The identical overlap argument, run for BC and for AC, produces the other five relationships in the formula box above — none of them need a separate derivation from scratch.
Five Expressions That Collapse to Almost Nothing
| Expression | Rewritten | Result |
|---|---|---|
| tan36°/cot54° | cot54°/cot54° | 1 |
| cos12°−sin78° | sin78°−sin78° | 0 |
| cosec31°−sec59° | sec59°−sec59° | 0 |
| sin15°·sec75° | cos75°·(1/cos75°) | 1 |
| tan26°·tan64° | cot64°·tan64° | 1 |
Every row follows the identical single move: spot that the two angles inside the expression add to exactly 90°, rewrite one of them in terms of the other's complementary ratio, and watch two now-identical terms cancel or combine. None of these five answers depend on knowing the actual numeric value of any of the angles involved — 36°, 54°, 12°, 78°, and every other angle here stays completely unevaluated throughout.
Spotting the 90°-sum pairing is really the only genuine skill each row is testing — once that pairing is noticed, the actual algebra afterward is nothing more than substituting one ratio for its equal and simplifying. It's worth scanning any unfamiliar-looking trigonometric expression for a complementary pair before assuming it needs a longer identity or a numeric table lookup; a surprising number of seemingly complicated expressions turn out to collapse this same way.
Pairing Angles Across a Longer Expression
Two proof problems apply the identical cancellation trick, just spread across more than one pair of angles at once.
- tan48°·tan16°·tan42°·tan74° = 1: pair 48° with 42° (sum 90°) and 16° with 74° (sum 90°). tan48°=cot42° and tan16°=cot74°, so the product becomes (cot42°·tan42°)·(cot74°·tan74°) = 1×1 = 1.
- cos36°cos54° − sin36°sin54° = 0: since cos36°=sin54° and cos54°=sin36°, substituting turns the expression into sin54°sin36° − sin36°sin54° — the identical product subtracted from itself.
The second result is also cos(36°+54°)=cos90°=0 in disguise — the exact cosine addition formula tested numerically back in Exercise 11.2, here appearing as the reason a seemingly complicated expression collapses to zero.
The first result is worth a second look for how it scales the single-pair trick up to four angles at once. Rather than solving the whole product in one step, it splits cleanly into two independent complementary pairs — 48°/42° and 16°/74° — each pair resolving to exactly 1 on its own, before the two 1's are multiplied together at the very end. Recognising that a longer expression is really just two (or more) shorter, already-familiar patterns glued together is often the fastest route through it, rather than trying to simplify everything in one continuous pass.
Solving for an Unknown Angle
If tan2A=cot(A−18°), with 2A itself given as an acute angle, find A. Rewriting cot(A−18°) as tan(90°−(A−18°))=tan(108°−A) turns the equation into a statement about two equal tangents:
tan2A = tan(108°−A) ⟹ 2A = 108°−A ⟹ 3A = 108° ⟹ A = 36°Converting a cotangent into a tangent-of-a-complementary-angle is what makes this solvable at all — with one side as tan and the other as cot, there's no direct way to set the arguments equal, since a tangent and a cotangent are never the same value at the same angle in general.
Once both sides of the equation are expressed as the same ratio (tangent, in this case), setting the two angle arguments equal to each other is valid precisely because tangent is one-to-one over the acute-angle range — a fact already established back in Exercise 11.1 for cosine, and equally true here for tangent. Without that one-to-one property, "tan(x)=tan(y) implies x=y" wouldn't be a safe step to take at all.
A Two-Line General Proof
If tanA=cotB for acute angles A and B, prove A+B=90°. Rewrite cotB as tan(90°−B): tanA=tan(90°−B) forces A=90°−B directly, which rearranges immediately to A+B=90°. This is really the exact converse of the six relationships this whole exercise opened with — instead of starting from two known complementary angles and deriving a ratio equality, it starts from a ratio equality and derives that the two angles must have been complementary all along.
A related triangle fact follows the same shape: in △ABC, since A+B+C=180°, (A+B)/2=90°−C/2, and substituting into the tangent function gives tan[(A+B)/2] = cot(C/2) — a genuine relationship between a triangle's angles that holds for every triangle, not just special ones, since it's built from nothing but the angle-sum property every triangle already satisfies.
Both results in this section share the same underlying template: start from a fact about how two angles relate (A+B=90°, or A+B+C=180°), divide or rearrange it into a complementary-angle statement, then translate that statement into matching trigonometric ratios. The triangle-angle version is worth remembering specifically, since half-angle relationships like this one show up again in later trigonometry work well beyond this particular chapter.
Rewriting an Expression Inside a Smaller Angle Range
Express sin75°+cos65° using only angles that fall between 0° and 45°. Since 75°=90°−15° and 65°=90°−25°:
sin75° = sin(90°−15°) = cos15°
cos65° = cos(90°−25°) = sin25°
∴ sin75° + cos65° = cos15° + sin25°Nothing about the expression's actual value changes here — cos15°+sin25° is exactly the same number as sin75°+cos65°, just written using two angles under 45° instead of two angles over 45°. This kind of rewriting is genuinely useful whenever a reference table or calculation only covers a limited angle range, letting any angle above 45° be re-expressed through its complement below it.
This same 45°-splitting trick is a big part of why the specific-angle table in the previous exercise only needed to go up to 90° in the first place — every angle beyond that range, and every angle between 45° and 90° too if a table only listed angles up to 45°, can always be rewritten through its complement into the covered range. Sine and cosine, tangent and cotangent, secant and cosecant: every complementary pair works this same way, letting a single half-sized table effectively describe the whole range.
From Complementary Pairs to General Identities
Every result in this exercise connected one specific angle to its specific complement, whether numeric or left as an unevaluated letter. Exercise 11.4 moves to relationships that hold for literally every angle at once, no complementary pairing required — built entirely from Pythagoras' theorem rather than from the angle-sum property this exercise leaned on throughout. For the six ratio definitions every complementary relationship here was built from, revisit the chapter introduction once more.