Intuition The big idea in one breathIn a right triangle, the two non-right angles always add up to 90 ° 90° 90° (they are complementary ). So one angle "sees" the triangle exactly the way the other angle sees it — but with opposite and adjacent swapped. That swap is why the sine of one angle equals the cosine of the other. Every co-function identity is just this single swap, dressed differently.
"Co-" literally means complement : co sine = co mplement's sine, co tangent = co mplement's tangent, co secant = co mplement's secant.
Definition Co-function identities (degrees)sin ( 90 ° − θ ) = cos θ cos ( 90 ° − θ ) = sin θ \sin(90°-\theta)=\cos\theta \qquad \cos(90°-\theta)=\sin\theta sin ( 90° − θ ) = cos θ cos ( 90° − θ ) = sin θ tan ( 90 ° − θ ) = cot θ cot ( 90 ° − θ ) = tan θ \tan(90°-\theta)=\cot\theta \qquad \cot(90°-\theta)=\tan\theta tan ( 90° − θ ) = cot θ cot ( 90° − θ ) = tan θ sec ( 90 ° − θ ) = csc θ csc ( 90 ° − θ ) = sec θ \sec(90°-\theta)=\csc\theta \qquad \csc(90°-\theta)=\sec\theta sec ( 90° − θ ) = csc θ csc ( 90° − θ ) = sec θ 90 ° 90° 90° π 2 \dfrac{\pi}{2} 2 π
Each trig function has a co-function partner:
sin ↔ cos \sin \leftrightarrow \cos sin ↔ cos tan ↔ cot \tan \leftrightarrow \cot tan ↔ cot sec ↔ csc \sec \leftrightarrow \csc sec ↔ csc
The rule: the function of an angle equals the co-function of its complement.
Take a right triangle with the right angle at C C C θ \theta θ A A A ϕ \phi ϕ B B B
Intuition Why the angles must be complementaryAngles of any triangle sum to 180 ° 180° 180° 90 ° 90° 90° θ + ϕ = 180 ° − 90 ° = 90 ° ⟹ ϕ = 90 ° − θ . \theta+\phi = 180°-90° = 90° \quad\Longrightarrow\quad \phi = 90°-\theta. θ + ϕ = 180° − 90° = 90° ⟹ ϕ = 90° − θ .
Now label the sides relative to angle θ \theta θ
opposite to θ \theta θ a a a adjacent to θ \theta θ b b b hypotenuse = h h h
Here is the key observation: the side opposite θ \theta θ adjacent to ϕ \phi ϕ
sin θ = a h = opp of θ h , cos ϕ = a h = adj of ϕ h . \sin\theta=\frac{a}{h}=\frac{\text{opp of }\theta}{h}, \qquad \cos\phi=\frac{a}{h}=\frac{\text{adj of }\phi}{h}. sin θ = h a = h opp of θ , cos ϕ = h a = h adj of ϕ .
Both equal a h \dfrac{a}{h} h a sin θ = cos ϕ = cos ( 90 ° − θ ) . ■ \sin\theta=\cos\phi=\cos(90°-\theta). \qquad\blacksquare sin θ = cos ϕ = cos ( 90° − θ ) . ■
Why this step? We didn't invent a formula — we noticed the same physical side (a a a θ \theta θ ϕ \phi ϕ
Once sin ( 90 ° − θ ) = cos θ \sin(90°-\theta)=\cos\theta sin ( 90° − θ ) = cos θ cos ( 90 ° − θ ) = sin θ \cos(90°-\theta)=\sin\theta cos ( 90° − θ ) = sin θ dividing :
tan ( 90 ° − θ ) = sin ( 90 ° − θ ) cos ( 90 ° − θ ) = cos θ sin θ = cot θ . \tan(90°-\theta)=\frac{\sin(90°-\theta)}{\cos(90°-\theta)}=\frac{\cos\theta}{\sin\theta}=\cot\theta. tan ( 90° − θ ) = c o s ( 90° − θ ) s i n ( 90° − θ ) = s i n θ c o s θ = cot θ .
Why this step? tan = sin / cos \tan = \sin/\cos tan = sin / cos definition ; we just substitute the two identities we already proved. No new geometry needed.
Similarly
sec ( 90 ° − θ ) = 1 cos ( 90 ° − θ ) = 1 sin θ = csc θ . \sec(90°-\theta)=\frac{1}{\cos(90°-\theta)}=\frac{1}{\sin\theta}=\csc\theta. sec ( 90° − θ ) = c o s ( 90° − θ ) 1 = s i n θ 1 = csc θ .
Worked example Example 1 — Convert to the co-functionSimplify sin 63 ° \sin 63° sin 63°
Step 1. Write 63 ° = 90 ° − 27 ° 63° = 90° - 27° 63° = 90° − 27° Why this step? We want the complement so the co-function rule fires.
Step 2. sin ( 90 ° − 27 ° ) = cos 27 ° \sin(90°-27°)=\cos 27° sin ( 90° − 27° ) = cos 27° Answer: sin 63 ° = cos 27 ° \sin 63° = \cos 27° sin 63° = cos 27°
Worked example Example 2 — Prove an expression collapsesShow sin 2 40 ° + sin 2 50 ° = 1 \sin^2 40° + \sin^2 50° = 1 sin 2 40° + sin 2 50° = 1
Step 1. 50 ° = 90 ° − 40 ° 50° = 90°-40° 50° = 90° − 40° sin 50 ° = cos 40 ° \sin 50° = \cos 40° sin 50° = cos 40° Why? Turns two different angles into ONE angle.
Step 2. sin 2 40 ° + cos 2 40 ° = 1 \sin^2 40° + \cos^2 40° = 1 sin 2 40° + cos 2 40° = 1 Answer: = 1 =1 = 1
Worked example Example 3 — RadiansEvaluate tan ( π 2 − π 5 ) \tan\!\left(\dfrac{\pi}{2}-\dfrac{\pi}{5}\right) tan ( 2 π − 5 π )
Step 1. Pattern: tan ( π 2 − θ ) = cot θ \tan(\frac{\pi}{2}-\theta)=\cot\theta tan ( 2 π − θ ) = cot θ θ = π 5 \theta=\frac{\pi}{5} θ = 5 π Why? π 2 \frac{\pi}{2} 2 π Answer: cot π 5 \cot\dfrac{\pi}{5} cot 5 π
Worked example Example 4 — Mixed with valuesSimplify cos 20 ° sin 70 ° \dfrac{\cos 20°}{\sin 70°} sin 70° cos 20°
Step 1. sin 70 ° = sin ( 90 ° − 20 ° ) = cos 20 ° \sin 70° = \sin(90°-20°)=\cos 20° sin 70° = sin ( 90° − 20° ) = cos 20° Why? Make numerator and denominator identical.
Step 2. cos 20 ° cos 20 ° = 1 \dfrac{\cos 20°}{\cos 20°}=1 cos 20° cos 20° = 1 Answer: 1 1 1
Recall Predict BEFORE reading the answerGuess cot ( 90 ° − θ ) = ? \cot(90°-\theta)=? cot ( 90° − θ ) = ? cot \cot cot tan θ \tan\theta tan θ cot θ \cot\theta cot θ flips the function.
Common mistake Mistake 1:sin ( 90 ° − θ ) = sin 90 ° − sin θ = 1 − sin θ \sin(90°-\theta) = \sin 90° - \sin\theta = 1-\sin\theta sin ( 90° − θ ) = sin 90° − sin θ = 1 − sin θ Why it feels right: we treat sin \sin sin The fix: sin \sin sin function , not a multiplier — it does not distribute. Use the identity: sin ( 90 ° − θ ) = cos θ \sin(90°-\theta)=\cos\theta sin ( 90° − θ ) = cos θ θ = 30 ° \theta=30° θ = 30° cos 30 ° = 0.866 \cos30°=0.866 cos 30° = 0.866 1 − 0.5 = 0.5 1-0.5=0.5 1 − 0.5 = 0.5
Common mistake Mistake 2: Forgetting the functionchanges
Writing cos ( 90 ° − θ ) = cos θ \cos(90°-\theta)=\cos\theta cos ( 90° − θ ) = cos θ Why it feels right: the angle "looks close" to θ \theta θ The fix: the complement rule always swaps to the co-function : cos ( 90 ° − θ ) = sin θ \cos(90°-\theta)=\sin\theta cos ( 90° − θ ) = sin θ
Common mistake Mistake 3: Using90 ° − θ 90°-\theta 90° − θ sin 120 ° \sin 120° sin 120° not a co-function case because 120 ° > 90 ° 120°>90° 120° > 90° Fix: use 120 ° = 180 ° − 60 ° 120°=180°-60° 120° = 180° − 60° supplementary / reference-angle rule) instead. Co-function needs the pair to sum to exactly 90 ° 90° 90°
Recall Feynman: explain to a 12-year-oldImagine two kids on opposite corners of a slide (the third corner is a perfect square corner). Whatever is "in front of" the first kid is "beside" the second kid, and vice versa. So when kid A measures how tall the slide looks (sine), kid B measures the exact same thing but calls it how wide it looks (cosine). Their two angles always add up to a square corner (90 ° 90° 90°
"Co- means 'complement's'." Add a co- and subtract from 90 ° 90° 90° s in ‾ ( 90 − θ ) = co s ‾ ⇒ in → co-s \text{s}\underline{\text{in}}(90-\theta)=\text{co}\underline{\text{s}} \;\; \Rightarrow \text{in}\to \text{co-s} s in ( 90 − θ ) = co s ⇒ in → co-s SCT ↔ CST swap — Sin/Cos, seC/Csc, Tan/coT trade places at the complement.
sin ( 90 ° − θ ) = ? \sin(90°-\theta)=? sin ( 90° − θ ) = ? cos ( 90 ° − θ ) = ? \cos(90°-\theta)=? cos ( 90° − θ ) = ? tan ( 90 ° − θ ) = ? \tan(90°-\theta)=? tan ( 90° − θ ) = ? sec ( 90 ° − θ ) = ? \sec(90°-\theta)=? sec ( 90° − θ ) = ? Why are the two acute angles of a right triangle complementary? They sum with the
90 ° 90° 90° angle to
180 ° 180° 180° , leaving
90 ° 90° 90° between them.
Geometrically, why does sin θ = cos ( 90 ° − θ ) \sin\theta=\cos(90°-\theta) sin θ = cos ( 90° − θ ) The side opposite
θ \theta θ is the side adjacent to its complement, so opp/hyp = adj/hyp.
Simplify cos 20 ° sin 70 ° \dfrac{\cos 20°}{\sin 70°} sin 70° cos 20° 1 1 1 (since
sin 70 ° = cos 20 ° \sin70°=\cos20° sin 70° = cos 20° ).
Value of sin 2 25 ° + sin 2 65 ° \sin^2 25°+\sin^2 65° sin 2 25° + sin 2 65° 1 1 1 (because
sin 65 ° = cos 25 ° \sin65°=\cos25° sin 65° = cos 25° ).
Radian form of the sine co-function identity? sin ( π 2 − θ ) = cos θ \sin(\tfrac{\pi}{2}-\theta)=\cos\theta sin ( 2 π − θ ) = cos θ .
Steel-man: why is sin ( 90 ° − θ ) = 1 − sin θ \sin(90°-\theta)=1-\sin\theta sin ( 90° − θ ) = 1 − sin θ sin \sin sin doesn't distribute over subtraction; correct value is
cos θ \cos\theta cos θ .
f of angle = co-f of complement
Simplify sin 63 as cos 27
Intuition Hinglish mein samjhoDekho, co-function identities ka core idea bahut simple hai: kisi bhi right-angle triangle mein jo do baaki angles hote hain woh hamesha milke 90 ° 90° 90° complementary angles kehte hain. Matlab agar ek angle θ \theta θ 90 ° − θ 90°-\theta 90° − θ
Isi swap ki wajah se sin θ = cos ( 90 ° − θ ) \sin\theta = \cos(90°-\theta) sin θ = cos ( 90° − θ ) opp hyp \frac{\text{opp}}{\text{hyp}} hyp opp adj hyp \frac{\text{adj}}{\text{hyp}} hyp adj tan \tan tan cot \cot cot sec \sec sec csc \csc csc
Yeh cheez exams mein bahut kaam aati hai jab expression simplify karna ho, jaise cos 20 ° sin 70 ° \frac{\cos 20°}{\sin 70°} s i n 70° c o s 20° sin 70 ° = cos 20 ° \sin70° = \cos20° sin 70° = cos 20° 1 1 1 sin 2 40 ° + sin 2 50 ° \sin^2 40° + \sin^2 50° sin 2 40° + sin 2 50° sin 2 40 ° + cos 2 40 ° = 1 \sin^2 40° + \cos^2 40° = 1 sin 2 40° + cos 2 40° = 1
Ek warning: sin ( 90 ° − θ ) \sin(90°-\theta) sin ( 90° − θ ) 1 − sin θ 1 - \sin\theta 1 − sin θ sin \sin sin badalta hai: cos ( 90 ° − θ ) = sin θ \cos(90°-\theta)=\sin\theta cos ( 90° − θ ) = sin θ cos θ \cos\theta cos θ f ( 90 ° − θ ) = co- f ( θ ) f(90°-\theta) = \text{co-}f(\theta) f ( 90° − θ ) = co- f ( θ )
