Co-function identities — Hinglish notes

3.1.11 · Maths › Advanced Trigonometry

Co-function identities KYA hain?

Har trig function ka ek co-function partner hota hai:

$sin \leftrightarrow cos$
$tan \leftrightarrow cot$
$sec \leftrightarrow csc$

Rule yeh hai: kisi angle ka function, uske complement ke co-function ke barabar hota hai.

YEH SACH KYUN HAI? — Right Triangle se Derivation

Ek right triangle lo jisme right angle $C$ par ho. Baaki do angles ko $θ$ ( $A$ par) aur $ϕ$ ( $B$ par) kaho.

Ab sides ko angle $θ$ ke relative label karo:

$θ$ ke opposite = side $a$
$θ$ ke adjacent = side $b$
hypotenuse = $h$

Yahan key observation yeh hai: $θ$ ke opposite wali side $ϕ$ ke adjacent hai, aur vice versa.

$sin θ = \frac{a}{h} = \frac{opp of θ}{h}, cos ϕ = \frac{a}{h} = \frac{adj of ϕ}{h} .$

Dono $\frac{a}{h}$ ke barabar hain! Isliye $sin θ = cos ϕ = cos (90° - θ) . ■$

Yeh step kyun? Humne koi formula invent nahi kiya — humne notice kiya ki wohi physical side ( $a$ ) $θ$ ke nazariye se "opposite" hai lekin $ϕ$ ke nazariye se "adjacent" hai. Sine aur cosine ki definitions phir equality ko force karti hain.

Baaki identities free mein milna

Jab ek baar $sin (90° - θ) = cos θ$ aur $cos (90° - θ) = sin θ$ mil gayi, toh baaki sab divide karke milti hain:

$tan (90° - θ) = \frac{s i n ( 90° - θ )}{c o s ( 90° - θ )} = \frac{c o s θ}{s i n θ} = cot θ .$

Yeh step kyun? $tan = sin / cos$ definition hai; hum bas woh do identities substitute karte hain jo hum already prove kar chuke hain. Koi nayi geometry nahi chahiye.

Isi tarah $sec (90° - θ) = \frac{1}{c o s ( 90° - θ )} = \frac{1}{s i n θ} = csc θ .$

USE KAISE KAREIN — Worked examples

Forecast-then-Verify

Common mistakes (Steel-manned)

Recall Feynman: 12-saal ke bachche ko samjhao

Socho do bachche ek slide ke opposite corners par hain (teesra corner perfect square corner hai). Pehle bachche ke "saamne" jo hai, wohi doosre bachche ke "bagal mein" hai, aur vice versa. Toh jab bachcha A dekhta hai ki slide kitni oonchi lagti hai (sine), toh bachcha B wohi cheez dekhta hai lekin use chaudi kehta hai (cosine). Unke do angles hamesha ek square corner ( $90°$ ) tak add hote hain, aur isliye ek ki sine doosre ki cosine hoti hai. "Co-" bas matlab hai "tere dost ka version."

Active-recall flashcards

sin (90° - θ) = ?

cos θ

cos (90° - θ) = ?

sin θ

tan (90° - θ) = ?

cot θ

sec (90° - θ) = ?

csc θ

Right triangle ke do acute angles complementary kyun hote hain?

Woh

90°

angle ke saath

180°

tak add hote hain, toh unke beech

90°

bacha rehta hai.

Geometrically,

sin θ = cos (90° - θ)

kyun hota hai?

θ

ke opposite wali side uske complement ke adjacent hoti hai, toh opp/hyp = adj/hyp.

\frac{cos 20°}{sin 70°}

simplify karo.

1

(kyunki

sin 70° = cos 20°

sin^{2} 25° + sin^{2} 65°

ki value?

1

(kyunki

sin 65° = cos 25°

Sine co-function identity ka radian form?

sin (\frac{π}{2} - θ) = cos θ

Steel-man:

sin (90° - θ) = 1 - sin θ

galat kyun hai?

sin

subtraction par distribute nahi hota; sahi value

cos θ

hai.

Connections

Right Triangle Trigonometry (SOH-CAH-TOA) — jahan opp/adj/hyp define hote hain.
Pythagorean Identities — khoobsurti se combine hota hai ( $sin^{2} + cos^{2} = 1$ ).
Reference Angles and Supplementary Identities — $180° - θ$ wala cousin.
Angle Sum and Difference Formulas — co-function identities special case hain $sin (90° - θ)$ ka.
Radian Measure — $π /2$ versions ke liye.
Complementary and Supplementary Angles — geometric backbone.