Intuition The one-line ideaA point on the unit circle always sits at distance 1 1 1 ( cos θ , sin θ ) (\cos\theta,\sin\theta) ( cos θ , sin θ ) forces cos 2 θ + sin 2 θ = 1 \cos^2\theta + \sin^2\theta = 1 cos 2 θ + sin 2 θ = 1 divided by something.
Definition The three Pythagorean identitiessin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1 1 + tan 2 θ = sec 2 θ 1 + \tan^2\theta = \sec^2\theta 1 + tan 2 θ = sec 2 θ 1 + cot 2 θ = csc 2 θ 1 + \cot^2\theta = \csc^2\theta 1 + cot 2 θ = csc 2 θ
These hold for every θ \theta θ tan , sec \tan,\sec tan , sec cos θ ≠ 0 \cos\theta\neq 0 cos θ = 0
The word "Pythagorean " is literal: each one is the theorem a 2 + b 2 = c 2 a^2+b^2=c^2 a 2 + b 2 = c 2
sin \sin sin cos \cos cos 1 1 1 θ \theta θ x x x defined to be ( cos θ , sin θ ) (\cos\theta,\ \sin\theta) ( cos θ , sin θ ) definition of cosine and sine — the x x x y y y
HOW we get identity 1:
Point on circle: P = ( cos θ , sin θ ) P=(\cos\theta,\sin\theta) P = ( cos θ , sin θ ) Why this step? By definition of the unit-circle coordinates.
Drop a perpendicular to the x x x
horizontal leg = ∣ cos θ ∣ =|\cos\theta| = ∣ cos θ ∣
vertical leg = ∣ sin θ ∣ =|\sin\theta| = ∣ sin θ ∣
hypotenuse = = = = 1 =1 = 1 Why this step? The radius is the straight-line distance from origin to P P P
Apply Pythagoras : ( leg 1 ) 2 + ( leg 2 ) 2 = ( hyp ) 2 (\text{leg}_1)^2+(\text{leg}_2)^2=(\text{hyp})^2 ( leg 1 ) 2 + ( leg 2 ) 2 = ( hyp ) 2 cos 2 θ + sin 2 θ = 1 2 = 1. \cos^2\theta + \sin^2\theta = 1^2 = 1. cos 2 θ + sin 2 θ = 1 2 = 1. Why this step? Squaring kills the absolute values, so it works in every quadrant, not just the first.
The trick: take sin 2 θ + cos 2 θ = 1 \sin^2\theta+\cos^2\theta=1 sin 2 θ + cos 2 θ = 1
Divide by cos 2 θ \cos^2\theta cos 2 θ cos θ ≠ 0 \cos\theta\neq 0 cos θ = 0
sin 2 θ cos 2 θ + cos 2 θ cos 2 θ = 1 cos 2 θ \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} c o s 2 θ s i n 2 θ + c o s 2 θ c o s 2 θ = c o s 2 θ 1
Why this step? We chose cos 2 θ \cos^2\theta cos 2 θ sin θ cos θ = tan θ \dfrac{\sin\theta}{\cos\theta}=\tan\theta cos θ sin θ = tan θ 1 cos θ = sec θ \dfrac{1}{\cos\theta}=\sec\theta cos θ 1 = sec θ
tan 2 θ + 1 = sec 2 θ ⟹ 1 + tan 2 θ = sec 2 θ \tan^2\theta + 1 = \sec^2\theta \quad\Longrightarrow\quad \boxed{1+\tan^2\theta=\sec^2\theta} tan 2 θ + 1 = sec 2 θ ⟹ 1 + tan 2 θ = sec 2 θ
Divide by sin 2 θ \sin^2\theta sin 2 θ sin θ ≠ 0 \sin\theta\neq 0 sin θ = 0
sin 2 θ sin 2 θ + cos 2 θ sin 2 θ = 1 sin 2 θ \frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta} s i n 2 θ s i n 2 θ + s i n 2 θ c o s 2 θ = s i n 2 θ 1
Why this step? Dividing by sin 2 θ \sin^2\theta sin 2 θ cot = cos sin \cot=\dfrac{\cos}{\sin} cot = sin cos csc = 1 sin \csc=\dfrac{1}{\sin} csc = sin 1
1 + cot 2 θ = csc 2 θ ⟹ 1 + cot 2 θ = csc 2 θ 1 + \cot^2\theta = \csc^2\theta \quad\Longrightarrow\quad \boxed{1+\cot^2\theta=\csc^2\theta} 1 + cot 2 θ = csc 2 θ ⟹ 1 + cot 2 θ = csc 2 θ
Mnemonic "Same family, divide to move"Sin–Cos identity is the parent. Divide by cos 2 \cos^2 cos 2 "Tan–Sec" house. Divide by sin 2 \sin^2 sin 2 "Cot–Csc" house.
Memory hook: "1 + TAN² is SEC²" and "1 + COT² is CSC²" — the function whose name starts with the co- of the divisor sits on the left. (Divide by sin \sin sin co t, c sc.)
cos θ \cos\theta cos θ sin θ = 3 5 \sin\theta=\tfrac35 sin θ = 5 3 θ \theta θ cos 2 θ = 1 − sin 2 θ = 1 − 9 25 = 16 25 \cos^2\theta = 1-\sin^2\theta = 1-\tfrac{9}{25}=\tfrac{16}{25} cos 2 θ = 1 − sin 2 θ = 1 − 25 9 = 25 16 Why? Rearranged the master identity to isolate cos 2 \cos^2 cos 2 cos θ = + 4 5 \cos\theta = +\tfrac45 cos θ = + 5 4 Why the +? Acute angle ⇒ first quadrant ⇒ cosine positive.
Worked example 2) Simplifysec 2 θ − tan 2 θ \sec^2\theta-\tan^2\theta sec 2 θ − tan 2 θ 1 + tan 2 θ = sec 2 θ 1+\tan^2\theta=\sec^2\theta 1 + tan 2 θ = sec 2 θ tan 2 θ \tan^2\theta tan 2 θ sec 2 θ − tan 2 θ = 1. \sec^2\theta-\tan^2\theta = 1. sec 2 θ − tan 2 θ = 1. Why this is instant: the identity is sec 2 − tan 2 = 1 \sec^2-\tan^2=1 sec 2 − tan 2 = 1
tan θ = 2 \tan\theta = 2 tan θ = 2 sec θ \sec\theta sec θ θ \theta θ sec 2 θ = 1 + tan 2 θ = 1 + 4 = 5 ⇒ sec θ = 5 \sec^2\theta = 1+\tan^2\theta = 1+4 = 5 \Rightarrow \sec\theta=\sqrt5 sec 2 θ = 1 + tan 2 θ = 1 + 4 = 5 ⇒ sec θ = 5 Why +√? Q1 ⇒ cos > 0 \cos>0 cos > 0 sec > 0 \sec>0 sec > 0 cos θ = 1 5 \cos\theta=\tfrac1{\sqrt5} cos θ = 5 1 sin θ = tan θ cos θ = 2 5 \sin\theta=\tan\theta\cos\theta=\tfrac{2}{\sqrt5} sin θ = tan θ cos θ = 5 2 Check: sin 2 + cos 2 = 4 5 + 1 5 = 1. \sin^2+\cos^2=\tfrac45+\tfrac15=1. sin 2 + cos 2 = 5 4 + 5 1 = 1.
1 1 + cos θ + 1 1 − cos θ = 2 csc 2 θ \dfrac{1}{1+\cos\theta}+\dfrac{1}{1-\cos\theta}=2\csc^2\theta 1 + cos θ 1 + 1 − cos θ 1 = 2 csc 2 θ = ( 1 − cos θ ) + ( 1 + cos θ ) ( 1 + cos θ ) ( 1 − cos θ ) = 2 1 − cos 2 θ =\dfrac{(1-\cos\theta)+(1+\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}=\dfrac{2}{1-\cos^2\theta} = ( 1 + cos θ ) ( 1 − cos θ ) ( 1 − cos θ ) + ( 1 + cos θ ) = 1 − cos 2 θ 2 Why? Common denominator, difference of squares in the bottom.
1 − cos 2 θ = sin 2 θ 1-\cos^2\theta=\sin^2\theta 1 − cos 2 θ = sin 2 θ = 2 sin 2 θ = 2 csc 2 θ . =\dfrac{2}{\sin^2\theta}=2\csc^2\theta. = sin 2 θ 2 = 2 csc 2 θ .
sin 2 θ + cos 2 θ = 1 \sin^2\theta+\cos^2\theta=1 sin 2 θ + cos 2 θ = 1 sin θ + cos θ = 1 \sin\theta+\cos\theta=1 sin θ + cos θ = 1 Why it feels right: it looks like you can just "take the square root of each piece."
Why it's wrong: a 2 + b 2 ≠ a + b \sqrt{a^2+b^2}\neq a+b a 2 + b 2 = a + b θ = 45 ° \theta=45° θ = 45° sin + cos = 2 2 + 2 2 = 2 ≈ 1.41 ≠ 1 \sin+\cos=\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}=\sqrt2\approx1.41\neq1 sin + cos = 2 2 + 2 2 = 2 ≈ 1.41 = 1 Fix: never split a squared sum; square only after isolating one term.
sin 2 θ \sin^2\theta sin 2 θ sin ( θ 2 ) \sin(\theta^2) sin ( θ 2 ) Why it feels right: the exponent floats near the argument.
Fix: sin 2 θ ≡ ( sin θ ) 2 \sin^2\theta \equiv (\sin\theta)^2 sin 2 θ ≡ ( sin θ ) 2 sin ( θ 2 ) \sin(\theta^2) sin ( θ 2 )
Common mistake Dropping the± \pm ± Why it feels right: cos 2 θ = k \cos^2\theta=k cos 2 θ = k cos θ = k \cos\theta=\sqrt k cos θ = k Fix: cos θ = ± k \cos\theta=\pm\sqrt k cos θ = ± k θ \theta θ
1 + tan 2 = sec 2 1+\tan^2=\sec^2 1 + tan 2 = sec 2 cos θ = 0 \cos\theta=0 cos θ = 0 Why it feels right: the identity "always" holds.
Fix: at θ = 90 ° \theta=90° θ = 90° tan \tan tan sec \sec sec
Recall Predict before revealingQ: What is csc 2 θ − cot 2 θ \csc^2\theta-\cot^2\theta csc 2 θ − cot 2 θ Forecast , then check.
... It equals 1 1 1 1 + cot 2 = csc 2 1+\cot^2=\csc^2 1 + cot 2 = csc 2
Recall Feynman: explain to a 12-year-oldImagine standing at the centre of a round trampoline of radius 1 metre. You walk out to the edge in some direction. How far right you went is called "cos", how far up you went is "sin". No matter which direction you pick, you're always exactly 1 metre from the centre. Pythagoras (the corner-triangle rule) then says: (right amount)² + (up amount)² = (1 metre)². That's the whole secret — cos 2 + sin 2 = 1 \cos^2+\sin^2=1 cos 2 + sin 2 = 1 same sentence, just after you shrink everything by "cos" or by "sin".
What is the master Pythagorean identity? sin 2 θ + cos 2 θ = 1 \sin^2\theta+\cos^2\theta=1 sin 2 θ + cos 2 θ = 1 How do you get 1 + tan 2 θ = sec 2 θ 1+\tan^2\theta=\sec^2\theta 1 + tan 2 θ = sec 2 θ Divide
sin 2 θ + cos 2 θ = 1 \sin^2\theta+\cos^2\theta=1 sin 2 θ + cos 2 θ = 1 by
cos 2 θ \cos^2\theta cos 2 θ How do you get 1 + cot 2 θ = csc 2 θ 1+\cot^2\theta=\csc^2\theta 1 + cot 2 θ = csc 2 θ Divide
sin 2 θ + cos 2 θ = 1 \sin^2\theta+\cos^2\theta=1 sin 2 θ + cos 2 θ = 1 by
sin 2 θ \sin^2\theta sin 2 θ What does sec 2 θ − tan 2 θ \sec^2\theta-\tan^2\theta sec 2 θ − tan 2 θ What does csc 2 θ − cot 2 θ \csc^2\theta-\cot^2\theta csc 2 θ − cot 2 θ Geometric source of sin 2 + cos 2 = 1 \sin^2+\cos^2=1 sin 2 + cos 2 = 1 Pythagoras on a right triangle with hypotenuse = unit-circle radius
Given sin θ = 3 / 5 \sin\theta=3/5 sin θ = 3/5 θ \theta θ cos θ \cos\theta cos θ Why can sin θ + cos θ ≠ 1 \sin\theta+\cos\theta\neq1 sin θ + cos θ = 1 sin 2 + cos 2 = 1 \sin^2+\cos^2=1 sin 2 + cos 2 = 1 a 2 + b 2 ≠ a + b \sqrt{a^2+b^2}\neq a+b a 2 + b 2 = a + b ; e.g. at
45 ° 45° 45° the sum is
2 \sqrt2 2 When does 1 + tan 2 θ = sec 2 θ 1+\tan^2\theta=\sec^2\theta 1 + tan 2 θ = sec 2 θ When
cos θ = 0 \cos\theta=0 cos θ = 0 (functions undefined), e.g.
θ = 90 ° \theta=90° θ = 90° Rearrange master identity for cos 2 θ \cos^2\theta cos 2 θ cos 2 θ = 1 − sin 2 θ \cos^2\theta=1-\sin^2\theta cos 2 θ = 1 − sin 2 θ Point cos theta, sin theta
Right triangle legs and hypotenuse
Intuition Hinglish mein samjhoDekho, ye teenon identities ek hi baat ke roop hain. Ek unit circle socho — radius exactly 1. Jis bhi angle θ \theta θ x x x cos θ \cos\theta cos θ y y y sin θ \sin\theta sin θ cos θ \cos\theta cos θ sin θ \sin\theta sin θ 1 1 1 cos 2 θ + sin 2 θ = 1 \cos^2\theta+\sin^2\theta=1 cos 2 θ + sin 2 θ = 1
Ab baaki do identities koi nayi cheez nahi hain — same equation ko divide karke banti hain. Agar poori identity ko cos 2 θ \cos^2\theta cos 2 θ tan \tan tan sec \sec sec 1 + tan 2 θ = sec 2 θ 1+\tan^2\theta=\sec^2\theta 1 + tan 2 θ = sec 2 θ sin 2 θ \sin^2\theta sin 2 θ cot \cot cot csc \csc csc 1 + cot 2 θ = csc 2 θ 1+\cot^2\theta=\csc^2\theta 1 + cot 2 θ = csc 2 θ
Do sabse common galtiyan: pehli, sin 2 + cos 2 = 1 \sin^2+\cos^2=1 sin 2 + cos 2 = 1 sin + cos = 1 \sin+\cos=1 sin + cos = 1 45 ° 45° 45° 2 \sqrt2 2 ± \pm ±
