Noesis
3.1.14Advanced Trigonometry

Half angle formulas — derivations from double angle

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🇮🇳 Yeh note Hinglish mein padho →

1. The seed: double-angle cosine (three faces)

Why derive these two? Because each isolates a single squared function — exactly what we need to solve for cos⁡2\cos^2 or sin⁡2\sin^2 later.

Derivation of the second form (Why this step?): cos⁡2θ=cos⁡2θ−sin⁡2θ=(1−sin⁡2θ)−sin⁡2θ=1−2sin⁡2θ.\cos 2\theta = \cos^2\theta - \sin^2\theta = (1-\sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta. Here we replaced cos⁡2θ\cos^2\theta by 1−sin⁡2θ1-\sin^2\theta because we want an expression in sin⁡\sin only.

2. Deriving the half-angle formulas (from scratch)

The trick: let A=2θA = 2\theta, so θ=A2\theta = \dfrac{A}{2}. Substitute into the two one-function forms.

(a) Half-angle for sine

Start from cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta. Replace 2θ→A2\theta \to A: cos⁡A=1−2sin⁡2A2\cos A = 1 - 2\sin^2\frac{A}{2} Solve for sin⁡2A2\sin^2\frac{A}{2} (Why? we want the half angle alone): 2sin⁡2A2=1−cos⁡A  ⇒  sin⁡2A2=1−cos⁡A22\sin^2\frac{A}{2} = 1 - \cos A \;\Rightarrow\; \boxed{\sin^2\frac{A}{2} = \frac{1-\cos A}{2}}

(b) Half-angle for cosine

Start from cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1. Replace 2θ→A2\theta \to A: cos⁡A=2cos⁡2A2−1  ⇒  cos⁡2A2=1+cos⁡A2\cos A = 2\cos^2\frac{A}{2} - 1 \;\Rightarrow\; \boxed{\cos^2\frac{A}{2} = \frac{1+\cos A}{2}}

(c) Half-angle for tangent

Divide (a) by (b) — Why? tan⁡=sin⁡/cos⁡\tan = \sin/\cos, so tan⁡2=sin⁡2/cos⁡2\tan^2 = \sin^2/\cos^2: tan⁡2A2=1−cos⁡A1+cos⁡A\tan^2\frac{A}{2} = \frac{1-\cos A}{1+\cos A}

Deriving the "no-radical" tangent forms

Multiply top and bottom of 1−cos⁡A1+cos⁡A\dfrac{1-\cos A}{1+\cos A} inside the square-root by a conjugate to remove the sign ambiguity: tan⁡2A2=(1−cos⁡A)(1+cos⁡A)⋅(1+cos⁡A)(1+cos⁡A)=1−cos⁡2A(1+cos⁡A)2=sin⁡2A(1+cos⁡A)2.\tan^2\frac{A}{2}=\frac{(1-\cos A)}{(1+\cos A)}\cdot\frac{(1+\cos A)}{(1+\cos A)}=\frac{1-\cos^2 A}{(1+\cos A)^2}=\frac{\sin^2 A}{(1+\cos A)^2}. Taking the root: tan⁡A2=sin⁡A1+cos⁡A\tan\frac{A}{2}=\dfrac{\sin A}{1+\cos A} — no ±\pm needed because sin⁡A\sin A carries the correct sign automatically. Similarly multiplying by (1−cos⁡A)(1-\cos A) gives tan⁡A2=1−cos⁡Asin⁡A\tan\frac{A}{2}=\dfrac{1-\cos A}{\sin A}.

3. Worked examples

4. Common mistakes

5. Active recall

Recall Can you reproduce these without looking?
  • The three faces of cos⁡2θ\cos 2\theta.
  • The substitution that converts double→half.
  • Both radical-free forms of tan⁡A2\tan\frac A2 and why they need no ±\pm.
  • How to decide the sign in sin⁡A2\sin\frac A2.
Which double-angle form gives sin⁡2A2\sin^2\frac{A}{2}?
cos⁡A=1−2sin⁡2θ\cos A = 1-2\sin^2\theta with 2θ=A2\theta=A, giving sin⁡2A2=1−cos⁡A2\sin^2\frac A2=\frac{1-\cos A}{2}.
Half-angle formula for cos⁡2A2\cos^2\frac{A}{2}?
1+cos⁡A2\dfrac{1+\cos A}{2}.
Half-angle formula for sin⁡2A2\sin^2\frac{A}{2}?
1−cos⁡A2\dfrac{1-\cos A}{2}.
Two radical-free forms of tan⁡A2\tan\frac{A}{2}?
1−cos⁡Asin⁡A\dfrac{1-\cos A}{\sin A} and sin⁡A1+cos⁡A\dfrac{\sin A}{1+\cos A}.
The ±\pm sign in half-angle formulas depends on the quadrant of which angle?
A/2A/2 (not AA).
Derive cos⁡A=2cos⁡2A2−1\cos A = 2\cos^2\frac A2 -1 from where?
From cos⁡2θ=2cos⁡2θ−1\cos2\theta=2\cos^2\theta-1, set 2θ=A2\theta=A.
Exact value of cos⁡15∘\cos 15^\circ?
2+32≈0.9659\frac{\sqrt{2+\sqrt3}}{2}\approx0.9659.
Why does tan⁡A2=sin⁡A1+cos⁡A\tan\frac A2=\frac{\sin A}{1+\cos A} need no ±\pm?
sin⁡A\sin A already carries the correct sign; the denominator 1+cos⁡A≥01+\cos A\ge 0.
Value of tan⁡22.5∘\tan 22.5^\circ?
2−1≈0.4142\sqrt2-1\approx0.4142.
Recall Feynman: explain to a 12-year-old

Imagine a machine that tells you facts about an angle. We already have a machine that, when you give it an angle, tells you the cosine of double that angle. Now suppose someone hands you the cosine of a full pizza slice and asks about half a slice. We just run the machine backwards: rearrange its equation until "half slice" is alone on one side. Two catches: (1) halving can flip which "corner of the map" (quadrant) you're in, so you must check whether the answer should be plus or minus; (2) you can't just chop the sine in half like the angle — trig doesn't work like cutting a cake.

Connections

  • Double angle formulas — the direct parent identities.
  • Weierstrass substitution — t=tan⁡x2t=\tan\frac{x}{2} built entirely on these.
  • Pythagorean identity — used to split cos⁡2θ\cos2\theta into one-function forms.
  • Exact trig values — 15∘,22.5∘,75∘15^\circ, 22.5^\circ, 75^\circ come from half-angles.
  • Product-to-sum formulas — sibling manipulations of the same seed identities.

Concept Map

use sin^2+cos^2=1

use sin^2+cos^2=1

relabel 2t to A, solve

relabel 2t to A, solve

divide sin^2 by cos^2

divide sin^2 by cos^2

take root with plus-minus

take root with plus-minus

multiply by conjugate

no plus-minus needed

sign by quadrant of A over 2

sign by quadrant of A over 2

cos 2t = cos^2 - sin^2

cos 2t = 2cos^2 - 1

cos 2t = 1 - 2sin^2

sin^2 half = 1-cosA over 2

cos^2 half = 1+cosA over 2

tan^2 half = 1-cosA over 1+cosA

sin half formula

cos half formula

tan half = sinA over 1+cosA

Weierstrass sub and exact values

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, half-angle formulas ka poora idea bahut simple hai: humare paas double-angle wali identity hai — cos⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ\cos 2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta. Ab isko "ulta padho". Agar main 2θ2\theta ki jagah AA likh doon, toh θ\theta automatically A/2A/2 ban jaata hai. Bas equation ko rearrange karke sin⁡2(A/2)\sin^2(A/2) ya cos⁡2(A/2)\cos^2(A/2) ko alag kar lo. Isliye cos⁡2A2=1+cos⁡A2\cos^2\frac A2=\frac{1+\cos A}{2} aur sin⁡2A2=1−cos⁡A2\sin^2\frac A2=\frac{1-\cos A}{2} mil jaate hain. Koi rattaa nahi — sirf double-angle se derive.

Yeh matter kyun karta hai? Kyunki exam mein cos⁡15∘\cos 15^\circ, tan⁡22.5∘\tan 22.5^\circ jaise "non-standard" angles ki exact value poochhi jaati hai. 15=30/215 = 30/2, aur 30∘30^\circ standard hai, toh half-angle formula laga do — kaam khatam. Integration mein bhi t=tan⁡x2t=\tan\frac x2 substitution poori tarah inhi formulas par khadi hai.

Do baatein hamesha yaad rakhna. Pehli: square-root ke saath ±\pm aata hai, aur sign ka faisla A/2A/2 ke quadrant se hota hai, na ki AA ke — kyunki angle aadha karne par woh doosre quadrant mein ja sakta hai. Doosri: sin⁡A2\sin\frac A2 kabhi bhi 12sin⁡A\frac12\sin A ke barabar nahi hota, trig nonlinear hai. Tangent ke liye safe form yaad rakho: tan⁡A2=sin⁡A1+cos⁡A\tan\frac A2=\frac{\sin A}{1+\cos A} — isme ±\pm ki tension hi nahi, sign khud aa jaata hai.

Mnemonic simple: "cosine ke saath PLUS, sine ke saath MINUS" — bottom dono mein 22, sirf upar ka sign badalta hai.

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Connections

Double angle formulas — sin 2A, cos 2A (three forms), tan 2AMaths · 3.1.13Product-to-sum formulasMaths · 3.1.15Double angle formulas — sin 2A, cos 2A (three forms), tan 2AMaths · 3.1.13