Half angle formulas — derivations from double angle: Exercises

Definition Where the

$\pm$ comes from — and the domain rules The first two formulas give the square $sin^{2} \frac{A}{2}$ and $cos^{2} \frac{A}{2}$ . To get $sin \frac{A}{2}$ or $cos \frac{A}{2}$ themselves you must take a square root, and every square root has two possible signs: $sin \frac{A}{2} = \pm \frac{1 - c o s A}{2}, cos \frac{A}{2} = \pm \frac{1 + c o s A}{2} .$ The symbol $\pm$ ("plus or minus") flags exactly this choice. You resolve it by asking which quadrant $\frac{A}{2}$ lives in — never $A$ . The tangent forms $\frac{1 - cos A}{sin A}$ and $\frac{sin A}{1 + cos A}$ need no $\pm$ because $sin A$ already carries the sign.

Domain rules (state these before using the formulas):

$tan \frac{A}{2} = \frac{1 - cos A}{sin A}$ requires $sin A \neq = 0$ , i.e. $A$ is not a multiple of $18 0^{\circ}$ ( $A \neq = k \cdot 18 0^{\circ}$ , $π$ in radians).
$tan \frac{A}{2} = \frac{sin A}{1 + cos A}$ requires $1 + cos A \neq = 0$ , i.e. $A \neq = 18 0^{\circ} + k \cdot 36 0^{\circ}$ ( $cos A \neq = - 1$ ).
Both fail together exactly where $\frac{A}{2} = 9 0^{\circ}$ , and there $tan \frac{A}{2}$ is genuinely undefined — no formula's fault.

Angle measure: we work in degrees throughout for concreteness. To convert, $18 0^{\circ} = π$ radians, so e.g. $3 0^{\circ} = \frac{π}{6}$ , $4 5^{\circ} = \frac{π}{4}$ . Every formula and quadrant rule reads identically in radians.

The quadrant map you will use every time

Before any sign question, you must know the four quadrants — the four quarters of a full turn ( $36 0^{\circ}$ ) — and which trig function is positive in each. Look at the figure: the unit circle (radius $1$ ) is split into four wedges. In wedge I ( $0^{\circ}$ – $9 0^{\circ}$ ) everything is positive; in II ( $9 0^{\circ}$ – $18 0^{\circ}$ ) only $sin$ ; in III ( $18 0^{\circ}$ – $27 0^{\circ}$ ) only $tan$ ; in IV ( $27 0^{\circ}$ – $36 0^{\circ}$ ) only $cos$ .

Figure — Half angle formulas — derivations from double angle

Level 1 — Recognition

Goal: pick the correct formula and plug in. No sign traps yet.

Recall Solution L1.1

WHAT: we want $sin^{2}$ , so we use the "minus on top" formula. WHY that one: the mnemonic — sine stays alone (MINUS). $sin^{2} \frac{A}{2} = \frac{1 - c o s A}{2} = \frac{1 - \frac{1}{2}}{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4} .$ Answer: $\frac{1}{4}$ .

Recall Solution L1.2

WHAT: $cos^{2}$ → "plus on top" formula (cosine likes company, PLUS). $cos^{2} \frac{A}{2} = \frac{1 + c o s A}{2} = \frac{1 + \frac{1}{2}}{2} = \frac{\frac{3}{2}}{2} = \frac{3}{4} .$ Answer: $\frac{3}{4}$ . (Sanity check: $\frac{1}{4} + \frac{3}{4} = 1$ , matching $sin^{2} + cos^{2} = 1$ from the Pythagorean identity.)

Recall Solution L1.3

Plug straight in — no sign decision needed because $sin A$ carries its own sign (and $sin A = \frac{4}{5} \neq = 0$ , so the formula is defined): $tan \frac{A}{2} = \frac{1 - \frac{3}{5}}{\frac{4}{5}} = \frac{\frac{2}{5}}{\frac{4}{5}} = \frac{2}{4} = \frac{1}{2} .$ Answer: $\frac{1}{2}$ .

Level 2 — Application

Goal: full computation, choose the sign by locating $\frac{A}{2}$ .

Recall Solution L2.1

Step 1 — locate $\frac{A}{2}$ . If $9 0^{\circ} < A < 18 0^{\circ}$ then $4 5^{\circ} < \frac{A}{2} < 9 0^{\circ}$ : quadrant I, so $cos \frac{A}{2} > 0$ → sign is $+$ . Step 2 — apply the formula. $cos^{2} \frac{A}{2} = \frac{1 + c o s A}{2} = \frac{1 - \frac{7}{25}}{2} = \frac{\frac{18}{25}}{2} = \frac{9}{25} .$ Step 3 — root with the chosen sign. $cos \frac{A}{2} = + \frac{9}{25} = \frac{3}{5} .$ Answer: $\frac{3}{5}$ .

Recall Solution L2.2

Step 1 — locate $\frac{A}{2}$ . $27 0^{\circ} < A < 36 0^{\circ} \Rightarrow 13 5^{\circ} < \frac{A}{2} < 18 0^{\circ}$ : quadrant II, where $sin > 0$ → sign $+$ . Step 2 & 3: $sin^{2} \frac{A}{2} = \frac{1 - \frac{7}{25}}{2} = \frac{\frac{18}{25}}{2} = \frac{9}{25} \Rightarrow sin \frac{A}{2} = + \frac{3}{5} .$ Answer: $\frac{3}{5}$ . Notice $A$ itself is in QIV (sin negative), yet the half angle sits in QII (sin positive) — proof that the quadrant of $A$ is a false guide.

Recall Solution L2.3

Use the safe form $tan \frac{A}{2} = \frac{1 - cos A}{sin A}$ (no $\pm$ ; here $sin A = \frac{4}{5} \neq = 0$ , so it is defined): $tan \frac{A}{2} = \frac{1 - ( - \frac{3}{5} )}{\frac{4}{5}} = \frac{\frac{8}{5}}{\frac{4}{5}} = \frac{8}{4} = 2.$ Answer: $2$ .

Level 3 — Analysis

Goal: manipulate the formulas, not just plug numbers.

Recall Solution L3.1

Take $A = 3 0^{\circ}$ so $\frac{A}{2} = 1 5^{\circ}$ , in quadrant I → sign $+$ . Use $cos 3 0^{\circ} = \frac{3}{2}$ from Exact trig values: $sin^{2} 1 5^{\circ} = \frac{1 - c o s 3 0 ^{\circ}}{2} = \frac{1 - \frac{3}{2}}{2} = \frac{2 - 3}{4} .$ $sin 1 5^{\circ} = \frac{2 - 3}{2} \approx 0.2588.$ Answer: $\frac{2 - 3}{2}$ .

Recall Solution L3.2

Scope: we exclude $A = k \cdot 18 0^{\circ}$ so that neither denominator ( $sin A$ , nor $1 + cos A$ ) is zero — outside that set both sides are well defined. Cross-multiply: $(1 - cos A) (1 + cos A) = ? sin A \cdot sin A .$ Left side is a difference of squares: $1 - cos^{2} A$ . By the Pythagorean identity, $1 - cos^{2} A = sin^{2} A$ . Right side is $sin^{2} A$ . Equal ✓. Both therefore equal $tan \frac{A}{2}$ .

Recall Solution L3.3

Divide numerator and denominator by $2$ : $\frac{1 - c o s A}{1 + c o s A} = \frac{\frac{1 - c o s A}{2}}{\frac{1 + c o s A}{2}} = \frac{s i n ^{2} \frac{A}{2}}{c o s ^{2} \frac{A}{2}} = tan^{2} \frac{A}{2} .$ Answer: $tan^{2} \frac{A}{2}$ . (This holds wherever $cos A \neq = - 1$ ; the sympy check confirms it as an identity in that domain.)

Level 4 — Synthesis

Goal: chain half-angle with double-angle or Pythagorean reasoning.

Recall Solution L4.1

Step 1 — get $cos A$ . In QII, $cos A < 0$ . By Pythagoras $cos A = - 1 - sin^{2} A = - 1 - \frac{576}{625} = - \frac{49}{625} = - \frac{7}{25}$ . Step 2 — safe tangent form (uses $sin A = \frac{24}{25} \neq = 0$ , no $\pm$ ): $tan \frac{A}{2} = \frac{1 - c o s A}{s i n A} = \frac{1 - ( - \frac{7}{25} )}{\frac{24}{25}} = \frac{\frac{32}{25}}{\frac{24}{25}} = \frac{32}{24} = \frac{4}{3} .$ Answer: $\frac{4}{3}$ .

Recall Solution L4.2

$tan 3 0^{\circ} = \frac{1}{3}$ , so $t = \frac{1}{3}$ and $t^{2} = \frac{1}{3}$ . $\frac{1 - t ^{2}}{1 + t ^{2}} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{2}{4} = \frac{1}{2} .$ And indeed $cos 6 0^{\circ} = \frac{1}{2}$ ✓. Answer: matches.

Recall Solution L4.3

(i) $cos^{2} 1 5^{\circ} = \frac{1 + cos 3 0 ^{\circ}}{2} = \frac{1 + \frac{3}{2}}{2} = \frac{2 + 3}{4}$ . (ii) From the parent note $cos 1 5^{\circ} = \frac{2 + 3}{2}$ , so squaring: $(\frac{2 + 3}{2})^{2} = \frac{2 + 3}{4}$ . Both give $\frac{2 + 3}{4} \approx 0.9330$ ✓.

Level 5 — Mastery

Goal: multi-step proofs and a full case sweep.

Recall Solution L5.1

Common denominator $sin A (1 + cos A)$ (nonzero in our domain): $\frac{s i n ^{2} A + ( 1 + c o s A ) ^{2}}{s i n A ( 1 + c o s A )} .$ Expand the top: $sin^{2} A + 1 + 2 cos A + cos^{2} A$ . Group $sin^{2} A + cos^{2} A = 1$ : $= 1 + 1 + 2 cos A = 2 + 2 cos A = 2 (1 + cos A) .$ So the fraction is $\frac{2 ( 1 + cos A )}{sin A ( 1 + cos A )} = \frac{2}{sin A}$ ✓. (The first term is $tan \frac{A}{2}$ and the second is $cot \frac{A}{2}$ , defined just above — their sum is what we simplified.)

Recall Solution L5.2

Halve each interval, then read $sin$ 's sign from the quadrant map (figure s01):

Quadrant of $A$	interval of $A$	interval of $\frac{A}{2}$	quadrant of $\frac{A}{2}$	sign of $sin \frac{A}{2}$
I	$0^{\circ}$ – $9 0^{\circ}$	$0^{\circ}$ – $4 5^{\circ}$	I	$+$
II	$9 0^{\circ}$ – $18 0^{\circ}$	$4 5^{\circ}$ – $9 0^{\circ}$	I	$+$
III	$18 0^{\circ}$ – $27 0^{\circ}$	$9 0^{\circ}$ – $13 5^{\circ}$	II	$+$
IV	$27 0^{\circ}$ – $36 0^{\circ}$	$13 5^{\circ}$ – $18 0^{\circ}$	II	$+$

For $A \in (0^{\circ}, 36 0^{\circ})$ , $\frac{A}{2} \in (0^{\circ}, 18 0^{\circ})$ — the whole upper half-plane — so $sin \frac{A}{2} > 0$ always here. The picture makes this vivid: no matter which wedge $A$ picks, $\frac{A}{2}$ can never fall below the horizontal axis on one turn.

Recall Solution L5.3

Locate $\frac{A}{2}$ . QIII means $18 0^{\circ} < A < 27 0^{\circ}$ , so $9 0^{\circ} < \frac{A}{2} < 13 5^{\circ}$ : quadrant II → $sin > 0$ , $cos < 0$ , $tan < 0$ . Sine: $sin^{2} \frac{A}{2} = \frac{1 - ( - \frac{1}{2} )}{2} = \frac{\frac{3}{2}}{2} = \frac{3}{4} \Rightarrow sin \frac{A}{2} = + \frac{3}{2}$ . Cosine: $cos^{2} \frac{A}{2} = \frac{1 + ( - \frac{1}{2} )}{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4} \Rightarrow cos \frac{A}{2} = - \frac{1}{2}$ . Tangent: $tan \frac{A}{2} = \frac{sin \frac{A}{2}}{cos \frac{A}{2}} = \frac{3 /2}{- 1/2} = - 3$ . Check: $sin^{2} + cos^{2} = \frac{3}{4} + \frac{1}{4} = 1$ ✓. Answers: $sin \frac{A}{2} = \frac{3}{2}, cos \frac{A}{2} = - \frac{1}{2}, tan \frac{A}{2} = - 3 .$

Connections

Parent: Half-angle derivations — where every formula here is built with pictures.
Double angle formulas — the seed identities read backwards.
Pythagorean identity — powers L1.2, L3.2, L4.1, L5.1 checks.
Exact trig values — supplies $cos 3 0^{\circ}$ , $tan 3 0^{\circ}$ used above.
Weierstrass substitution — L4.2's $t = tan \frac{A}{2}$ machinery.
Product-to-sum formulas — sibling identity manipulations like L5.1.