Half angle formulas — derivations from double angle: Worked examples

The scenario matrix

Here $A$ is the "full" angle we are given, and $A /2$ is the half we want facts about. A quadrant is one of the four corners of the circle: QI is $0^{\circ}$ – $9 0^{\circ}$ (both $sin, cos > 0$ ), QII is $9 0^{\circ}$ – $18 0^{\circ}$ ( $sin > 0, cos < 0$ ), QIII is $18 0^{\circ}$ – $27 0^{\circ}$ (both $< 0$ ), QIV is $27 0^{\circ}$ – $36 0^{\circ}$ ( $sin < 0, cos > 0$ ).

Cell	Case class	What can go wrong	Example
C1	$A$ in QI → $A /2$ in QI	signs all $+$ (baseline)	Ex 1
C2	$A$ in QII → $A /2$ in QI	mixing given-quadrant with half-quadrant	Ex 2
C3	$A$ in QIII → $A /2$ in QII	$sin \frac{A}{2} > 0$ but $cos \frac{A}{2} < 0$	Ex 3
C4	$A$ in QIV → $A /2$ in QII	reproduce parent's QIV case, plus $cos \frac{A}{2}$	Ex 4
C5	Degenerate $A = 18 0^{\circ}$	denominator $1 + cos A = 0$ → which $tan$ form survives	Ex 5
C6	Degenerate $A = 0^{\circ}$ / limit	denominator $sin A = 0$ → the other form dies	Ex 6
C7	Exact value (word: geometry)	half of a known angle → exact surd	Ex 7
C8	Real-world word problem	a slope/reflection angle bisection	Ex 8
C9	Exam twist: solve an equation	half-angle turns a nasty equation linear	Ex 9

The figure below is the map every example refers back to — memorise the four coloured corners.

Ex 1 — C1: the baseline (everything positive)

Ex 2 — C2: $A$ in QII, but $A /2$ in QI

Ex 3 — C3: $A$ in QIII → the sign split

Worked example

$cos A = - \frac{4}{5}$ , $A$ in QIII. Find $sin \frac{A}{2}$ and $cos \frac{A}{2}$ . Forecast: QIII is $18 0^{\circ} < A < 27 0^{\circ}$ , so $9 0^{\circ} < A /2 < 13 5^{\circ}$ → QII. In QII sine is $+$ and cosine is $-$ . This is the first case where the two answers get different signs. Predict them.

Step 1. Locate $A /2$ : $18 0^{\circ} < A < 27 0^{\circ} \Rightarrow 9 0^{\circ} < \frac{A}{2} < 13 5^{\circ}$ = QII (see s01, coral corner). Why this step? We need two different sign decisions, so pin down the quadrant carefully.

Step 2. Sine (positive in QII): $sin \frac{A}{2} = + \frac{1 - ( - \frac{4}{5} )}{2} = \frac{9/5}{2} = \frac{9}{10} = \frac{3}{10} .$ Step 3. Cosine (negative in QII): $cos \frac{A}{2} = - \frac{1 + ( - \frac{4}{5} )}{2} = - \frac{1/5}{2} = - \frac{1}{10} = - \frac{1}{10} .$ Why opposite signs? QII gives $sin > 0, cos < 0$ — read straight off the map.

Verify: Pythagorean check (Pythagorean identity): $(\frac{3}{10})^{2} + (- \frac{1}{10})^{2} = \frac{9}{10} + \frac{1}{10} = 1$ . ✓ And $sin \frac{A}{2} cos \frac{A}{2} = - \frac{3}{10}$ ; doubling, $2 sin \frac{A}{2} cos \frac{A}{2} = - \frac{3}{5} = sin A$ — matches QIII ( $sin A < 0$ ). ✓

Ex 4 — C4: the parent's QIV case, extended

Worked example

$cos A = \frac{7}{25}$ , $A$ in QIV. Find $sin \frac{A}{2}$ and $tan \frac{A}{2}$ . Forecast: QIV is $27 0^{\circ} < A < 36 0^{\circ}$ so $13 5^{\circ} < A /2 < 18 0^{\circ}$ → QII. The parent found $sin \frac{A}{2} = \frac{3}{5}$ . Predict whether $tan \frac{A}{2}$ is positive or negative before reading. (Hint: QII → $tan < 0$ .)

Step 1. $A /2 \in (13 5^{\circ}, 18 0^{\circ})$ = QII. Why this step? Sign hunting, as always, comes first.

Step 2. Sine (positive in QII): $sin \frac{A}{2} = + \frac{1 - \frac{7}{25}}{2} = \frac{18/25}{2} = \frac{9}{25} = \frac{3}{5} .$ Step 3. For $tan \frac{A}{2}$ use the radical-free form $\frac{1 - cos A}{sin A}$ . We need $sin A$ : in QIV, $sin A < 0$ , and $sin^{2} A = 1 - cos^{2} A = 1 - \frac{49}{625} = \frac{576}{625}$ , so $sin A = - \frac{24}{25}$ . Why this form? It carries the sign automatically — no $\pm$ to agonise over. $tan \frac{A}{2} = \frac{1 - \frac{7}{25}}{- \frac{24}{25}} = \frac{18/25}{- 24/25} = - \frac{18}{24} = - \frac{3}{4} .$ Result: negative, as QII demanded. ✓

Verify: $tan \frac{A}{2} = \frac{sin ( A /2 )}{cos ( A /2 )}$ . We have $sin \frac{A}{2} = \frac{3}{5}$ ; from $cos^{2} \frac{A}{2} = \frac{1 + 7/25}{2} = \frac{16}{25}$ and QII gives $cos \frac{A}{2} = - \frac{4}{5}$ . Then $\frac{3/5}{- 4/5} = - \frac{3}{4}$ . ✓ Both routes agree.

Ex 5 — C5: degenerate $A = 18 0^{\circ}$ (a denominator dies)

Worked example Compute

$tan \frac{A}{2}$ at $A = 18 0^{\circ}$ . Which formula survives? Forecast: $A /2 = 9 0^{\circ}$ , and $tan 9 0^{\circ}$ is undefined (infinite). Which of the two radical-free forms blows up correctly, and which gives a fake $0/0$ ? Guess.

Step 1. Values: $cos 18 0^{\circ} = - 1$ , $sin 18 0^{\circ} = 0$ . Why this step? We must see which denominator hits zero.

Step 2. Try $\frac{sin A}{1 + cos A} = \frac{0}{1 + ( - 1 )} = \frac{0}{0}$ — indeterminate, useless here. Step 3. Try $\frac{1 - cos A}{sin A} = \frac{1 - ( - 1 )}{0} = \frac{2}{0}$ — diverges to $\pm \infty$ , which is the correct behaviour: $tan 9 0^{\circ}$ is undefined/infinite. Why the difference? The two forms share a value everywhere except at their respective bad points. At $A = 18 0^{\circ}$ the form with $sin A$ on the bottom is the honest one.

Step 4 (limit check). Approach $A \to 18 0^{\circ}$ from below with $A = 17 9^{\circ}$ : $cos 17 9^{\circ} \approx - 0.99985$ , $sin 17 9^{\circ} \approx 0.01745$ . Then $\frac{1 - cos A}{sin A} \approx \frac{1.99985}{0.01745} \approx 114.6$ , i.e. growing large — consistent with $tan 89. 5^{\circ} \approx 114.6$ . ✓

Verify: $tan 89. 5^{\circ} \approx 114.589$ , matching the limit. So the $\frac{1 - c o s A}{s i n A}$ form is the safe one near $A = 18 0^{\circ}$ .

Ex 6 — C6: degenerate $A = 0^{\circ}$ (the other denominator dies)

Ex 7 — C7: exact value from geometry ( $cos 7 5^{\circ}$ )

Ex 8 — C8: a real-world bisection (light on a wedge)

Ex 9 — C9: exam twist (half-angle linearises an equation)

Worked example Solve

$1 + cos θ = sin θ$ for $θ \in [0^{\circ}, 36 0^{\circ})$ . Forecast: Both sides carry $θ$ nonlinearly. But $\frac{sin θ}{1 + cos θ} = tan \frac{θ}{2}$ turns this into a single tangent equation. Guess how many solutions.

Step 1. Rearrange to $\frac{sin θ}{1 + cos θ} = 1$ provided $1 + cos θ \neq = 0$ . Why this step? The left side is exactly a radical-free half-angle tangent — that's the whole trick. Step 2. So $tan \frac{θ}{2} = 1$ , giving $\frac{θ}{2} = 4 5^{\circ}$ (within the valid range $θ \in [0, 360)$ → $θ /2 \in [0, 180)$ ), hence $θ = 9 0^{\circ}$ . Why only $4 5^{\circ}$ ? Over $θ /2 \in [0^{\circ}, 18 0^{\circ})$ , $tan = 1$ occurs only at $4 5^{\circ}$ ( $13 5^{\circ}$ gives $tan = - 1$ ). Step 3. Check the excluded case $1 + cos θ = 0$ , i.e. $θ = 18 0^{\circ}$ : then LHS $= 1 + (- 1) = 0$ but RHS $= sin 18 0^{\circ} = 0$ too — it is a solution the algebra dropped! Add $θ = 18 0^{\circ}$ . Why this matters: dividing by $1 + cos θ$ silently assumed it was nonzero; degenerate inputs (Ex 5) must be reinstated by hand.

Verify: $θ = 9 0^{\circ}$ : $1 + cos 9 0^{\circ} = 1 + 0 = 1$ , $sin 9 0^{\circ} = 1$ ✓. $θ = 18 0^{\circ}$ : $1 + (- 1) = 0 = sin 18 0^{\circ}$ ✓. Two solutions: ${9 0^{\circ}, 18 0^{\circ}}$ .

Recall

Recall Which

$tan \frac{A}{2}$ form near which degenerate angle? Near $A = 0^{\circ}$ use ::: $\frac{sin A}{1 + cos A}$ (denominator $= 2 \neq = 0$ ). Near $A = 18 0^{\circ}$ use ::: $\frac{1 - cos A}{sin A}$ (the other one gives $0/0$ ). $A$ in QIII puts $A /2$ in ::: QII, so $sin \frac{A}{2} > 0$ but $cos \frac{A}{2} < 0$ . $cos 7 5^{\circ}$ exact ::: $\frac{2 - 3}{2} \approx 0.2588$ . When you write $\frac{s i n θ}{1 + c o s θ} = tan \frac{θ}{2}$ you must recheck ::: $θ = 18 0^{\circ}$ (where $1 + cos θ = 0$ ).

Connections

Double angle formulas — the seed identities behind every formula used here.
Weierstrass substitution — Ex 5/6 explain why $t = tan \frac{x}{2}$ needs both tangent forms.
Pythagorean identity — the verification tool in Ex 3.
Exact trig values — Ex 1, 7 produce $3 0^{\circ}, 7 5^{\circ}$ surds.
Product-to-sum formulas — sibling manipulations of the same double-angle seed.