Double angle formulas — sin 2A, cos 2A (three forms), tan 2A: Worked examples

Before we build the matrix, we fix two pieces of shared language: the three named forms of $cos 2 A$ , and the word quadrant.

Figure 1 below is the master map for this whole page. Each amber label names a quadrant; underneath it in white is the sign pattern $(cos A, sin A)$ you must read off before touching any formula. The amber arrow shows a sample angle $A$ in Q1. Every worked example starts by locating its angle on exactly this picture, so keep it in view.

Figure — Double angle formulas — sin 2A, cos 2A (three forms), tan 2A

Figure 1 — the four quadrants and their sign patterns.

Why does this matter for double angles? Because to use $sin 2 A = 2 sin A cos A$ you often first recover a missing ratio (say $cos A$ from $sin A$ ), and $cos A = \pm 1 - sin^{2} A$ — the $\pm$ is decided by the quadrant (read the $x$ -part sign off Figure 1), nothing else. Get the quadrant wrong and every later number flips sign.

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. Each row is worked below and tagged with its cell code.

Cell	Scenario class	What is tricky	Worked in
C1	$A$ in Q1 (all ratios positive)	baseline, no sign traps	Ex 1
C2	$A$ in Q2 ( $sin > 0, cos < 0$ )	picking the negative root	Ex 2
C3	$A$ in Q3 ( $sin < 0, cos < 0$ )	two negatives → $sin 2 A > 0$	Ex 3
C4	$A$ in Q4 ( $sin < 0, cos > 0$ )	$sin 2 A < 0$ but $2 A$ can leave Q4	Ex 4
Z	Zero / degenerate input ( $sin A = 0$ , or $tan A$ undefined)	limits, $0/0$ , blow-ups	Ex 5
T	$tan 2 A$ undefined ( $tan^{2} A = 1$ )	the denominator hits $0$	Ex 6
W	Word problem (real-world)	translating physics → angle	Ex 7
X	Exam twist (identity + given range)	choose the right form, respect the range	Ex 8

Ex 1 — Cell C1: everything positive (baseline)

Worked example

$A$ acute, $cos A = \frac{12}{13}$ . Find $sin 2 A$ , $cos 2 A$ , $tan 2 A$ . Forecast: $cos A = \frac{12}{13} \approx 0.923$ means $A \approx 22. 6^{\circ}$ , so $2 A \approx 45. 2^{\circ}$ . Guess $sin 2 A \approx 0.71$ , $cos 2 A \approx 0.70$ , $tan 2 A \approx 1.0$ .

Step 1 — recover $sin A$ . $sin A = + 1 - cos^{2} A = 1 - \frac{144}{169} = \frac{25}{169} = \frac{5}{13}$ . Why this step? $sin 2 A$ needs both $sin A$ and $cos A$ ; we only had $cos A$ . Why the $+$ root? C1 = Q1, where $sin A > 0$ .

Step 2 — $sin 2 A$ . $2 sin A cos A = 2 \cdot \frac{5}{13} \cdot \frac{12}{13} = \frac{120}{169}$ . Why? Direct formula; both factors known and positive.

Step 3 — $cos 2 A$ . We already have $cos A$ , so pick Form 2: $2 cos^{2} A - 1 = 2 \cdot \frac{144}{169} - 1 = \frac{288 - 169}{169} = \frac{119}{169}$ . Why Form 2? It reuses the number we were given, minimising fresh algebra.

Step 4 — $tan 2 A$ . $\frac{sin 2 A}{cos 2 A} = \frac{120/169}{119/169} = \frac{120}{119}$ . Why this route? Once $sin 2 A, cos 2 A$ are known, their ratio is $tan 2 A$ for free — no separate formula needed.

Verify: $(\frac{120}{169})^{2} + (\frac{119}{169})^{2} = \frac{14400 + 14161}{28561} = \frac{28561}{28561} = 1$ . ✓ Consistent with $sin^{2} (2 A) + cos^{2} (2 A) = 1$ . And $\frac{120}{119} \approx 1.008 \approx tan 45. 2^{\circ}$ . ✓

Ex 2 — Cell C2: $A$ in Q2 (negative cosine)

Worked example

$sin A = \frac{3}{5}$ with $9 0^{\circ} < A < 18 0^{\circ}$ . Find $sin 2 A$ and $cos 2 A$ ; say which quadrant $2 A$ lands in. Forecast: Q2 means $A$ between $9 0^{\circ}$ and $18 0^{\circ}$ , so $2 A$ is between $18 0^{\circ}$ and $36 0^{\circ}$ — the lower half. Sine there is often negative. Expect $sin 2 A < 0$ .

Step 1 — recover $cos A$ with the RIGHT sign. $cos A = - 1 - \frac{9}{25} = - \frac{4}{5}$ . Why the minus? In Q2 the $x$ -part is negative (read it off the master map, Figure 1), so $cos A < 0$ . This is the single most common slip — the square root offers $\pm$ , the quadrant chooses.

Step 2 — $sin 2 A$ . $2 sin A cos A = 2 \cdot \frac{3}{5} \cdot (- \frac{4}{5}) = - \frac{24}{25}$ . Why negative? One positive ( $sin A$ ) times one negative ( $cos A$ ) → negative. Matches the forecast.

Step 3 — $cos 2 A$ . Use Form 3 (we were given $sin A$ ): $1 - 2 sin^{2} A = 1 - 2 \cdot \frac{9}{25} = \frac{7}{25}$ . Why Form 3? It never touches $cos A$ , so the sign worry in Step 1 can't leak in here.

Step 4 — where is $2 A$ ? $sin 2 A < 0$ and $cos 2 A > 0$ ⇒ $2 A$ is in Q4. Why? Q4 is exactly the $(+, -)$ region for $(cos, sin)$ .

Figure 2 below makes this concrete: the cyan arrow is $A$ sitting in Q2, and the amber arrow is $2 A$ — doubling has swung it right around into Q4, exactly where a negative $sin 2 A$ and positive $cos 2 A$ must live. The dashed white arc traces the doubling.

Verify: $(- \frac{24}{25})^{2} + (\frac{7}{25})^{2} = \frac{576 + 49}{625} = \frac{625}{625} = 1$ . ✓ A concrete check: $A = 143.1 3^{\circ}$ gives $2 A = 286.2 6^{\circ}$ , $sin \approx - 0.96$ , $cos \approx 0.28$ . ✓

Figure 2 — $A$ in Q2 doubles to $2 A$ in Q4.

Ex 3 — Cell C3: $A$ in Q3 (both negative → $sin 2 A$ positive)

Worked example

$tan A = \frac{3}{4}$ with $18 0^{\circ} < A < 27 0^{\circ}$ . Find $sin 2 A$ , $cos 2 A$ , $tan 2 A$ . Forecast: $tan A > 0$ can happen in Q1 or Q3 (tan repeats every $18 0^{\circ}$ ). We are told Q3, so both $sin A$ and $cos A$ are negative. Two negatives multiplied give a positive, so expect $sin 2 A > 0$ .

Step 1 — build the triangle, then fix signs. $tan A = \frac{opp}{adj} = \frac{3}{4}$ , so a $3 - 4 - 5$ triangle. In Q3 both legs point negative: $sin A = - \frac{3}{5}$ , $cos A = - \frac{4}{5}$ . Why this step? $tan A$ alone doesn't give $sin, cos$ — only their ratio. The hypotenuse $5$ (always positive) and the Q3 sign pattern pin them down.

Step 2 — $sin 2 A$ . $2 \cdot (- \frac{3}{5}) (- \frac{4}{5}) = 2 \cdot \frac{12}{25} = \frac{24}{25}$ . Why positive? $(-) \times (-) = +$ . The forecast holds.

Step 3 — $cos 2 A$ . Form 1 is cleanest here since we have both squares: $cos^{2} A - sin^{2} A = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$ . Why Form 1? We already know both squares directly; no trading needed.

Step 4 — $tan 2 A$ from the formula. $\frac{2 tan A}{1 - tan ^{2} A} = \frac{2 \cdot \frac{3}{4}}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2} \cdot \frac{16}{7} = \frac{24}{7}$ . Why the tan-formula and not $sin 2 A / cos 2 A$ ? Both give $\frac{24/25}{7/25} = \frac{24}{7}$ — showing the two routes agree is itself a check.

Verify: $\frac{24/25}{7/25} = \frac{24}{7}$ ✓ matches Step 4. And $(\frac{24}{25})^{2} + (\frac{7}{25})^{2} = 1$ ✓.

Ex 4 — Cell C4: $A$ in Q4 ( $sin 2 A < 0$ )

Ex 5 — Cell Z: zero and degenerate inputs

Worked example Three degenerate checks: (a)

$A = 0^{\circ}$ , (b) $A = 9 0^{\circ}$ , (c) the limit of $tan 2 A$ as $A \to 9 0^{\circ}$ . Forecast: At the "corners" of the circle, formulas should still behave — either give a clean number or a clearly-flagged infinity. Guess: at $0^{\circ}$ everything is $0$ or $1$ ; at $9 0^{\circ}$ , $sin 2 A$ collapses.

(a) $A = 0^{\circ}$ . $sin A = 0, cos A = 1$ . $sin 2 A = 2 \cdot 0 \cdot 1 = 0$ ; $cos 2 A = 2 \cdot 1^{2} - 1 = 1$ ; $tan 2 A = \frac{2 \cdot 0}{1 - 0} = 0$ . Why check this? $2 A = 0^{\circ}$ too, and indeed $sin 0 = 0, cos 0 = 1, tan 0 = 0$ . The formulas pass the trivial gate.

(b) $A = 9 0^{\circ}$ . $sin A = 1, cos A = 0$ . $sin 2 A = 2 \cdot 1 \cdot 0 = 0$ ; $cos 2 A = 2 \cdot 0^{2} - 1 = - 1$ ; $tan 2 A = \frac{sin 2 A}{cos 2 A} = \frac{0}{- 1} = 0$ . Why interesting? $2 A = 18 0^{\circ}$ : $sin 18 0^{\circ} = 0$ ✓, $cos 18 0^{\circ} = - 1$ ✓, $tan 18 0^{\circ} = 0$ ✓. Note $sin 2 A = 0$ arises not because $sin A = 0$ but because $cos A = 0$ — the product structure matters. And here $tan 2 A$ is perfectly defined ( $= 0$ ) because the denominator $cos 2 A = - 1 \neq = 0$ ; contrast this with case (c).

(c) $tan 2 A$ as $A \to 9 0^{\circ}$ . Here $tan A \to \infty$ , so plugging into $\frac{2 tan A}{1 - tan ^{2} A}$ gives $\frac{\infty}{- \infty}$ — a $0/0$ -style indeterminate. Divide top and bottom by $tan^{2} A$ : $tan 2 A = \frac{2/ t a n A}{1/ t a n ^{2} A - 1} t a n A \to \infty \frac{0}{0 - 1} = 0.$ Why divide by $tan^{2} A$ ? It converts the runaway $tan A$ into the shrinking quantity $1/ tan A \to 0$ , turning nonsense into a clean limit. And $tan (2 \cdot 9 0^{\circ}) = tan 18 0^{\circ} = 0$ ✓ — reassuringly the same value the direct route (b) gave.

Verify: all sub-results ( $0, 1, 0$ ; then $0, - 1, 0$ ; then limit $0$ ) are machine-checked in the page's verification block.

Ex 6 — Cell T: $tan 2 A$ genuinely undefined (ALL angles in $[0^{\circ}, 36 0^{\circ})$ )

Worked example Find

every $A$ in $0^{\circ} \leq A < 36 0^{\circ}$ where $tan 2 A$ is undefined, and interpret it. Forecast: The formula's denominator is $1 - tan^{2} A$ . It vanishes when $tan^{2} A = 1$ , i.e. $tan A = \pm 1$ . In one full turn, $tan A = \pm 1$ happens at the four diagonal angles $4 5^{\circ}, 13 5^{\circ}, 22 5^{\circ}, 31 5^{\circ}$ — not just the first two. Expect a division-by-zero at all four.

Step 1 — solve $tan^{2} A = 1$ over the whole turn. $tan A = \pm 1$ . Because $tan$ repeats every $18 0^{\circ}$ :

$tan A = 1$ at $A = 4 5^{\circ}$ and $A = 22 5^{\circ}$ (add $18 0^{\circ}$ ).
$tan A = - 1$ at $A = 13 5^{\circ}$ and $A = 31 5^{\circ}$ (add $18 0^{\circ}$ ).

So the full list is $A \in {4 5^{\circ}, 13 5^{\circ}, 22 5^{\circ}, 31 5^{\circ}}$ . Why this step? A fraction is undefined exactly where its denominator is $0$ ; that's the only place the formula can fail, and $tan A = \pm 1$ recurs four times per turn.

Step 2 — interpret geometrically. Doubling each angle: $2 A = 9 0^{\circ}, 27 0^{\circ}, 45 0^{\circ} \equiv 9 0^{\circ}, 63 0^{\circ} \equiv 27 0^{\circ}$ . Every one lands on $9 0^{\circ}$ or $27 0^{\circ}$ , where $cos 2 A = 0$ , and $tan 2 A = \frac{sin 2 A}{cos 2 A}$ divides by zero — a vertical asymptote. Why does this match? "Denominator $1 - tan^{2} A = 0$ " and " $cos 2 A = 0$ " are the same event: indeed $cos 2 A = cos^{2} A - sin^{2} A = cos^{2} A (1 - tan^{2} A)$ , which is $0$ exactly when $1 - tan^{2} A = 0$ .

Verify: at each of $4 5^{\circ}, 13 5^{\circ}, 22 5^{\circ}, 31 5^{\circ}$ we get $cos 2 A = 0$ (checked in the verification block), so $tan 2 A$ is undefined at all four — no case in the interval is missed.

Ex 7 — Cell W: a real-world word problem

Worked example Projectile range. A ball is launched at speed

$v$ and angle $A$ above flat ground. Physics gives horizontal range $R = \frac{v ^{2}}{g} sin 2 A$ . If $v = 20 m/s$ , $g = 10 m/s^{2}$ and the launch angle is $A = 3 0^{\circ}$ , find $R$ . Then explain why $4 5^{\circ}$ is the best angle. Forecast: $sin 2 A = sin 6 0^{\circ} = \frac{3}{2} \approx 0.87$ , and $\frac{v ^{2}}{g} = \frac{400}{10} = 40 m$ . So $R \approx 40 \times 0.87 \approx 35 m$ .

Step 1 — substitute. $R = \frac{2 0 ^{2}}{10} sin (2 \cdot 3 0^{\circ}) = 40 \cdot sin 6 0^{\circ} = 40 \cdot \frac{3}{2} = 203 m \approx 34.6 m$ . Why does $sin 2 A$ appear at all? The range depends on up-speed $\times$ across-speed $\propto sin A cos A$ , and $2 sin A cos A = sin 2 A$ packs that product into one clean term. That is the whole reason the double-angle formula lives in this equation.

Step 2 — why $4 5^{\circ}$ maximises range. $R$ is largest when $sin 2 A$ is largest, i.e. $sin 2 A = 1$ , i.e. $2 A = 9 0^{\circ}$ , i.e. $A = 4 5^{\circ}$ . Why this step? $sin$ maxes out at $1$ ; solving $sin 2 A = 1$ is trivial because we compressed the messy product into a single sine.

Step 3 — units check. $\frac{v ^{2}}{g} = \frac{( m/s ) ^{2}}{m/s ^{2}} = m$ , and $sin 2 A$ is dimensionless, so $R$ is in metres. ✓

Verify: $203 \approx 34.64$ m. At $4 5^{\circ}$ : $R = 40 \cdot sin 9 0^{\circ} = 40$ m $> 34.6$ m, confirming $4 5^{\circ}$ beats $3 0^{\circ}$ . ✓

Ex 8 — Cell X: exam twist (identity + range constraint)

Worked example Solve

$cos 2 A + 3 sin A = 2$ for $0^{\circ} \leq A < 36 0^{\circ}$ . Forecast: There's a $cos 2 A$ and a $sin A$ . To solve, everything must be in one ratio. Since $sin A$ is already here, convert $cos 2 A$ to sine using Form 3. Expect a quadratic in $sin A$ .

Step 1 — choose the form that unifies the equation. Form 3: $cos 2 A = 1 - 2 sin^{2} A$ . Substitute: $1 - 2 sin^{2} A + 3 sin A = 2.$ Why Form 3? It leaves the equation in $sin A$ only — no stray $cos A$ to worry about. Form 2 would introduce $cos A$ and force another conversion.

Step 2 — tidy into a standard quadratic. Let $s = sin A$ : $- 2 s^{2} + 3 s + 1 - 2 = 0 \Rightarrow - 2 s^{2} + 3 s - 1 = 0 \Rightarrow 2 s^{2} - 3 s + 1 = 0.$ Why let $s = sin A$ ? Renaming exposes an ordinary quadratic we can factor. (I multiplied through by $- 1$ to make the leading coefficient positive — easier to factor.)

Step 3 — factor and solve for $s$ . $2 s^{2} - 3 s + 1 = (2 s - 1) (s - 1) = 0 \Rightarrow s = \frac{1}{2}$ or $s = 1$ . Why factor? A product of factors is zero only when a factor is zero — the fastest route to both roots.

Step 4 — back-substitute $s = sin A$ and find every $A$ in range. We must return from $s$ to the actual angles inside $0^{\circ} \leq A < 36 0^{\circ}$ .

$sin A = \frac{1}{2}$ : $sin$ is positive in Q1 and Q2. The reference angle is $3 0^{\circ}$ , so $A = 3 0^{\circ}$ (Q1) or $A = 18 0^{\circ} - 3 0^{\circ} = 15 0^{\circ}$ (Q2). Both lie in range → keep both.
$sin A = 1$ : this is the single peak of sine, at $A = 9 0^{\circ}$ only. (There is no second solution, because $sin A = 1$ touches the top of the circle exactly once per turn.)

Why two answers for $\frac{1}{2}$ but one for $1$ ? A horizontal line $sin A = c$ with $- 1 < c < 1$ cuts the sine wave twice per turn (once rising in Q1, once falling in Q2 for positive $c$ ); at the extreme $c = 1$ the line is tangent to the peak and touches once.

Solution set: $A \in {3 0^{\circ}, 9 0^{\circ}, 15 0^{\circ}}$ .

Verify (plug each root into the ORIGINAL equation):

$A = 3 0^{\circ}$ : $cos 6 0^{\circ} + 3 sin 3 0^{\circ} = \frac{1}{2} + 3 \cdot \frac{1}{2} = 2$ ✓
$A = 15 0^{\circ}$ : $cos 30 0^{\circ} + 3 sin 15 0^{\circ} = \frac{1}{2} + 3 \cdot \frac{1}{2} = 2$ ✓
$A = 9 0^{\circ}$ : $cos 18 0^{\circ} + 3 sin 9 0^{\circ} = - 1 + 3 = 2$ ✓

All three satisfy the original equation and lie in the required range, so the solution set is complete.

Recall Recall — pick the right cell fast

Given only $sin A$ , how do you get the sign of $cos A$ ? ::: The quadrant of $A$ decides it; the only gives the magnitude. Given $tan A > 0$ , which quadrants could $A$ be in? ::: Q1 or Q3 (tan repeats every $18 0^{\circ}$ ). Which $cos 2 A$ form makes $cos 2 A + 3 sin A = 2$ into a quadratic in $sin A$ ? ::: Form 3, $1 - 2 sin^{2} A$ . In $[0^{\circ}, 36 0^{\circ})$ , at which angles is $tan 2 A$ undefined? ::: $4 5^{\circ}, 13 5^{\circ}, 22 5^{\circ}, 31 5^{\circ}$ (all where $tan^{2} A = 1$ ). Why does $sin 2 A$ appear in projectile range? ::: Range $\propto$ up-speed $\times$ across-speed $\propto sin A cos A = \frac{1}{2} sin 2 A$ .

Connections

Parent topic — the formulas these examples exercise.
Compound (Addition) Angle Formulas — where all three formulas are born.
Pythagorean Identity — the sign-fixing and verification tool used in every example.
Power-Reduction / Half-Angle Formulas — the Form-3 rearrangement behind Ex 8's strategy.
Integration of sin²x and cos²x — the natural next home for these identities.
Triple Angle Formulas — the same "substitute and simplify" game, one step further.
Weierstrass t = tan(A/2) Substitution — pushes the $tan 2 A$ idea into integration.