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

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Step 0 — The three words we need first

Before a single formula, let us agree on what our symbols mean on a picture. Draw a circle of radius $1$ (a unit circle) centred at the origin. Pick a point on it and draw the line from the centre to that point. The angle that line makes with the flat positive $x$-axis, measured anticlockwise, we call $A$.

Everything below is just watching this point move — and asking what happens when the angle doubles to $2A$. That word "doubles" is the whole chapter.

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Step 1 — Why we don't need new magic: $2A=A+A$

WHAT. We notice that a double angle is nothing exotic. Doubling means adding the angle to itself: $2A=\underset{\text{first copy}}{\underbrace{A}}+\underset{\text{second copy}}{\underbrace{A}}.$

WHY. We already have machines that eat two angles $X$ and $Y$ and spit out the sine and cosine of their sum — the Compound (Addition) Angle Formulas. If we feed the same angle into both slots, the machine will hand us the double-angle result for free. No new tool required — that is the cheapest possible move.

PICTURE. In the figure, the ray at angle $A$ is drawn, and then the same rotation again stacks on top to land at $2A$. The second wedge is a carbon copy of the first.

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Step 2 — Building $\sin 2A$

WHAT. Put $X=A$ and $Y=A$ into the sine machine: $\sin 2A=\sin (A+A)=\underset{\text{up}}{\underbrace{\sin A}}\underset{\text{right}}{\underbrace{\cos A}}+\underset{\text{right}}{\underbrace{\cos A}}\underset{\text{up}}{\underbrace{\sin A}}.$

WHY. The two terms are the same product written in the opposite order (multiplication doesn't care about order). "Same thing twice" is just "twice the thing": $\sin A\cos A+\cos A\sin A=2\sin A\cos A.$

PICTURE. The area of the shaded rectangle with sides $\sin A$ and $\cos A$ appears twice — once for each term. Two identical rectangles $\Rightarrow$ the factor of $2$ out front. That is the whole formula, made of two equal tiles.

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Step 3 — Building $\cos 2A$, Form 1 (the raw one)

WHAT. Same substitution, now in the cosine machine: $\cos 2A=\cos (A+A)=\underset{\text{right}\times \text{right}}{\underbrace{\cos A\cos A}}-\underset{\text{up}\times \text{up}}{\underbrace{\sin A\sin A}}={\cos }^{2}A-{\sin }^{2}A.$

WHY. The cosine machine has a minus sign (unlike sine's plus). With $X=Y=A$, $\cos A\cos A={\cos }^{2}A$ and $\sin A\sin A={\sin }^{2}A$. So the sum turns into a difference of two squares.

PICTURE. Think of two squares: a big blue one of side $\cos A$ (area ${\cos }^{2}A$) and a pink one of side $\sin A$ (area ${\sin }^{2}A$). $\cos 2A$ is the blue area minus the pink area. When $A$ is small, blue dominates and $\cos 2A$ is near $1$; as $A$ grows past $45^{\circ }$, pink overtakes blue and $\cos 2A$ goes negative — you can literally watch the sign flip.

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Step 4 — The swap trick gives Forms 2 and 3

WHAT. We use the Pythagorean Identity ${\sin }^{2}A+{\cos }^{2}A=1$ to trade one square for the other, producing two more faces of the same formula.

WHY. Real problems often hand you only $\cos A$ or only $\sin A$. Rather than compute the missing one, we rewrite $\cos 2A$ so it uses only the letter we already own. The identity says the two squares always add to $1$, so either can stand in for the other.

Trade ${\sin }^{2}A=1-{\cos }^{2}A$: $\cos 2A={\cos }^{2}A-(1-{\cos }^{2}A)=2{\cos }^{2}A-1.$ Trade ${\cos }^{2}A=1-{\sin }^{2}A$: $\cos 2A=(1-{\sin }^{2}A)-{\sin }^{2}A=1-2{\sin }^{2}A.$

PICTURE. Picture the unit square of area $1$ split into a blue ${\cos }^{2}A$ slab and a pink ${\sin }^{2}A$ slab (they must fill it exactly — that is Pythagoras). Removing one colour and doubling the other is the "swap": each rearrangement below is the same total, re-tiled.

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Step 5 — Building $\tan 2A$ from steepness

WHAT. Divide what we already found: $\tan 2A=\frac{\sin 2A}{\cos 2A}=\frac{2\sin A\cos A}{{\cos }^{2}A-{\sin }^{2}A}$. Now divide top and bottom by ${\cos }^{2}A$ to turn every $\sin$/$\cos$ into a $\tan$: $\tan 2A=\frac{\frac{2\sin A\cos A}{{\cos }^{2}A}}{\frac{{\cos }^{2}A-{\sin }^{2}A}{{\cos }^{2}A}}=\frac{2\cdot \frac{\sin A}{\cos A}}{1-\frac{{\sin }^{2}A}{{\cos }^{2}A}}=\frac{2\tan A}{1-{\tan }^{2}A}.$

WHY divide by ${\cos }^{2}A$? Because $\tan$ is steepness $=\frac{\sin }{\cos }$, and we want the answer to speak only in steepness. Dividing by ${\cos }^{2}A$ is the exact bookkeeping that converts every term into a $\tan$. We choose this and not, say, dividing by ${\sin }^{2}A$, because ${\cos }^{2}A$ cleanly makes the constant $1$ appear in the denominator.

PICTURE. The ray at $A$ has slope $\tan A$; the ray at $2A$ has slope $\tan 2A$. The formula shows the new steepness is not double the old one — the $-{\tan }^{2}A$ in the floor makes it grow faster and faster, then snap to a vertical wall when the denominator hits zero.

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Step 6 — The edge case: when $\tan 2A$ explodes

WHAT. The denominator $1-{\tan }^{2}A$ becomes $0$ exactly when ${\tan }^{2}A=1$, i.e. $\tan A=\pm 1$, i.e. $A=45^{\circ },135^{\circ },\dots$

WHY it must blow up. At $A=45^{\circ }$ we have $2A=90^{\circ }$, and a $90^{\circ }$ line is perfectly vertical — infinite steepness, zero "run". Dividing by zero is the algebra's honest way of saying "this ray is straight up, there is no finite slope." The formula didn't break; it correctly refused to name infinity.

PICTURE. As $A$ creeps toward $45^{\circ }$, the doubled ray swings toward vertical. The value $\tan 2A$ races off the top of the board — a vertical asymptote (dashed wall).

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Step 7 — Sign check across a full turn (all three formulas)

WHAT. We confirm every double-angle formula behaves for every position of $A$ around a full $360^{\circ }$ turn — tracking the signs and the zeros of $\sin 2A$, $\cos 2A$ and $\tan 2A$.

WHY. $\sin A$ and $\cos A$ change sign in different quadrants, so we must check the products and quotients don't lie. Each formula has its own sign-flip schedule, and we want no gap left uncovered.

$\sin 2A=2\sin A\cos A$ — signs from the product, split into the four equal quadrants of $A$:

Quadrant of $A$	$\sin A$	$\cos A$	$\sin 2A$	Reason
I ($0$–$90^{\circ }$)	$+$	$+$	$+$	up × right
II ($90$–$180^{\circ }$)	$+$	$-$	$-$	up × left
III ($180$–$270^{\circ }$)	$-$	$-$	$+$	$-\times -$
IV ($270$–$360^{\circ }$)	$-$	$+$	$-$	down × right

$\sin 2A$ zeros at $A=0,90^{\circ },180^{\circ },270^{\circ },360^{\circ }$ (every $90^{\circ }$), because then either $\sin A$ or $\cos A$ is $0$. Notice the sign pattern $+,-,+,-$ has period $180^{\circ }$, as promised.

$\cos 2A={\cos }^{2}A-{\sin }^{2}A$ — signs from which square wins. Its sign flips at $2A=90^{\circ },270^{\circ },450^{\circ },630^{\circ }$, i.e. at $A=45^{\circ },135^{\circ },225^{\circ },315^{\circ }$. Those four points chop the full circle into eight equal $45^{\circ }$ slices, so we list all eight:

Range of $A$	dominant square	$\cos 2A$
$0$–$45^{\circ }$	${\cos }^{2}A$	$+$
$45$–$90^{\circ }$	${\sin }^{2}A$	$-$
$90$–$135^{\circ }$	${\sin }^{2}A$	$-$
$135$–$180^{\circ }$	${\cos }^{2}A$	$+$
$180$–$225^{\circ }$	${\cos }^{2}A$	$+$
$225$–$270^{\circ }$	${\sin }^{2}A$	$-$
$270$–$315^{\circ }$	${\sin }^{2}A$	$-$
$315$–$360^{\circ }$	${\cos }^{2}A$	$+$

Read down the sign column: $+,-,-,+,+,-,-,+$ — the first half ($0$–$180^{\circ }$) is exactly repeated in the second half ($180$–$360^{\circ }$), confirming period $180^{\circ }$. The final arc $315$–$360^{\circ }$ is now explicitly covered: $\cos 2A$ is positive there (blue wins again as $A$ heads back toward $360^{\circ }$).

$\tan 2A=\frac{\sin 2A}{\cos 2A}$ — sign is the two above divided. Because its period is $90^{\circ }$, the pattern on $0$–$90^{\circ }$ simply repeats on $90$–$180^{\circ }$, $180$–$270^{\circ }$, and $270$–$360^{\circ }$. We give all of them:

Range of $A$	$\sin 2A$	$\cos 2A$	$\tan 2A$
$0$–$45^{\circ }$	$+$	$+$	$+$
$45$–$90^{\circ }$	$+$	$-$	$-$
$90$–$135^{\circ }$	$-$	$-$	$+$
$135$–$180^{\circ }$	$-$	$+$	$-$
$180$–$225^{\circ }$	$+$	$+$	$+$
$225$–$270^{\circ }$	$+$	$-$	$-$
$270$–$315^{\circ }$	$-$	$-$	$+$
$315$–$360^{\circ }$	$-$	$+$	$-$

$\tan 2A$ is zero at $A=0,90^{\circ },180^{\circ },270^{\circ },360^{\circ }$ (where $\sin 2A=0$) and explodes to a vertical wall at $A=45^{\circ },135^{\circ },225^{\circ },315^{\circ }$ (where $\cos 2A=0$) — matching Step 6. The pattern $+,-,+,-$ repeats every $90^{\circ }$, so $180$–$360^{\circ }$ is a carbon copy of $0$–$180^{\circ }$.

PICTURE. The doubled point sweeps the circle twice as fast. When $A$ finishes half a lap ($180^{\circ }$), the $2A$ point has done a full lap — so every sign pattern above cycles twice as often as the ordinary single-angle one.

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The one-picture summary

Everything on this page is one substitution ($X=Y=A$) fed into the addition machines, then tidied with Pythagoras. This final board compresses all seven steps: the seed, the three results, the swap trick, and the asymptote.

Recall Feynman: the whole walkthrough in plain words

Draw a circle of size one. A point on it has an across-value (that's cosine) and an up-value (that's sine). "Double the angle" just means "spin the point twice as far", and that is the same as adding the angle to itself. So I feed the same angle into both slots of the add-two-angles machine I already trust. For sine, the machine gives me the same little rectangle twice, so I write $2\sin A\cos A$. For cosine, the machine has a minus, giving big blue square minus little pink square: ${\cos }^{2}A-{\sin }^{2}A$; and because the blue and pink squares must fill a unit square, I can swap one for the other to get two more faces of the same fact. Tangent is just up-over-across of the doubled point, and cleaning it up needs dividing by ${\cos }^{2}A$ — which is only allowed when $\cos A$ isn't zero, so $\tan A$ must exist. The tidy result $\frac{2\tan A}{1-{\tan }^{2}A}$ sensibly explodes to infinity right when the doubled ray stands perfectly vertical at $90^{\circ }$. Nothing new was invented; I only reused old spells with the same number typed twice, and I keep track of signs by watching the doubled point race twice as fast around the circle, so the whole pattern just repeats — $\sin 2A$ and $\cos 2A$ every $180^{\circ }$, $\tan 2A$ every $90^{\circ }$.

Connections

• Parent topic note
• Compound (Addition) Angle Formulas — the seed machines of Step 1.
• Pythagorean Identity — the swap trick of Step 4.
• Power-Reduction / Half-Angle Formulas — Form 3 rearranged.
• Integration of sin²x and cos²x — where these get used.
• Triple Angle Formulas — double plus single angle.
• Weierstrass t = tan(A/2) Substitution — extends the $\tan 2A$ idea.

Concept Map