Visual walkthrough — Half angle formulas — derivations from double angle

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Step 0 — The words before the symbols

Before any equation, let us agree on the pictures behind three words.

The tool we start from is the double-angle cosine. We are not going to prove it here (that is the Double angle formulas page's job) — we take it as our seed and grow the half-angle formulas out of it.

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Step 1 — The seed identity, drawn

WHAT. We start from the single fact: $\underset{\text{cosine of the doubled angle}}{\underbrace{\cos 2\theta }}=\underset{(\cos \theta {)}^2}{\underbrace{{\cos }^2\theta }}-\underset{(\sin \theta {)}^2}{\underbrace{{\sin }^2\theta }}$

Read each piece: the left side is the $x$-coordinate of the point at $2\theta$. The right side mixes only the $x$ and $y$ of the point at the single angle $\theta$. So this equation is a bridge from "angle $\theta$" information to "angle $2\theta$" information.

WHY. This is the only input we trust. Everything below is just algebra done to this one line — plus one more fact, next.

PICTURE. The figure shows the point at $\theta$ (its $x$ is $\cos \theta$, its $y$ is $\sin \theta$) and, further round, the point at $2\theta$ whose $x$-coordinate is $\cos 2\theta$.

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Step 2 — One extra fact: the Pythagorean identity

WHAT. For any point $(\cos \theta ,\sin \theta )$ on the unit circle, its distance to the origin is $1$. Distance${}^2=x^2+y^2$, so: ${\cos }^2\theta +{\sin }^2\theta =1$

Each symbol: ${\cos }^2\theta$ is the squared horizontal leg, ${\sin }^2\theta$ the squared vertical leg, and $1$ is the squared radius (the hypotenuse). This is just Pythagoras drawn on the circle — see Pythagorean identity.

WHY. Because $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$ contains two different functions. To later solve for a single one, we need a way to trade ${\sin }^2$ for ${\cos }^2$ (or the reverse). This identity is exactly that trade-in slip.

PICTURE. A right triangle whose hypotenuse is the radius $1$; its horizontal leg $\cos \theta$ and vertical leg $\sin \theta$ satisfy ${\cos }^2\theta +{\sin }^2\theta =1$.

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Step 3 — Split the seed into two one-function faces

WHAT. Substitute the trade-in slip into the seed, two different ways.

Kill the ${\sin }^2$ (replace ${\sin }^2\theta =1-{\cos }^2\theta$): $\cos 2\theta ={\cos }^2\theta -(1-{\cos }^2\theta )=2{\cos }^2\theta -1$

Kill the ${\cos }^2$ (replace ${\cos }^2\theta =1-{\sin }^2\theta$): $\cos 2\theta =(1-{\sin }^2\theta )-{\sin }^2\theta =1-2{\sin }^2\theta$

Annotating the first: we swapped the $-{\sin }^2\theta$ term for $-(1-{\cos }^2\theta )$, so two ${\cos }^2\theta$'s pile up into $2{\cos }^2\theta$ and a lone $-1$ falls out.

WHY. Each new form contains only one trig function. That is the whole point — soon we will solve for that single function, and you cannot cleanly solve an equation that mixes ${\sin }^2$ and ${\cos }^2$.

PICTURE. A "prism" splitting the one seed beam into two coloured beams: an orange $2{\cos }^2\theta -1$ beam and a teal $1-2{\sin }^2\theta$ beam.

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Step 4 — The relabel: read the equation backwards

WHAT. Nothing forces the letter to be $\theta$. Set $A=2\theta ⟺\theta =\frac{A}{2}.$ Wherever we see $2\theta$, write $A$; wherever we see $\theta$, write $\frac{A}{2}$. The two faces become: $\cos A=2{\cos }^2\frac{A}{2}-1\text{and}\cos A=1-2{\sin }^2\frac{A}{2}.$

Here $\cos A$ is a known number (the cosine of your given full angle), and ${\cos }^2\frac{A}{2}$, ${\sin }^2\frac{A}{2}$ are the unknowns we want.

WHY. The double-angle formula predicts the doubled angle from the single one. We want the reverse: we are handed the "full" angle $A$ and want its "half". Relabelling turns the same true equation into a machine that runs the other direction — no new maths, just a change of names.

PICTURE. The same number line, once labelled with $\theta$ and $2\theta$, then relabelled with $\frac{A}{2}$ and $A$ — identical geometry, new names.

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Step 5 — Solve for the half-angle squares

WHAT. Isolate the unknown in each relabelled face.

From $\cos A=1-2{\sin }^2\frac{A}{2}$, add $2{\sin }^2\frac{A}{2}$ and subtract $\cos A$: $2{\sin }^2\frac{A}{2}=1-\cos A⟹\boxed{{\sin }^2\frac{A}{2}=\frac{1-\cos A}{2}}$

From $\cos A=2{\cos }^2\frac{A}{2}-1$, add $1$ and divide by $2$: $\boxed{{\cos }^2\frac{A}{2}=\frac{1+\cos A}{2}}$

Term-by-term: same denominator $2$ in both; the only difference is the sign on top — a $-$ for sine, a $+$ for cosine. That single sign is the whole "personality" of these formulas.

WHY. These squared forms are the raw results; the actual $\sin \frac{A}{2}$ and $\cos \frac{A}{2}$ come from square-rooting (Step 7). We keep them as squares first because squares hide the sign — which we deal with deliberately, not by accident.

PICTURE. Two "balance scales": on the left pan the unknown square, on the right pan $\frac{1\pm \cos A}{2}$, with the $\pm$ highlighted so you see the mirrored signs.

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Step 6 — The tangent, and why it needs no $\pm$

WHAT. Since $\tan =\frac{\sin }{\cos }$, we get ${\tan }^2=\frac{{\sin }^2}{{\cos }^2}$. Divide Step 5's two results: ${\tan }^2\frac{A}{2}=\frac{\frac{1-\cos A}{2}}{\frac{1+\cos A}{2}}=\frac{1-\cos A}{1+\cos A}.$

Now the clever move — multiply top and bottom by $(1+\cos A)$: ${\tan }^2\frac{A}{2}=\frac{(1-\cos A)(1+\cos A)}{(1+\cos A{)}^2}=\frac{1-{\cos }^{2}A}{(1+\cos A{)}^2}=\frac{{\sin }^{2}A}{(1+\cos A{)}^2},$ using $1-{\cos }^{2}A={\sin }^{2}A$ (the Pythagorean identity again). Take the square root: $\boxed{\tan \frac{A}{2}=\frac{\sin A}{1+\cos A}}$

WHY no $\pm$? Look at the two ingredients. The denominator $1+\cos A$ is never negative (cosine never drops below $-1$, so $1+\cos A\ge 0$). The numerator $\sin A$ already carries the correct sign all by itself. So the ratio automatically comes out with the right sign — the ambiguity vanishes. Multiplying instead by $(1-\cos A)$ gives the twin form $\tan \frac{A}{2}=\frac{1-\cos A}{\sin A}$.

PICTURE. Two little sign-strips: $1+\cos A$ shaded entirely non-negative (a solid teal bar), while $\sin A$ swings positive/negative — showing the ratio inherits $\sin A$'s sign.

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Step 7 — The sign of the root lives in the quadrant of $A/2$

WHAT. Square-rooting the boxes in Step 5 gives $\sin \frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{2}},\cos \frac{A}{2}=\pm \sqrt{\frac{1+\cos A}{2}}.$ The  ​ symbol always outputs a non-negative number, but $\sin \frac{A}{2}$ or $\cos \frac{A}{2}$ can genuinely be negative. So we must choose the $\pm$ ourselves.

WHY the quadrant of $A/2$, not $A$? The formula is about the angle $\frac{A}{2}$. Its sign is decided by where $\frac{A}{2}$ lands on the circle, and halving can move it into a different quadrant than $A$. You must locate $\frac{A}{2}$ first, then read off:

• In quadrant I $(0^{\circ }$–$90^{\circ })$: $\sin +$, $\cos +$.
• In quadrant II $(90^{\circ }$–$180^{\circ })$: $\sin +$, $\cos -$.
• In quadrant III $(180^{\circ }$–$270^{\circ })$: $\sin -$, $\cos -$.
• In quadrant IV $(270^{\circ }$–$360^{\circ })$: $\sin -$, $\cos +$.

Worked case (all covered). Given $\cos A=\frac{7}{25}$ with $A$ in quadrant IV ($270^{\circ }<A<360^{\circ }$). Halving: $135^{\circ }<\frac{A}{2}<180^{\circ }$, i.e. quadrant II, where $\sin >0$. So: $\sin \frac{A}{2}=+\sqrt{\frac{1-\frac{7}{25}}{2}}=\sqrt{\frac{18/25}{2}}=\sqrt{\frac{9}{25}}=\frac{3}{5}.$

PICTURE. The circle split into its four quadrants, each stamped with the $(\pm ,\pm )$ sign of $(\sin ,\cos )$, and an arrow showing $A$ (in IV) halving down into II.

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Step 8 — Degenerate and boundary checks (never hit an unshown case)

WHAT. Test the formulas at the extreme inputs to be sure nothing breaks.

• $A=0^{\circ }$: $\cos A=1$. Then ${\sin }^2\frac{A}{2}=\frac{1-1}{2}=0$, so $\sin 0^{\circ }=0$. ✓ And ${\cos }^2\frac{A}{2}=\frac{1+1}{2}=1$, so $\cos 0^{\circ }=1$. ✓
• $A=180^{\circ }$: $\cos A=-1$. Then ${\sin }^2\frac{A}{2}=\frac{1-(-1)}{2}=1$, so $\sin 90^{\circ }=1$. ✓ This also kills the tempting wrong rule $\sin \frac{A}{2}=\frac{1}{2}\sin A$: here $\frac{1}{2}\sin 180^{\circ }=0\ne 1$.
• $A=360^{\circ }$: $\cos A=1$, so $\sin \frac{A}{2}=\sin 180^{\circ }=0$ ✓ — but note $\frac{A}{2}=180^{\circ }$ is in quadrant II/III boundary; the formula still gives the correct $0$.
• Tangent blow-up: $\tan \frac{A}{2}=\frac{\sin A}{1+\cos A}$ has denominator $0$ exactly when $\cos A=-1$, i.e. $A=180^{\circ }$, where $\frac{A}{2}=90^{\circ }$ and $\tan 90^{\circ }$ is genuinely undefined. The formula's blow-up matches reality — it is honest, not broken. (Use the twin form $\frac{1-\cos A}{\sin A}$ near this point.)

WHY. A formula you cannot trust at $0$, $180^{\circ }$, $360^{\circ }$ is a formula you cannot trust anywhere. These pass.

PICTURE. A strip showing $\cos A$ sweeping from $1$ down to $-1$ and back, with the resulting ${\sin }^2\frac{A}{2}=\frac{1-\cos A}{2}$ sweeping oppositely from $0$ to $1$ — the mirror motion made visible.

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The one-picture summary

Everything above, compressed into a single flow: seed $\to$ split $\to$ relabel $\to$ solve $\to$ tangent $\to$ sign.

Recall Feynman: retell the whole walkthrough in plain words

We had one trustworthy fact: the machine that turns an angle $\theta$ into the cosine of double that angle, written $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$. That fact mixed two functions, so we used the circle's own Pythagoras rule (${\cos }^2+{\sin }^2=1$) to rewrite it two ways — one pure-cosine, one pure-sine. Then we simply renamed the doubled angle "$A$", which made the single angle "$A/2$" — and suddenly the machine ran backwards: give it the full angle's cosine, get the half angle's square. Solving each renamed line put the half-angle square alone, sharing a bottom of $2$ and differing only by a plus (cosine) or minus (sine) on top. Dividing sine-square by cosine-square gave the tangent; a conjugate trick then handed us a tangent form so clean it needs no $\pm$, because $\sin A$ already knows its own sign and the bottom $1+\cos A$ can never go negative. Last, we remembered the square root only spits out positives, so we pick the sign — by looking at which quadrant the half angle sits in, not the full one. Checking the corners ($0^{\circ }$, $180^{\circ }$, $360^{\circ }$) proved the formulas honest, including the one spot where the tangent legitimately blows up.

Connections

• Double angle formulas — the seed we grew everything from (Steps 1–3).
• Pythagorean identity — the trade-in slip used in Steps 2, 3 and 6.
• Weierstrass substitution — $t=\tan \frac{x}{2}$ is exactly Step 6's clean tangent.
• Exact trig values — $15^{\circ }$, $22.5^{\circ }$ fall out of these (parent's examples).
• Product-to-sum formulas — sibling algebra on the same seed identities.