The half angle formulas — also called half angle identities — give:
$\sin\frac{\theta}{2}$ from $\cos\theta$,
$\cos\frac{\theta}{2}$ from $\cos\theta$,
$\tan\frac{\theta}{2}$ from $\sin\theta$ and $\cos\theta$.
Each one carries a $\pm$ that you resolve by asking a single question: which quadrant does $\frac{\theta}{2}$ live in? Get that one habit right and these identities stop being a trap.
What Is the Half Angle Formula?
The half angle formula is a set of three trigonometric identities that express the sine, cosine, and tangent of half an angle, $\frac{\theta}{2}$, in terms of the trig functions of the full angle $\theta$. They are the reverse direction of the double-angle identities, and they let you find exact values for angles like $15°$, $22.5°$, and $\frac{\pi}{12}$ that never appear on the standard unit circle.
$$\boxed{;\sin\frac{\theta}{2} = \pm\sqrt{\dfrac{1 - \cos\theta}{2}};}$$
$$\boxed{;\cos\frac{\theta}{2} = \pm\sqrt{\dfrac{1 + \cos\theta}{2}};}$$
$$\boxed{;\tan\frac{\theta}{2} = \pm\sqrt{\dfrac{1 - \cos\theta}{1 + \cos\theta}} = \dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1 + \cos\theta};}$$
The tangent identity has two sign-free forms (the last two), and they are the ones worth using — because they decide the sign for you, no quadrant lookup needed. More on that below.
How Is the Half Angle Formula Derived?
The cleanest proof starts from a double-angle identity for cosine and works backwards. Cosine has three double-angle forms; two of them are built for exactly this.
The cosine half-angle. Start from $\cos 2x = 2\cos^2 x - 1$. Let $x = \frac{\theta}{2}$, so $2x = \theta$:
$$\cos\theta = 2\cos^2\frac{\theta}{2} - 1.$$
Solve for $\cos^2\frac{\theta}{2}$:
$$\cos^2\frac{\theta}{2} = \frac{1 + \cos\theta}{2}, \qquad \cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}.$$
The square root is where the $\pm$ enters — squaring lost the sign, so we recover it from the quadrant.
The sine half-angle. Use the other form, $\cos 2x = 1 - 2\sin^2 x$, again with $x = \frac{\theta}{2}$:
$$\cos\theta = 1 - 2\sin^2\frac{\theta}{2} ;\Rightarrow; \sin^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2} ;\Rightarrow; \sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}.$$
The tangent half-angle. Divide sine by cosine:
$$\tan\frac{\theta}{2} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \pm\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}.$$
Rationalising this radical produces the two sign-free forms. Multiply top and bottom inside the root by $(1 + \cos\theta)$, then use $\sin^2\theta = 1 - \cos^2\theta$:
$$\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta} = \frac{\sin\theta}{1 + \cos\theta}.$$
These two carry no ambiguous sign because $\sin\theta$ and $(1 - \cos\theta)$ already encode it. That is the version that saves students from quadrant errors.
How Do You Choose the ± Sign by Quadrant?
The sign on $\sin\frac{\theta}{2}$ and $\cos\frac{\theta}{2}$ is not part of the algebra — it comes from where $\frac{\theta}{2}$ sits. Halve the angle first, then read the quadrant.
Quadrant of $\frac{\theta}{2}$ | $\sin\frac{\theta}{2}$ | $\cos\frac{\theta}{2}$ | $\tan\frac{\theta}{2}$ |
|---|---|---|---|
I ($0$ to $90°$) | $+$ | $+$ | $+$ |
II ($90°$ to $180°$) | $+$ | $-$ | $-$ |
III ($180°$ to $270°$) | $-$ | $-$ | $+$ |
IV ($270°$ to $360°$) | $-$ | $+$ | $-$ |
The mnemonic "All Students Take Calculus" marks which functions are positive in each quadrant (All, Sine, Tangent, Cosine). The subtle part: if $\theta$ is in Quadrant II, then $\frac{\theta}{2}$ is in Quadrant I — halving moves the angle backward toward the axis. Always evaluate $\frac{\theta}{2}$, never $\theta$, before picking the sign.
Examples of the Half Angle Formula
Example 1
Find $\sin 15°$ exactly using the half angle formula.
Write $15° = \frac{30°}{2}$, so $\theta = 30°$ and $\cos 30° = \frac{\sqrt{3}}{2}$. Since $15°$ is in Quadrant I, take the positive root.
$$\sin 15° = \sqrt{\frac{1 - \cos 30°}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}.$$
In radians, $15° = \frac{\pi}{12}$. Final answer: $\sin 15° = \dfrac{\sqrt{2 - \sqrt{3}}}{2} \approx 0.259$.
Example 2
Find $\cos\dfrac{\theta}{2}$ when $\cos\theta = \dfrac{3}{5}$ and $\dfrac{\theta}{2}$ is in Quadrant II.
A natural first move is to drop the $\pm$ and just take the positive root, the way most square roots behave by default.
Wrong path. Writing $\cos\frac{\theta}{2} = +\sqrt{\frac{1 + 3/5}{2}} = +\sqrt{\frac{4}{5}} = \frac{2}{\sqrt5}$. The arithmetic is correct, but the sign is wrong. A cosine in Quadrant II must be negative — and $+\frac{2}{\sqrt5}$ is positive. The answer contradicts the quadrant it was supposed to live in.
The break. The radical only ever returns a non-negative number. The $\pm$ in front is not decoration; it is the entire job of the quadrant rule. Skipping it means the formula can never produce a negative cosine, even when the angle demands one.
The rescue. Quadrant II forces a negative cosine:
$$\cos\frac{\theta}{2} = -\sqrt{\frac{1 + \frac{3}{5}}{2}} = -\sqrt{\frac{8/5}{2}} = -\sqrt{\frac{4}{5}} = -\frac{2}{\sqrt5} = -\frac{2\sqrt5}{5}.$$
Final answer: $\cos\frac{\theta}{2} = -\dfrac{2\sqrt5}{5} \approx -0.894$.
Example 3
Find $\tan 22.5°$ exactly.
Write $22.5° = \frac{45°}{2}$, so $\theta = 45°$. Use the sign-free form $\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}$ to skip the quadrant lookup entirely. With $\cos 45° = \sin 45° = \frac{\sqrt2}{2}$:
$$\tan 22.5° = \frac{1 - \frac{\sqrt2}{2}}{\frac{\sqrt2}{2}} = \frac{2 - \sqrt2}{\sqrt2} = \frac{2}{\sqrt2} - 1 = \sqrt2 - 1.$$
In radians, $22.5° = \frac{\pi}{8}$. Final answer: $\tan 22.5° = \sqrt2 - 1 \approx 0.414$.
Example 4
Find $\cos\dfrac{\pi}{8}$ exactly (radian form).
Here $\frac{\pi}{8} = \frac{1}{2}\cdot\frac{\pi}{4}$, so $\theta = \frac{\pi}{4}$ and $\cos\frac{\pi}{4} = \frac{\sqrt2}{2}$. Since $\frac{\pi}{8}$ is in Quadrant I, take the positive root.
$$\cos\frac{\pi}{8} = \sqrt{\frac{1 + \frac{\sqrt2}{2}}{2}} = \sqrt{\frac{2 + \sqrt2}{4}} = \frac{\sqrt{2 + \sqrt2}}{2}.$$
Final answer: $\cos\frac{\pi}{8} = \dfrac{\sqrt{2 + \sqrt2}}{2} \approx 0.924$. (In degrees this is $\cos 22.5°$ — same value, two notations.)
Example 5
Given $\sin\theta = \dfrac{4}{5}$ with $\theta$ in Quadrant I, find $\sin\dfrac{\theta}{2}$.
The sine half-angle needs $\cos\theta$, not $\sin\theta$. Get it from the Pythagorean identity — see our Pythagorean identities walk-through for the full derivation. Since $\theta$ is in Quadrant I, $\cos\theta = +\sqrt{1 - \frac{16}{25}} = \frac{3}{5}$. With $\theta$ in Quadrant I, $\frac{\theta}{2}$ is also in Quadrant I, so the root is positive:
$$\sin\frac{\theta}{2} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt5} = \frac{\sqrt5}{5}.$$
Final answer: $\sin\frac{\theta}{2} = \dfrac{\sqrt5}{5} \approx 0.447$.
Example 6
Prove the identity $\dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1 + \cos\theta}$ (the two tangent half-angle forms agree).
Cross-multiply and check both sides expand to the same thing:
$$(1 - \cos\theta)(1 + \cos\theta) = 1 - \cos^2\theta = \sin^2\theta = \sin\theta \cdot \sin\theta.$$
Both sides equal $\sin^2\theta$, so the two forms are identical wherever $\sin\theta \neq 0$. This is why you can pick whichever form has the friendlier arithmetic. Final answer: identity verified.
Where the Half Angle Formula Earns Its Keep
These identities are not a textbook curiosity. Wherever a calculation needs an angle that isn't on the standard table, halving is the route in.
Astronomy and the first trig tables. Ptolemy's Almagest (c. 150 CE) built a chord table in half-degree steps by repeatedly halving known chords — the geometric ancestor of $\sin\frac{\theta}{2}$. Without it, no almanac could predict where a planet would rise.
Calculus — the Weierstrass substitution. Setting $t = \tan\frac{x}{2}$ turns any rational function of $\sin x$ and $\cos x$ into a rational function of $t$, which integrates by ordinary algebra. The tangent half-angle is the engine of that substitution.
Computer graphics — quaternion rotation. A rotation by angle $\theta$ about an axis is encoded as a quaternion using $\cos\frac{\theta}{2}$ and $\sin\frac{\theta}{2}$. Every smooth camera turn in a game or simulation runs on half-angles.
Surveying and navigation. The haversine formula, used to compute great-circle distances between two points on Earth, is built directly from $\sin^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2}$.
For Class 11 and high-school students, the everyday context is just exact-value computation — but the same identity carries straight into JEE-level integration and the rotation math behind 3D engines.
Tripping Points to Avoid
Mistake 1: Checking the quadrant of θ instead of θ/2
Where it slips in: When $\theta$ is given in Quadrant III or IV and a student reads the sign off $\theta$ rather than the halved angle.
Don't do this: Assume $\frac{\theta}{2}$ shares $\theta$'s quadrant. It almost never does — halving an angle moves it.
The correct way: Compute the numeric value of $\frac{\theta}{2}$, locate that quadrant, then pick the sign.
Mistake 2: Confusing the half-angle formula with the double-angle formula
Where it slips in: Reaching for $\sin\frac{\theta}{2} = 2\sin\frac{\theta}{4}\cos\frac{\theta}{4}$ — applying double-angle structure to a half-angle problem.
Don't do this: Mix the two families. Double-angle goes from $\theta$ up to $2\theta$; half-angle goes from $\theta$ down to $\frac{\theta}{2}$, and the half-angle identities live under a square root.
The correct way: Half-angle uses $\cos\theta$ inside a radical: $\sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}$. If there is no square root in your line of work, you are not using the half-angle formula. The memorizer archetype — fluent at reciting both formula sets — is the one who freezes here, because the cue ($\frac{\theta}{2}$ vs $2\theta$) and the formula have come unlinked.
Mistake 3: Forgetting that the tangent sign-free forms already carry the sign
Where it slips in: Slapping a $\pm$ in front of $\frac{1 - \cos\theta}{\sin\theta}$ out of habit.
Don't do this: Add a manual sign to the rationalised tangent forms. They are exact, not ambiguous.
The correct way: Use $\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}$ as written — the sign of $\sin\theta$ in the denominator delivers the correct result automatically. Reserve the $\pm$ for the radical form only.
Conclusion
The half angle formula gives $\sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}$, $\cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}$, and $\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}$.
The $\pm$ sign is decided by the quadrant of $\frac{\theta}{2}$ — always halve first, then read the sign.
The two rationalised tangent forms carry the sign automatically, so they sidestep the most common error.
Each identity comes from a double-angle cosine formula solved backwards under a square root.
The half-angle is the foundation of the Weierstrass integration substitution, quaternion 3D rotation, and the haversine distance formula.
Practice These Before Moving On
Find $\cos 15°$ exactly using the half angle formula.
Find $\tan\frac{\pi}{8}$ exactly, and confirm it equals $\sqrt2 - 1$.
Given $\cos\theta = -\frac{7}{25}$ with $\frac{\theta}{2}$ in Quadrant II, find $\sin\frac{\theta}{2}$ and $\cos\frac{\theta}{2}$.
If Problem 3 gave you a positive cosine, return to Mistake 1 and re-check the quadrant of $\frac{\theta}{2}$.
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