Half Angle Formula — Sin, Cos, Tan, Proof

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.

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

1. Find $\cos 15°$ exactly using the half angle formula.

2. Find $\tan\frac{\pi}{8}$ exactly, and confirm it equals $\sqrt2 - 1$.

3. 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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