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When you first dive into trigonometry, you’ll encounter a powerful set of tools called half‑angle identities. These formulas let you translate trigonometric expressions that involve θ/2 into expressions that use the more familiar angle θ. In practice, they help you either simplify an expression or compute the exact value of a trigonometric function when the argument is half of a well‑known angle.

Core Half‑Angle Identities

Below are the primary identities you’ll need. While many texts present them in one form, each can be algebraically transformed into several useful variations.

Half‑Angle Identity for Sine

\(\sin\bigg(\frac{θ}{2}\bigg) = ±\sqrt{\frac{1 – \cosθ}{2}}\)

Half‑Angle Identity for Cosine

\(\cos\bigg(\frac{θ}{2}\bigg) = ±\sqrt{\frac{1 + \cosθ}{2}}\)

Half‑Angle Identities for Tangent

\(\tan\bigg(\frac{θ}{2}\bigg) = ±\sqrt{\frac{1 -\cosθ}{1 + \cosθ}}\)

\(\tan\bigg(\frac{θ}{2}\bigg) = \frac{\sinθ}{1 + \cosθ}\)

\(\tan\bigg(\frac{θ}{2}\bigg) = \frac{1 – \cosθ}{\sinθ}\)

\(\tan\bigg(\frac{θ}{2}\bigg) = \cscθ – \cotθ\)

Half‑Angle Identities for Cotangent

\(\cot\bigg(\frac{θ}{2}\bigg) = ±\sqrt{\frac{1 + \cosθ}{1 – \cosθ}}\)

\(\cot\bigg(\frac{θ}{2}\bigg) = \frac{\sinθ}{1 – \cosθ}\)

\(\cot\bigg(\frac{θ}{2}\bigg) = \frac{1 + \cosθ}{\sinθ}\)

\(\cot\bigg(\frac{θ}{2}\bigg) = \cscθ + \cotθ\)

Practical Example: Computing sin 15°

Let’s walk through how to apply these identities to find the exact value of sin 15°, an angle that is not part of the standard 30°, 45°, or 60° family.

1. Express the angle as half of a known value

Set θ/2 = 15°, giving θ = 30°. Since 30° is a familiar angle, we can use the sine half‑angle identity.

2. Select the appropriate formula

Because we need sin, we use: \(\sin\bigg(\frac{θ}{2}\bigg) = ±\sqrt{\frac{1 – \cosθ}{2}}\)

3. Resolve the ± sign

The sign depends on the quadrant. Here θ = 30° lies in Quadrant I, where sine is positive, so we drop the negative option.

4. Substitute known values

Replace cos 30° with its exact value \(\sqrt{3}/2\): \(\sin(15°) = \sqrt{\frac{1 – \sqrt{3}/2}{2}}\)

5. Simplify

Multiply numerator and denominator inside the root by 2 to clear the fraction: \(\sin(15°) = \sqrt{\frac{2(1 – \sqrt{3}/2)}{4}}\)

Which simplifies to: \(\sin(15°) = \sqrt{\frac{2 – \sqrt{3}}{4}}\)

Finally, factor the square root of 4: \(\sin(15°) = \frac{1}{2}\sqrt{2 – \sqrt{3}}\)

Thus, the exact value of sin 15° is \(\frac{1}{2}\sqrt{2 – \sqrt{3}}\).

Quick Quadrant Reference for Sign Determination

• Quadrant I: all functions are positive.
• Quadrant II: sine and cosecant are positive.
• Quadrant III: tangent and cotangent are positive.
• Quadrant IV: cosine and secant are positive.

By following these steps, you can confidently apply half‑angle identities to any trigonometric problem, whether you’re simplifying an expression or finding an exact value.