Half-Angle Formulas

The half-angle formulas find the sine, cosine or tangent of \frac{\theta}{2} from \cos \theta: \sin \frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos \theta}{2}} and \cos \frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos \theta}{2}}. The quadrant of \frac{\theta}{2} decides the sign. They give exact values for angles like 15° and 22.5°, and they come from the double-angle formulas for cosine.

Before this lesson: You should know Double-Angle Formulas, Reference Angles and Simplifying Square Roots.

Halving an Angle

You know exact values for 30°, 45° and 60°. Halving those angles gives 15°, 22.5° and 30°. The half-angle formulas turn a known cosine into the sine or cosine of half the angle.

They come from the double-angle formula \cos 2x = 1 - 2\sin^2 x. Solve it for \sin^2 x to get \sin^2 x = \frac{1 - \cos 2x}{2}, then replace x with \frac{\theta}{2} and take the square root. The square root is why there is a \pm sign.

The Half-Angle Formulas

The tangent forms have no \pm, because their signs work out on their own. For sine and cosine, choose the sign from the quadrant of \frac{\theta}{2}, not of \theta.

1. Write the angle as half of a known angle, such as 15^\circ = \frac{30^\circ}{2}.
2. Find the quadrant of the half angle to choose the sign.
3. Substitute \cos \theta into the formula.
4. Simplify the fraction inside the root, then simplify the root.

Worked Examples

Example 1: Exact value of sin 15°

Find the exact value of \sin 15^\circ.

\sin 15^\circ = \sin \frac{30^\circ}{2}

\sin 15^\circ = \sqrt{\frac{1 - \cos 30^\circ}{2}}

\sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}

\sqrt{\frac{2 - \sqrt{3}}{4}}

\frac{\sqrt{2 - \sqrt{3}}}{2}

Answer: \sin 15^\circ = \frac{\sqrt{2 - \sqrt{3}}}{2} \approx 0.2588. This equals \frac{\sqrt{6} - \sqrt{2}}{4} from the sum and difference formulas, written another way.

Example 2: Exact value of cos 22.5°

Find the exact value of \cos 22.5^\circ.

\cos 22.5^\circ = \cos \frac{45^\circ}{2}

\sqrt{\frac{1 + \cos 45^\circ}{2}}

\sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}

\sqrt{\frac{2 + \sqrt{2}}{4}}

\frac{\sqrt{2 + \sqrt{2}}}{2}

Answer: \cos 22.5^\circ = \frac{\sqrt{2 + \sqrt{2}}}{2} \approx 0.9239

Example 3: Tangent of a half angle

Find the exact value of \tan \frac{\pi}{8}.

\tan \frac{\pi}{8} = \frac{1 - \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}

\frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}

\frac{2 - \sqrt{2}}{\sqrt{2}}

\frac{(2 - \sqrt{2})\sqrt{2}}{2} = \frac{2\sqrt{2} - 2}{2}

\sqrt{2} - 1

Answer: \tan \frac{\pi}{8} = \sqrt{2} - 1 \approx 0.414

Example 4: Choose the signs

If \cos \theta = -\frac{7}{25} and 180^\circ \lt \theta \lt 270^\circ, find \sin \frac{\theta}{2} and \cos \frac{\theta}{2}.

Divide the interval by 2 to locate the half angle: 90^\circ \lt \frac{\theta}{2} \lt 135^\circ, which is Quadrant II

\sin \frac{\theta}{2} = \sqrt{\frac{1 - (-\frac{7}{25})}{2}}

\sqrt{\frac{\frac{32}{25}}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}

\cos \frac{\theta}{2} = -\sqrt{\frac{1 + (-\frac{7}{25})}{2}}

-\sqrt{\frac{\frac{18}{25}}{2}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}

Answer: \sin \frac{\theta}{2} = \frac{4}{5} and \cos \frac{\theta}{2} = -\frac{3}{5}

Example 5: Word problem with a sonic boom

When a jet flies faster than sound, it makes a cone-shaped shock wave. If the cone’s full angle is \theta, the jet’s Mach number is M = \frac{1}{\sin \frac{\theta}{2}}. Use a half-angle formula to find M when \theta = 90^\circ.

\sin \frac{90^\circ}{2} = \sqrt{\frac{1 - \cos 90^\circ}{2}}

\sqrt{\frac{1 - 0}{2}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}

M = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}

\sqrt{2} \approx 1.41

Answer: The jet is flying at about Mach 1.41, about 1.41 times the speed of sound. A narrower cone means a faster jet.

Common Mistakes

Related Methods

Power-reducing formulas

Squaring the half-angle formulas gives \sin^2 x = \frac{1 - \cos 2x}{2} and \cos^2 x = \frac{1 + \cos 2x}{2}. Calculus uses these to integrate \sin^2 x and \cos^2 x.

Two ways to find the same value

\sin 15^\circ can come from 45^\circ - 30^\circ or from half of 30°. The answers look different, \frac{\sqrt{6} - \sqrt{2}}{4} and \frac{\sqrt{2 - \sqrt{3}}}{2}, but both are about 0.2588.

Halving again

You can apply the formulas more than once: 22.5° is half of 45°, and 11.25° is half of 22.5°. Each step adds another nested square root.

Practice Problems

1. Find the exact value of \sin 22.5^\circ.
2. Find the exact value of \cos 15^\circ.
3. Find the exact value of \tan 15^\circ.
4. Find the exact value of \sin 105^\circ using a half-angle formula.
5. Find the exact value of \cos 165^\circ using a half-angle formula.
6. If \cos \theta = \frac{3}{5} and \theta is in Quadrant I, find \sin \frac{\theta}{2}.
7. Use the angle in problem 6 to find \cos \frac{\theta}{2}.
8. Use the angle in problem 6 to find \tan \frac{\theta}{2}.
9. If \cos \theta = -\frac{5}{13} and \theta is in Quadrant II, find \sin \frac{\theta}{2}.
10. A jet’s shock cone has a full angle of 120°. Use M = \frac{1}{\sin \frac{\theta}{2}} and a half-angle formula to find its Mach number, to the nearest hundredth.

\sqrt{\frac{1 - \cos 45^\circ}{2}}

\sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{2}}{4}}

\frac{\sqrt{2 - \sqrt{2}}}{2}

\sqrt{\frac{1 + \cos 30^\circ}{2}}

\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}}

\frac{\sqrt{2 + \sqrt{3}}}{2}

\frac{1 - \cos 30^\circ}{\sin 30^\circ}

\frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}}

\frac{2 - \sqrt{3}}{1}

\sqrt{\frac{1 - \cos 210^\circ}{2}}

\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}}

\frac{\sqrt{2 + \sqrt{3}}}{2}

-\sqrt{\frac{1 + \cos 330^\circ}{2}}

-\sqrt{\frac{2 + \sqrt{3}}{4}}

-\frac{\sqrt{2 + \sqrt{3}}}{2}

\sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{1}{5}}

\frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}

\sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{8}{5}}{2}} = \sqrt{\frac{4}{5}}

\frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}

\sin \theta = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}

\frac{\frac{4}{5}}{1 + \frac{3}{5}} = \frac{\frac{4}{5}}{\frac{8}{5}}

\frac{1}{2}

45^\circ \lt \frac{\theta}{2} \lt 90^\circ

\sqrt{\frac{1 - (-\frac{5}{13})}{2}} = \sqrt{\frac{\frac{18}{13}}{2}} = \sqrt{\frac{9}{13}}

\frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}

\sin 60^\circ = \sqrt{\frac{1 - \cos 120^\circ}{2}}

\sqrt{\frac{1 + \frac{1}{2}}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}

M = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

\frac{2\sqrt{3}}{3} \approx 1.15

FAQ

What are the half-angle formulas?

\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}.

How do you choose the plus or minus sign?

Find the quadrant of the half angle \frac{\theta}{2}, then use the sign of sine or cosine in that quadrant. The quadrant of \theta itself does not decide the sign.

Where do the half-angle formulas come from?

They come from the double-angle formulas for cosine. Solving \cos 2x = 1 - 2\sin^2 x for \sin x and replacing x with \frac{\theta}{2} gives the sine formula.

Why does tan(θ/2) have no plus or minus sign?

The forms \frac{1 - \cos \theta}{\sin \theta} and \frac{\sin \theta}{1 + \cos \theta} have no square root, so their signs come out correctly by themselves.

Summary

• \sin \frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos \theta}{2}} and \cos \frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos \theta}{2}}.
• \tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta} = \frac{\sin \theta}{1 + \cos \theta}.
• The sign depends on the quadrant of \frac{\theta}{2}.
• They come from the double-angle formulas for cosine.
• Use them for exact values like 15°, 22.5° and 105°.

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