Half Angle Formulas

Running the Machine Backwards

Double angle took us from θ\theta up to 2θ2\theta. Half angle goes the other way: from a full angle xx down to x2\frac{x}{2}.

And we do not need anything new. The half angle formulas come straight out of rearranging the cosine double angle forms.

The green ray turns at xx. The yellow ray turns at x2\frac{x}{2}, half as fast. The bars show that the height sinx2\sin\frac{x}{2} exactly matches 1cosx2\sqrt{\frac{1 - \cos x}{2}} at every step. That match is the formula.


Deriving the Sine Half Angle

Recall one of the cosine double angle forms:

cos2θ=12sin2θ\cos 2\theta = 1 - 2\sin^2\theta

Solve it for sin2θ\sin^2\theta:

sin2θ=1cos2θ2\sin^2\theta = \frac{1 - \cos 2\theta}{2}

This is true for any angle. So replace θ\theta with x2\frac{x}{2}, which turns 2θ2\theta into xx:

sin2x2=1cosx2\sin^2\frac{x}{2} = \frac{1 - \cos x}{2}

Take the square root:

sinx2=±1cosx2\sin\frac{x}{2} = \pm\sqrt{\frac{1 - \cos x}{2}}


Deriving the Cosine Half Angle

Start from the other cosine form:

cos2θ=2cos2θ1\cos 2\theta = 2\cos^2\theta - 1

Solve for cos2θ\cos^2\theta, then swap θ\theta for x2\frac{x}{2}:

cos2x2=1+cosx2\cos^2\frac{x}{2} = \frac{1 + \cos x}{2}

cosx2=±1+cosx2\cos\frac{x}{2} = \pm\sqrt{\frac{1 + \cos x}{2}}

Same shape as the sine version, just a plus sign under the root instead of a minus.


What the ± Means

The square root forces a choice: is the answer positive or negative?

The sign is decided by the quadrant of x2\frac{x}{2}, not the quadrant of xx.

Work out where x2\frac{x}{2} lands, then pick the sign that sine (or cosine) has in that quadrant. The formula gives the size; the quadrant gives the sign.


The Tangent Half Angle

Dividing sine by cosine gives the tangent version:

tanx2=±1cosx1+cosx\tan\frac{x}{2} = \pm\sqrt{\frac{1 - \cos x}{1 + \cos x}}

But tangent has two cleaner forms that carry the correct sign on their own, so you never have to guess:

tanx2=1cosxsinx=sinx1+cosx\tan\frac{x}{2} = \frac{1 - \cos x}{\sin x} = \frac{\sin x}{1 + \cos x}

These are the ones to reach for in practice.


Example

Suppose cosx=35\cos x = \frac{3}{5} and xx is acute, so x2\frac{x}{2} is also in the first quadrant (everything positive).

Sine half angle:

sinx2=1352=2/52=15=55\begin{aligned} \sin\frac{x}{2} &= \sqrt{\frac{1 - \frac{3}{5}}{2}} \\[0.5em] &= \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \frac{\sqrt{5}}{5} \end{aligned}

Cosine half angle:

cosx2=1+352=8/52=45=255\begin{aligned} \cos\frac{x}{2} &= \sqrt{\frac{1 + \frac{3}{5}}{2}} \\[0.5em] &= \sqrt{\frac{8/5}{2}} = \sqrt{\frac{4}{5}} = \frac{2\sqrt{5}}{5} \end{aligned}

We took the positive root for both because x2\frac{x}{2} sits in the first quadrant.


The Set to Know

FormulaResult
sinx2\sin\frac{x}{2}±1cosx2\pm\sqrt{\dfrac{1 - \cos x}{2}}
cosx2\cos\frac{x}{2}±1+cosx2\pm\sqrt{\dfrac{1 + \cos x}{2}}
tanx2\tan\frac{x}{2}1cosxsinx=sinx1+cosx\dfrac{1 - \cos x}{\sin x} = \dfrac{\sin x}{1 + \cos x}

Minus under the root for sine, plus for cosine, and let the quadrant of x2\frac{x}{2} settle the sign.