Isolate the squares

We start with the identities $\cos2u=1-2\sin^2u$ and $\cos2u=2\cos^2u-1$. Making $u=x/2$ and isolating:

$$\sin^2\frac x2=\frac{1-\cos x}{2},\qquad \cos^2\frac x2=\frac{1+\cos x}{2}.$$

Consequently:

$$\sin\frac x2=\pm\sqrt{\frac{1-\cos x}{2}},\qquad \cos\frac x2=\pm\sqrt{\frac{1+\cos x}{2}}.$$

The signs are not a free choice: they are determined by the image quadrant of $x/2$. If $0

Half may be in another quadrant

If $x=300°$, we have $x/2=150°$. Then:

$$\sin150°=\sqrt{\frac{1-1/2}{2}}=\frac12,$$

$$\cos150°=-\sqrt{\frac{1+1/2}{2}}=-\frac{\sqrt3}{2}.$$

The negative sign of the cosine arises from the second quadrant.

Demonstration of the tangent of half

Using $\sin x=2\sin(x/2)\cos(x/2)$ and $1+\cos x=2\cos^2(x/2)$:

$$\frac{\sin x}{1+\cos x}=\tan\frac x2,$$

as long as $\cos(x/2)\ne0$. The condition is equivalent to $1+\cos x\ne0$.

Another way is:

$$\tan\frac x2=\frac{1-\cos x}{\sin x},$$

but it also requires $\sin x\ne0$. For $x=0$, the first form is valid 0 and the second does not exist. The expressions coincide in the common domain, without having exactly the same domain.

Squared:

$$\tan^2\frac x2=\frac{1-\cos x}{1+\cos x}.$$

Replacing with a parameter

If $t=\tan(x/2)$ exists, the previous identities give:

$$\sin x=\frac{2t}{1+t^2},\qquad \cos x=\frac{1-t^2}{1+t^2}.$$

For the tangent, $\tan x=2t/(1-t^2)$ also requires $t^2\ne1$. This substitution can transform trigonometric equations into algebraic equations; points where $t$ does not exist must be checked separately.

In everyday life

  • Distances on Earth. The haversine of an angle is $\operatorname{sen}^2\frac{\theta}{2}$, an arc-half formula used by map applications.
  • Computer design. CAD programs describe arcs with $t = \operatorname{tg}\frac{\theta}{2}$, which makes sine and cosine fractions of polynomials.
  • Cutting angles. To splice two boards at a corner of $90^\circ$, each is cut at $45^\circ$: half the angle.
Miter: To form a $90^\circ$ corner, each board is cut at half the angle, $45^\circ$.

Exercises with resolution

1. Calculate $\cos22{,}5°$.

Check resolution 1

The angle is in the first quadrant: $\cos22{,}5°=\sqrt{(1+\sqrt2/2)/2}=\sqrt{2+\sqrt2}/2$.

2. Calculate $\tan15°$ using the formula for half of $30°$.

Check resolution 2

$\tan15°=(1/2)/(1+\sqrt3/2)=1/(2+\sqrt3)=2-\sqrt3$.

In the interactive book

The same subjects, with all the demonstrations and the complete list of exercises, are in the interactive Trigonometry book:

This class is a shorter introduction to the topic; In the book, the theory is accompanied by solved and proposed exercises from the chapter.

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