Math Core

The double-angle formulas take you from $\theta$ to $2\theta$ . Running them backward takes you from $\theta$ to $\dfrac{\theta}{2}$ . That lets you find exact values like $\sin 22.5^\circ$ , and it lets you rewrite squared trig functions without the squares, a step you'll need often in calculus.

Power-reducing formulas

Start with two forms of $\cos 2x$ and solve each one for the squared term.

\cos 2x = 1 - 2\sin^2 x \;\Longrightarrow\; \sin^2 x = \frac{1 - \cos 2x}{2} \qquad\qquad \cos 2x = 2\cos^2 x - 1 \;\Longrightarrow\; \cos^2 x = \frac{1 + \cos 2x}{2}

Dividing the first by the second gives $\tan^2 x = \dfrac{1 - \cos 2x}{1 + \cos 2x}$ .

These are the power-reducing formulas. They trade a square for a first power of cosine at double the angle.

Worked example: Reducing a fourth power

Rewrite $\sin^4 x$ using only first powers of cosine.

\begin{aligned} \sin^4 x &= \left(\sin^2 x\right)^2 = \left(\frac{1 - \cos 2x}{2}\right)^2 \\ &= \frac{1 - 2\cos 2x + \cos^2 2x}{4} \\ &= \frac{1}{4}\left(1 - 2\cos 2x + \frac{1 + \cos 4x}{2}\right) && \text{reduce } \cos^2 2x \\ &= \frac{3 - 4\cos 2x + \cos 4x}{8} \end{aligned}

Check at $x = 90^\circ$ : $\sin^4 90^\circ = 1$ , and $\dfrac{3 - 4\cos 180^\circ + \cos 360^\circ}{8} = \dfrac{3 + 4 + 1}{8} = 1$ .

The half-angle formulas

Now substitute $x = \dfrac{\theta}{2}$ into the power-reducing formulas, so $2x = \theta$ , and take square roots.

Half-angle formulas

\sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}} \qquad\qquad \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 of sine or cosine is chosen by the quadrant of $\dfrac{\theta}{2}$ , not the quadrant of $\theta$ . The tangent formulas have no $\pm$ ; the sign takes care of itself.

Why no $\pm$ for tangent? Multiply the top and bottom of $\dfrac{\sin(\theta/2)}{\cos(\theta/2)}$ by $2\sin\dfrac{\theta}{2}$ . The numerator becomes $2\sin^2\dfrac{\theta}{2} = 1 - \cos\theta$ and the denominator becomes $2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2} = \sin\theta$ . No square root was ever taken.

Common mistake

Decide the sign using the half angle. If $\theta = 250^\circ$ (Quadrant III), then $\dfrac{\theta}{2} = 125^\circ$ (Quadrant II), so $\sin\dfrac{\theta}{2}$ is positive and $\cos\dfrac{\theta}{2}$ is negative. Using the quadrant of $\theta$ instead is the classic error.

Exact values

Worked example: Sine of 15°

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

$15^\circ$ is half of $30^\circ$ , and $15^\circ$ is in Quadrant I, so take the positive root:

\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}.

In the last lesson you found $\sin 15^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4}$ . Both are $\approx 0.2588$ . They look different but are the same number.

Worked example: A negative half-angle value

Find the exact value of $\cos 112.5^\circ$ .

$112.5^\circ$ is half of $225^\circ$ , and $\cos 225^\circ = -\dfrac{\sqrt{2}}{2}$ . Since $112.5^\circ$ is in Quadrant II, cosine is negative:

\cos 112.5^\circ = -\sqrt{\frac{1 + \cos 225^\circ}{2}} = -\sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = -\frac{\sqrt{2 - \sqrt{2}}}{2}.

Worked example: Tangent of 22.5°

Find the exact value of $\tan 22.5^\circ$ .

Use the form without a square root, with $\theta = 45^\circ$ :

\tan 22.5^\circ = \frac{1 - \cos 45^\circ}{\sin 45^\circ} = \frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2} - 2}{2} = \sqrt{2} - 1.

Half angles from given information

Worked example: Given a cosine and a quadrant

Suppose $\cos\theta = -\dfrac{7}{25}$ and $180^\circ < \theta < 270^\circ$ . Find $\sin\dfrac{\theta}{2}$ , $\cos\dfrac{\theta}{2}$ and $\tan\dfrac{\theta}{2}$ .

Dividing the inequality by $2$ gives $90^\circ < \dfrac{\theta}{2} < 135^\circ$ , which is Quadrant II. So sine is positive and cosine is negative.

\begin{aligned} \sin\frac{\theta}{2} &= +\sqrt{\frac{1 - \left(-\frac{7}{25}\right)}{2}} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \\ \cos\frac{\theta}{2} &= -\sqrt{\frac{1 + \left(-\frac{7}{25}\right)}{2}} = -\sqrt{\frac{18/25}{2}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5} \\ \tan\frac{\theta}{2} &= \frac{4/5}{-3/5} = -\frac{4}{3} \end{aligned}

Tip

Check a half-angle result with a Pythagorean identity: $\left(\tfrac{4}{5}\right)^2 + \left(-\tfrac{3}{5}\right)^2 = \tfrac{16}{25} + \tfrac{9}{25} = 1$ . Good.

Choosing a formula

With so many related formulas, it helps to know which one to reach for.

You want	Use
to remove a square such as $\cos^2 x$	a power-reducing formula
an exact value at half of a special angle ( $15^\circ$ , $22.5^\circ$ , $75^\circ$ , $112.5^\circ$ )	a half-angle formula
$\tan\dfrac{\theta}{2}$	$\dfrac{1 - \cos\theta}{\sin\theta}$ or $\dfrac{\sin\theta}{1 + \cos\theta}$ , no sign decision needed
$\sin\dfrac{\theta}{2}$ or $\cos\dfrac{\theta}{2}$ from given data	the half-angle formula, with the sign from the quadrant of $\dfrac{\theta}{2}$

Half-angle answers often come out as nested radicals like $\sqrt{2 - \sqrt{3}}$ . That is a perfectly good exact answer. Sometimes a nested radical can be rewritten more simply (as with $\sin 15^\circ$ ), but you are not expected to find those rewrites on your own. If you want to make sure two exact forms agree, compare their decimal values.

For the sign decision, always write the inequality for $\theta$ and divide every part by $2$ . That one line tells you exactly which quadrant the half angle is in, and it takes the guesswork out of choosing $+$ or $-$ .