The double-angle formulas can be run in reverse. Solving them for $\sin^2$ and $\cos^2$ turns a square into a first power of cosine, a trick calculus uses constantly, and it also gives values at half of a known angle.
Key idea
Power-reducing: $\sin^2 u = \dfrac{1 - \cos 2u}{2}$ and $\cos^2 u = \dfrac{1 + \cos 2u}{2}$.
Half-angle: $$\sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}, \quad \cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}, \quad \tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}.$$ The sign comes from the quadrant of $\frac{\theta}{2}$.
Half-angle: $$\sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}, \quad \cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}, \quad \tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}.$$ The sign comes from the quadrant of $\frac{\theta}{2}$.
Why it works
Start from $\cos 2u = 1 - 2\sin^2 u$ and solve for $\sin^2 u$: $\sin^2 u = \frac{1 - \cos 2u}{2}$. Now let $u = \frac{\theta}{2}$, so $2u = \theta$, and take a square root. The root gives only the size; the sign depends on where the half angle actually is. For tangent, $\tan\frac{\theta}{2} = \frac{\sin(\theta/2)}{\cos(\theta/2)}$; multiplying top and bottom by $2\sin\frac{\theta}{2}$ gives $\frac{1 - \cos\theta}{\sin\theta}$, with no sign to choose.
Worked examples
Worked example: A half of a special angle
Find $\cos 15^\circ$ exactly.
- $15^\circ = \frac{30^\circ}{2}$, and $15^\circ$ is in Quadrant I, so the sign is $+$.
- $\cos 15^\circ = \sqrt{\dfrac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\dfrac{2 + \sqrt{3}}{4}}$.
- $= \dfrac{\sqrt{2 + \sqrt{3}}}{2}$.
Answer: $\frac{\sqrt{2 + \sqrt{3}}}{2} \approx 0.966$ (the same as $\frac{\sqrt{6} + \sqrt{2}}{4}$)
Worked example: Signs from the half angle
$\cos\theta = -\frac{7}{25}$ with $\pi < \theta < \frac{3\pi}{2}$. Find $\cos\frac{\theta}{2}$.
- Halve the inequality: $\frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$. The half angle is in Quadrant II, where cosine is negative.
- $\cos\frac{\theta}{2} = -\sqrt{\dfrac{1 - \frac{7}{25}}{2}} = -\sqrt{\dfrac{9}{25}}$.
Answer: $-\frac{3}{5}$
Worked example: Power reducing
Rewrite $\sin^2(3x)$ with no squares.
- Use $\sin^2 u = \frac{1 - \cos 2u}{2}$ with $u = 3x$.
- The angle doubles: $2u = 6x$.
Answer: $\frac{1 - \cos 6x}{2}$
Common mistake
Choosing the sign from $\theta$ instead of from $\frac{\theta}{2}$. In the second example, $\theta$ is in Quadrant III, where cosine is negative, but that is not why the answer is negative: it is negative because $\frac{\theta}{2}$ is in Quadrant II. Always halve the interval for $\theta$ first and look where the half angle lands.