Half-Angle and Power-Reducing Formulas

Half-angle formulas let you find exact trig values for angles like 22.5° (half of 45°) or 15° (half of 30°) — angles that do not appear in the standard reference table but whose halved parents do. Power-reducing formulas convert squared trig functions into expressions with no exponents, which is essential for integration in calculus and for RMS (root mean square) calculations in electrical engineering. The two formula families are closely related: the half-angle formulas are derived directly from the power-reducing formulas.

Power-Reducing Formulas

Power-reducing formulas are the simpler family, so we start here. They rewrite squared trig functions using only first-power cosines:

$\sin^2\theta = \dfrac{1 - \cos(2\theta)}{2}$

$\cos^2\theta = \dfrac{1 + \cos(2\theta)}{2}$

$\tan^2\theta = \dfrac{1 - \cos(2\theta)}{1 + \cos(2\theta)}$

Where Do They Come From?

Start from the double-angle identity for cosine:

$\cos(2\theta) = 1 - 2\sin^2\theta$

Solve for $\sin^2\theta$: add $2\sin^2\theta$ to both sides, subtract $\cos(2\theta)$, then divide by 2:

$\sin^2\theta = \dfrac{1 - \cos(2\theta)}{2}$

Now use the other double-angle form $\cos(2\theta) = 2\cos^2\theta - 1$. Add 1 to both sides and divide by 2:

$\cos^2\theta = \dfrac{1 + \cos(2\theta)}{2}$

The tangent formula follows by dividing: $\tan^2\theta = \sin^2\theta / \cos^2\theta$.

Worked Example 1: Rewrite $\sin^4\theta$ Without Exponents

Write $\sin^4\theta$ as $(\sin^2\theta)^2$ and apply the power-reducing formula:

$\sin^4\theta = \left(\dfrac{1 - \cos(2\theta)}{2}\right)^2 = \dfrac{1 - 2\cos(2\theta) + \cos^2(2\theta)}{4}$

The $\cos^2(2\theta)$ term still has an exponent. Apply the power-reducing formula again, replacing $\theta$ with $2\theta$:

$\cos^2(2\theta) = \dfrac{1 + \cos(4\theta)}{2}$

Substitute back:

$\sin^4\theta = \dfrac{1 - 2\cos(2\theta) + \dfrac{1 + \cos(4\theta)}{2}}{4}$

Multiply numerator and denominator to clear the inner fraction:

$= \dfrac{2 - 4\cos(2\theta) + 1 + \cos(4\theta)}{8} = \dfrac{3 - 4\cos(2\theta) + \cos(4\theta)}{8}$

Answer: $\sin^4\theta = \dfrac{3 - 4\cos(2\theta) + \cos(4\theta)}{8}$

No exponents remain — only first-power cosines. This is exactly the form needed for integration in calculus.

Half-Angle Formulas

To obtain the half-angle formulas, replace $\theta$ with $\theta/2$ in the power-reducing formulas. In the sine formula, for instance, $\sin^2(\theta/2) = (1 - \cos\theta)/2$. Taking the square root of both sides:

$\sin\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1 - \cos\theta}{2}}$

$\cos\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1 + \cos\theta}{2}}$

$\tan\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1 - \cos\theta}{1 + \cos\theta}} = \dfrac{\sin\theta}{1 + \cos\theta} = \dfrac{1 - \cos\theta}{\sin\theta}$

Choosing the Sign

The $\pm$ means you must decide whether the result is positive or negative. The rule is straightforward: choose the sign based on the quadrant where $\theta/2$ lies, not where $\theta$ lies. For example, if $\theta = 300°$, then $\theta/2 = 150°$, which is in Quadrant II — so $\sin(\theta/2)$ is positive and $\cos(\theta/2)$ is negative.

The two rational forms of the tangent half-angle formula (without the square root) have a built-in sign — they automatically give the correct sign without needing to choose.

Worked Example 2: Find the Exact Value of $\sin(22.5°)$

Since $22.5° = 45°/2$, use the half-angle formula with $\theta = 45°$:

$\sin(22.5°) = +\sqrt{\dfrac{1 - \cos 45°}{2}}$

Positive because 22.5° is in Quadrant I. Substitute $\cos 45° = \dfrac{\sqrt{2}}{2}$:

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

Answer: $\sin(22.5°) = \dfrac{\sqrt{2 - \sqrt{2}}}{2} \approx 0.3827$

Worked Example 3: Find the Exact Value of $\cos(\pi/8)$

Since $\dfrac{\pi}{8} = \dfrac{\pi/4}{2}$, use the half-angle formula with $\theta = \pi/4$:

$\cos\!\left(\dfrac{\pi}{8}\right) = +\sqrt{\dfrac{1 + \cos(\pi/4)}{2}}$

Positive because $\pi/8$ is in Quadrant I. Substitute $\cos(\pi/4) = \dfrac{\sqrt{2}}{2}$:

$= \sqrt{\dfrac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\dfrac{\frac{2 + \sqrt{2}}{2}}{2}} = \sqrt{\dfrac{2 + \sqrt{2}}{4}} = \dfrac{\sqrt{2 + \sqrt{2}}}{2}$

Answer: $\cos\!\left(\dfrac{\pi}{8}\right) = \dfrac{\sqrt{2 + \sqrt{2}}}{2} \approx 0.9239$

Worked Example 4: Find $\tan(\theta/2)$ Given $\cos\theta = 3/5$

Assume $\theta$ is in Quadrant I. First find $\sin\theta$ using the Pythagorean identity:

$\sin^2\theta = 1 - \cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$

Since $\theta$ is in Q I, $\sin\theta = \dfrac{4}{5}$.

Now use the rational form of the tangent half-angle formula (no square root needed):

$\tan\!\left(\dfrac{\theta}{2}\right) = \dfrac{\sin\theta}{1 + \cos\theta} = \dfrac{4/5}{1 + 3/5} = \dfrac{4/5}{8/5} = \dfrac{4}{8} = \dfrac{1}{2}$

Answer: $\tan(\theta/2) = \dfrac{1}{2}$

The rational form is often the cleanest approach when you know both $\sin\theta$ and $\cos\theta$.

Real-World Connection: RMS Power in AC Circuits

The power-reducing formula is not just a calculus trick — it is how electrical engineers compute RMS (root mean square) values for alternating current.

AC voltage follows a sinusoidal wave: $v(t) = V_{\text{peak}} \sin(\omega t)$. To find average power, engineers need the average of $\sin^2(\omega t)$ over a full cycle. Apply the power-reducing formula:

$\sin^2(\omega t) = \dfrac{1 - \cos(2\omega t)}{2}$

When you average this over a complete cycle, the $\cos(2\omega t)$ term averages to zero (it completes full cycles), leaving:

$\text{average of } \sin^2(\omega t) = \dfrac{1}{2}$

This is why $V_{\text{rms}} = \dfrac{V_{\text{peak}}}{\sqrt{2}}$. For standard household current with a peak of about 170 V:

$V_{\text{rms}} = \dfrac{170}{\sqrt{2}} \approx 120 \text{ V}$

That single power-reducing identity is the reason your outlets are labeled 120 V.

Common Mistakes

1. Forgetting the $\pm$ sign on half-angle formulas. The square root always introduces $\pm$. You must determine the correct sign from the quadrant of $\theta/2$.
2. Using the wrong quadrant to determine the sign. The sign depends on where $\theta/2$ falls, not where $\theta$ falls. If $\theta = 300°$, then $\theta/2 = 150°$ (Q II), so cosine is negative even though $\cos 300°$ is positive.
3. Confusing half-angle with double-angle. The double-angle formulas multiply the angle ($\sin 2\theta$); the half-angle formulas divide it ($\sin(\theta/2)$). Mixing them up inverts the relationship.
4. Not simplifying the nested radical fully. An answer like $\sqrt{\dfrac{2 - \sqrt{2}}{4}}$ should be written as $\dfrac{\sqrt{2 - \sqrt{2}}}{2}$. Always simplify the denominator out of the radical.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Power-reducing formulas convert $\sin^2\theta$, $\cos^2\theta$, and $\tan^2\theta$ into expressions involving $\cos(2\theta)$ with no exponents
• The power-reducing formulas are derived directly from the double-angle identity for cosine
• Half-angle formulas are obtained by substituting $\theta/2$ into the power-reducing formulas and taking the square root
• The $\pm$ sign in half-angle formulas is determined by the quadrant of $\theta/2$, not of $\theta$
• The rational tangent half-angle forms — $\sin\theta/(1 + \cos\theta)$ and $(1 - \cos\theta)/\sin\theta$ — avoid the square root entirely and give the correct sign automatically
• Power-reducing formulas are essential in calculus (integrating even powers of trig functions) and electrical engineering (RMS calculations)

Return to Trigonometry for more topics in this section.