Trigonometry: Half Angle Formulas

Knowing the sine, cosine, or tangent of an angle is useful, but what about knowing the value for half of that angle? Half-angle formulas are powerful tools that allow you to compute the exact trigonometric values of angles like $15^{\circ }$, $22.5^{\circ }$, or $\frac{\pi }{8}$, which are not found on the standard unit circle. These formulas are indispensable in calculus for integrating trigonometric powers, in engineering for signal processing, and in physics for analyzing wave interference. Mastering their derivation, application, and the subtle sign rules transforms complex problems into manageable algebraic ones.

Derivation from Double Angle Formulas

The half-angle formulas do not appear from thin air; they are a direct and clever consequence of the double angle formulas you already know. This connection is the foundational concept. Let’s start with the double angle identity for cosine: $$\cos (2\theta )=1-2{\sin }^2(\theta )$$ If we let $2\theta =A$, then $\theta =\frac{A}{2}$. Substituting these into the identity gives: $$\cos (A)=1-2{\sin }^2\left(\frac{A}{2}\right)$$ Now, we simply solve algebraically for $\sin \left(\frac{A}{2}\right)$:

1. Rearrange: $2{\sin }^2\left(\frac{A}{2}\right)=1-\cos (A)$
2. Divide by 2: ${\sin }^2\left(\frac{A}{2}\right)=\frac{1-\cos (A)}{2}$
3. Take the square root: $\sin \left(\frac{A}{2}\right)=\pm \sqrt{\frac{1-\cos (A)}{2}}$

This is the half-angle formula for sine. The $\pm$ sign is crucial and will be discussed in the next section. We can follow a nearly identical process using the other form of the cosine double angle identity, $\cos (2\theta )=2{\cos }^2(\theta )-1$.

1. Substitute $\theta =\frac{A}{2}$: $\cos (A)=2{\cos }^2\left(\frac{A}{2}\right)-1$
2. Rearrange: $2{\cos }^2\left(\frac{A}{2}\right)=1+\cos (A)$
3. Solve: $\cos \left(\frac{A}{2}\right)=\pm \sqrt{\frac{1+\cos (A)}{2}}$

For the tangent, we can derive its formula directly from the sine and cosine results: $$\tan \left(\frac{A}{2}\right)=\frac{\sin \left(\frac{A}{2}\right)}{\cos \left(\frac{A}{2}\right)}=\frac{\pm \sqrt{\frac{1-\cos (A)}{2}}}{\pm \sqrt{\frac{1+\cos (A)}{2}}}=\pm \sqrt{\frac{1-\cos (A)}{1+\cos (A)}}$$ A more useful, sign-agnostic form comes from multiplying the numerator and denominator by $\sqrt{1+\cos (A)}$ (or using the sine double angle formula), yielding: $$\tan \left(\frac{A}{2}\right)=\frac{1-\cos (A)}{\sin (A)}=\frac{\sin (A)}{1+\cos (A)}$$ These last two forms are often preferred as they avoid the ambiguous $\pm$ sign.

Determining the Correct Sign Based on Quadrant

The most common error when using the half-angle formulas for sine and cosine is ignoring or misapplying the $\pm$ sign. The radical ​ only yields a non-negative number. However, the sine or cosine of $\frac{A}{2}$ can be positive or negative. Therefore, the sign is not arbitrary; you must determine the correct sign based on the quadrant where the half-angle $\frac{A}{2}$ lies.

Your process should always be:

1. Given angle $A$, compute the half-angle $\frac{A}{2}$.
2. Determine the quadrant in which $\frac{A}{2}$ terminates.
3. Recall the "ASTC" rule (All Students Take Calculus) to know if sine and cosine are positive or negative in that quadrant.
4. Attach that sign to the front of your square-root expression.

For example, suppose $A=300^{\circ }$. Then $\frac{A}{2}=150^{\circ }$. The angle $150^{\circ }$ lies in the second quadrant. In Quadrant II, sine is positive and cosine is negative. Therefore:

• $\sin (150^{\circ })=+\sqrt{\frac{1-\cos (300^{\circ })}{2}}$
• $\cos (150^{\circ })=-\sqrt{\frac{1+\cos (300^{\circ })}{2}}$

This step is non-negotiable for obtaining the correct answer. In engineering contexts, such as analyzing alternating current phases, using the wrong sign leads to a completely inverted or erroneous signal interpretation.

Evaluating Exact Values for Common Angles

The true power of these formulas is their ability to find exact trigonometric values for non-standard angles. Let's work through two classic examples: finding $\sin (15^{\circ })$ and $\cos (22.5^{\circ })$.

Example 1: Find the exact value of $\sin (15^{\circ })$ and $\cos (15^{\circ })$. We recognize $15^{\circ }$ is half of $30^{\circ }$. So, let $A=30^{\circ }$.

1. Find the half-angle: $\frac{A}{2}=15^{\circ }$. This lies in Quadrant I, so all signs are positive.
2. We know $\cos (30^{\circ })=\frac{\sqrt{3}}{2}$.
3. Apply the formulas:

$$\sin (15^{\circ })=+\sqrt{\frac{1-\cos (30^{\circ })}{2}}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}}=\sqrt{\frac{2-\sqrt{3}}{4}}=\frac{\sqrt{2-\sqrt{3}}}{2}$$ $$\cos (15^{\circ })=+\sqrt{\frac{1+\cos (30^{\circ })}{2}}=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}=\sqrt{\frac{2+\sqrt{3}}{4}}=\frac{\sqrt{2+\sqrt{3}}}{2}$$

Example 2: Find the exact value of $\cos (22.5^{\circ })$. We recognize $22.5^{\circ }$ is half of $45^{\circ }$. So, $A=45^{\circ }$ and $\frac{A}{2}=22.5^{\circ }$ (Quadrant I, positive).

1. We know $\cos (45^{\circ })=\frac{\sqrt{2}}{2}$.
2. Apply the cosine formula:

$$\cos (22.5^{\circ })=\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}=\sqrt{\frac{\frac{2+\sqrt{2}}{2}}{2}}=\sqrt{\frac{2+\sqrt{2}}{4}}=\frac{\sqrt{2+\sqrt{2}}}{2}$$

These exact values, often involving nested radicals, are more precise and useful for subsequent algebraic manipulation than any decimal approximation.

Common Pitfalls

1. Ignoring the $\pm$ Sign: The most frequent and critical error is dropping the $\pm$ or choosing its sign randomly. Correction: Always determine the quadrant of the half-angle $\frac{A}{2}$, not the original angle $A$, and use the ASTC rule to assign the correct sign to sine and cosine.

1. Using the Wrong Double Angle Formula: Attempting to derive the half-angle formula for sine from $\cos (2\theta )={\cos }^2\theta -{\sin }^2\theta$ is more algebraically messy. Correction: For the cleanest derivation, start with the power-reduction forms: $\cos (2\theta )=1-2{\sin }^2\theta$ for sine and $\cos (2\theta )=2{\cos }^2\theta -1$ for cosine.

1. Misidentifying the Half-Angle: Confusing which angle is the "whole" and which is the "half" leads to using the formula backwards. Correction: If you want the trig value of angle $B$, ask yourself: "Is there a known angle $A$ such that $B=\frac{A}{2}$?" For $15^{\circ }$, the whole angle $A$ is $30^{\circ }$.

1. Algebraic Errors in Simplification: Making mistakes when combining fractions under the radical is common. Correction: Work step-by-step. For a cosine value like $\cos A=\frac{a}{b}$, compute $1\pm \cos A=1\pm \frac{a}{b}=\frac{b\pm a}{b}$ before dividing by 2.

Summary

• Half-angle formulas express trigonometric functions of $\frac{\theta }{2}$ in terms of functions of $\theta$, and are derived by solving the double angle formulas for cosine algebraically.
• The formulas for sine and cosine require a $\pm$ sign before the radical, and you must determine the correct sign based on the quadrant in which the half-angle $\frac{\theta }{2}$ lies.
• These formulas are the primary tool for finding the exact values of trigonometric functions for angles like $15^{\circ }$ (half of $30^{\circ }$) and $22.5^{\circ }$ (half of $45^{\circ }$), yielding expressions with nested radicals.
• The tangent half-angle formula has convenient alternative forms, $\tan (\frac{A}{2})=\frac{1-\cos A}{\sin A}=\frac{\sin A}{1+\cos A}$, that avoid the sign ambiguity altogether.
• In engineering and physics applications, correct sign determination is critical for accurate modeling of waveforms, oscillations, and rotational systems.