Half Angle Formula

It is sometimes very crucial to determine the value of the trigonometric functions for half-angles. For instance, using some half-angle formula we can convert an expression with exponents to one without exponents, and whose angles are multiples of the original angle. It is to note that we get half-angle formulas from double-angle formulas. Both sin (2A) and cos (2A) are obtained from the double angle formula for the cosine.

The formula for half-angle identities is as below:

• \[\text{Sine of a Half Angle} : sin(\frac{a}{2}) = \pm \sqrt{\frac{(1 - cos a)}{2}}\]

• \[\text{Cosine of a Half Angle}: cos(\frac{a}{2}) = \pm \sqrt{\frac{(1+cosa)}{2}}\]

• \[\text{Tangent of a Half Angle}: tan(\frac{a}{2}) = \frac{1 - cosa}{sina} = \frac{sina}{1+cosa}\]

Half - Angle Formula

Here is a mathematical representation of trigonometry half-angle formulas:

Double‐Angle and Half‐Angle Identities

Certain cases of the sums and differences formulas for sine and cosine generate what are called the double‐angle identities and the half‐angle identities. We will first start by incorporating the sum identity for the sine as given in the reference, prove that 

\[\begin{vmatrix} sin \alpha & sin \beta  & sin \gamma \\cos \alpha  & cos \beta  & cos\gamma \\sin 2 \alpha & sin 2 \beta  & sin 2 \gamma \end{vmatrix}\]

In the same manner, for the cosine,

cos 2α = cos (α + α)

cos 2α = (cos α cos α - sin α sin α)

cos 2α = (cos²α – sin²α)

Applying the property of Pythagorean, sin 2α + cos 2α =1, we can obtain two additional cosine identities i.e.

Cos 2α = cos²α – sin²α

Cos 2α = [1- sin²α] - sin²α

Cos 2α = 1- 2 sin²α

Cos 2α = cos²α – sin²α

cos 2α = cos²α – [1 – cos²α]

cos 2α = 2cos²α – 1

The half‐angle identities for the sine and cosine are derived from two of the cosine identities explained above.

Cos 2α = 2cos²α – 1

Cos 2 (β/2) = 2cos² (β/2) – 1

Cosβ = 2cos² (β/2) – 1

2cos² (β/2) = 1 + cosβ

Cos 2α = 1 - 2 sin²α

Cos 2 (β/2) = 1 – 2 sin² (β/2)

Cosβ = 1 – 2 sin² (β/2)

2 sin₂ (β/2) = 1- Cosβ

\[Cos^{2} (\frac{\beta}{2}) = \frac{(1 + cos\beta )}{2}\]

\[Cos (\frac{\beta}{2}) = \pm \frac{\sqrt{(1+cos \beta)}}{2}\]

Similarly,

\[sin^{2} (\frac{\beta}{2}) = \frac{(1-cos \beta)}{2}\]

\[sin (\frac{\beta}{2}) = \frac{\sqrt{(1-cos \beta)}}{2}\]

The symbol of the two preceding functions is dependent on the quadrant in which the resulting angle is positioned.

For computing the tangent of the half-angle, tan (2A), we need to combine the identities for sine and cosine: \[tan ^{2} (A) = \frac{\frac{(1-cos (2A))}{2}}{\frac{(1+cos (2A))}{2}} = \frac{ 1-cos(2A)}{1 +cos(2A)}\]

Again replacing A by (1/2)A, we obtain

\[tan = [\frac{A}{2}] = \pm \sqrt{\frac{1 - cosA}{1 + cosA}}\]

Trigonometry Angles and their Notations

These are actually the use of Greek letters such as alpha (α), beta (β), theta (θ) and gamma (γ) to denote angles. It represents the signs of trigonometric functions in each quadrant. A couple of different units of angle measure are extensively used, including degree, radian, and gradian (gons): As an example,

1 complete circle (turn) = 360° = 2π radian = 400 gons.

However it is not particularly annotated by (°) for a degree or (g) for gradian, all values for angles are assumed to be given in radian.

The graphical representation shown below depicts some common angles and their values and the conversions of the basic trigonometric functions:

(Image will be uploaded soon)

Solved Examples

Example1: Find the absolute value for cosine 165° using the half angle identities.

Solution1:

In the given verification, note that 165 degrees is in the 2nd quadrant, and cosine functions in the 2nd quadrant are negative. Also, 330 degrees is in the 4th quadrant, and cosine functions in the 4th quadrant are positive.

That said, the reference triangle of 330° in the 4th quadrant is a 30°–60°–90° triangle. Thus, cos 330° = cos 30°.

Now, applying the half angle identity for the cosine:

We get,

\[Cos 165^{\circ} = \frac{cos 330^{\circ}}{2}\]

\[Cos 165^{\circ} = \sqrt{\frac{1 + cos 330^{\circ}}{2}}\]

\[Cos 165^{\circ} = \sqrt{\frac{(1 + cos 30^{\circ})}{2}}\]

\[Cos 165^{\circ} = - \sqrt{\frac{[1 + (-\frac{3}{2})]}{2}}\] (here the minus sign is chosen because 165° is in the 2nd quadrant)

\[Cos 165^{\circ} = - [2 + \sqrt{(\frac{3}{4})}]\]

= \[0.9659\]

So, our answer is Cos 165° = \[- [2+(\frac{\sqrt{3}}{2}]\]

(Images will be uploaded soon)

Example2: Calculate the value of sin15° using the sine half-angle relationship.

Solution2:

Given that α = 30°

Using the formula

\[sin \frac{\alpha}{2} = \pm  \sqrt{\frac{1 - cos \alpha}{2}}\]

We get,

\[sin 15^{\circ} = \pm \sqrt{\frac{1 - cos 30^{\circ}}{2}}\]

= \[\pm \sqrt{\frac{1 - 0.866}{2}}\] ...

= \[0.2588\]

Thus, our answer is sin 15° = 0.2588

​Remember here the positive value is taken since 15° is in the 1st quadrant.

Key Facts

• In the half-angle formula problems for sine and cosine, observe that a plus/minus sign occurs in front of each square root (radical).

• Whether your answer is negative or positive depends on which quadrant the new half-angle is in.

• The tan half-angle formula doesn’t have a ± sign in front, so the above doesn’t apply to tangents.

Conclusion 

The article is very useful for the students to understand the concept of half-angle formulas. All the important half-angle formulas and their conversion is provided. Solved examples will help students apply the half-angle formula in problems.