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:

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

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

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

Here is a mathematical representation of trigonometry half angle formulas:

Certain cases of the sums and differences formulas for sine and cosine generate what is 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 2 α + 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 – sin2α]

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

Cos2 (β/2) = 1 + cosβ/2

Cos (β/2) = ± √[1+ cos(β/2)]

Similarly,

sin2 (β/2) = 1 - cosβ/2

sin (β/2) = √[1 - cos(β/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:

tan2 (A) = 1 – cos (2A)/2 / 1 + cos (2A)/2 = [1 – cos (2A)]/ [1 + cos (2A)]

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

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

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 gon.

However it is not particularly annotated by (°) for 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)

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+ (√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.

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 tangent.

FAQ (Frequently Asked Questions)

Q1. What are Trigonometric Identities?

Ans. In mathematical terms, trigonometric identities demonstrate the equalities that involve trigonometric functions and are true for every value of the variables taking place where both sides of the equality are described. However geometrically, these are identities taking into account only specified functions of one or more angles. They are quite different from triangle identities, which are identities likely to involve angles in addition to involving side lengths or other lengths of a triangle.

These identities are of help whenever expressions involving trigonometric functions require to be simplified. An important application of using half angle identities is the integration of non-trigonometric functions: a general method entails first using the substitution law with a trigonometric function, and afterwards simplifying the resulting integral using a trigonometric identity.

Q2. What is the Use of Half Angle Formulas?

Ans. There are many applications of trigonometry half angle formulas to science and engineering with respect to light and sound. Many of these processes need equations involving the sine and cosine of x, 2x, 3x, 4x and more. Neither, doubling the sin x will provide you the value of sin 2x, nor will taking half of sin x, provide you sin (x/2). Moreover, we can create the double angle formulas just by using the addition formulas for sine, cosine and tangent.