Half-angle identities are trigonometric identities used to simplify trigonometric expressions and calculate the sine, cosine, or tangent of half-angles when we know the values of a given angle. These identities are obtained by using the double angle identities and performing a substitution.

Here, we will learn to derive the half-angle identities and apply them to solve some practice exercises.

## What are the half-angle identities?

Half-angle identities are trigonometric identities that are used to calculate or simplify half-angle expressions, such as $latex \sin(\frac{\theta}{2})$. These identities can also be used to transform trigonometric expressions with exponents to one without exponents.

The half-angle identity of the sine is:

$latex \sin(\frac{\theta}{2})=\pm \sqrt{\frac{1-\cos(\theta)}{2}}$ |

The half-angle identity of the cosine is:

$latex \cos(\frac{\theta}{2})=\pm \sqrt{\frac{1+\cos(\theta)}{2}}$ |

The half-angle identity of the tangent is:

$latex \tan(\frac{\theta}{2})=\frac{\sin(\theta)}{1+\cos(\theta)}$ $latex =\frac{1-\cos(\theta)}{\sin(\theta)}$ |

## Proof of the half-angle identities

The mean angle identities can be derived using the double angle identities.

To derive the formula for the identity of half-angle of sines, we start with the double angle identity of cosines:

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

If we use the relation $latex \theta=\frac{\alpha}{2}$, we have $latex 2\theta=\alpha$. Substituting these expressions in the identity above, we have:

$latex \cos(\alpha)=1-2{{\sin}^2}(\frac{\alpha}{2})$

Now, we solve this expression for $latex \sin(\frac{\alpha}{2})$:

$latex \cos(\alpha)=1-2{{\sin}^2}(\frac{\alpha}{2})$

$latex 2{{\sin}^2}(\frac{\alpha}{2})=1-\cos(\alpha)$

$latex {{\sin}^2}(\frac{\alpha}{2})=\frac{1-\cos(\alpha)}{2}$

$latex \sin(\frac{\alpha}{2})=\pm\sqrt{\frac{1-\cos(\alpha)}{2}}$

The sign of $latex \sin(\frac{\alpha}{2})$ depends on the quadrant in which $latex \frac{\alpha}{2}$ is located. If $latex \frac{\alpha}{2}$ is in the first or second quadrant, the formula uses the positive sign, and if $latex \frac{\alpha}{2}$ is in the third or fourth quadrant, the formula uses the negative sign.

We use a similar process to find the half-cosine angle identity. Therefore, we start with the double-angle identity of the cosine in the following form:

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

After making the substitutions, we get:

$latex \cos(\alpha)=2{{\cos}^2}(\frac{\alpha}{2})-1$

Now, we solve for $latex \cos(\frac{\alpha}{2})$:

$latex \cos(\alpha)=2{{\cos}^2}(\frac{\alpha}{2})-1$

$latex 2{{\cos}^2}(\frac{\alpha}{2})=1+\cos(\alpha)$

$latex {{\cos}^2}(\frac{\alpha}{2})=\frac{1+\cos(\alpha)}{2}$

$latex \cos(\frac{\alpha}{2})=\pm\sqrt{\frac{1+\cos(\alpha)}{2}}$

In this case, if $latex \frac{\alpha}{2}$ is in the first or fourth quadrant, the formula uses the positive sign and if $latex \frac{\alpha}{2}$ is in the second or third quadrant, the formula uses the negative sign.

## Half-angle identities – Example with answers

The following examples are solved using what you have learned about half-angle identities. Study and analyze these examples to understand the process used.

**EXAMPLE 1**

Use the half-angle identity of the sine to find the sine value of 15°.

##### Solution

We use the formula for the half-angle identity of the sine with the given value. Therefore, we have:

$latex \sin(\frac{\theta}{2})=\pm\sqrt{\frac{1-\cos(\theta)}{2}}$

$latex \sin(15^{\circ})=\pm\sqrt{\frac{1-\cos(30^{\circ} )}{2}}$

$latex =\pm\sqrt{\frac{1-0.866}{2}}$

$latex =0.259$

We use the positive value since 15° is in the first quadrant.

**EXAMPLE 2**

Determine the value of the cosine of 165° using the half-angle identity of the cosine.

##### Solution

To use the half-angle identity of cosine, we use the angle $latex \frac{\theta}{2} = 165$°. This means that we have $latex \theta = 330$°. Therefore, we use the formula with these values:

$latex \cos(\theta)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}$

$latex \cos(165^{\circ})=\pm\sqrt{\frac{1+\cos(330^{\circ})}{2}}$

$latex =\pm\sqrt{\frac{1+0.866}{2}}$

$latex =-0.966$

We chose the negative value since the angle 165° is in the second quadrant.

**EXAMPLE 3**

Check that the identity $latex 2{{\sin}^2}(\frac{x}{2})+\cos(x)= 1$.

##### Solution

We can use the identity of the half-angle of sine to substitute and simplify the expression. By doing this, we have:

$latex 2{{\sin}^2}(\frac{x}{2})+\cos(x)$

$latex =2{{(\sqrt{\frac{1-\cos(x)}{2}})}^2}+\cos(x)$

$latex =2(\frac{1-\cos(x)}{2})+\cos(x)$

$latex =1-\cos(x)+\cos(x)$

$latex =1$

After simplifying, we see that the left side of the identity is equal to the right side, so the identity is true.

## Half-angle identities – Practice problems

Solve the following practice problems using what you have learned about the half-angle identities of sine, cosine, and tangent. Select an answer and check it to see if you got the correct answer.

## See also

Interested in learning more about trigonometric identities? Take a look at these pages:

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