Half Angle Formula – Explanation and Examples

The half angle formula gives the output of a trigonometric function for half of a given angle.

These formulas are useful for finding exact trig function values for minor angles. The half angle formulas are important for physical sciences and engineering.

Before moving forward with this topic, review trigonometric ratios and trig identities.

What Is the Half Angle Formula?

The half angle formula is an equation that gives a trigonometric ratio for an angle that is half of an angle with a known trigonometric value. There is one half angle formula for sine and another for cosine. 

These formulae mean that one can figure out the trigonometric ratios for one-half, one-fourth, one-eighth, one-sixteenth, etc. of an angle as long as the cosine ratio for that angle is known.

Both the sine and cosine half angle formulas require only the cosine of the angle to be known. That is, it is possible to figure out the sine of half of an angle without knowing the sine of the angle. The half angle formulas for the other trig functions follow from the half angle formulas for sine and cosine.

Sine Half Angle Formula

The formula for the sine of half of a given angle, $x$, is:

$sin(\frac{x}{2}) = \pm \sqrt{\frac{1-cosx}{2}}$.

Take note that this formula actually gives two values. It is important to figure out which quadrant the half angle is in to determine which value to use. 

Cos Half Angle Formula

Given an angle, $x$, the cosine of half of the angle is:

$cos(\frac{x}{2}) = \pm \sqrt{\frac{1+cosx}{2}}$.

Determining the quadrant of the half-angle determines whether to use the positive or negative value.

Tangent Half Angle Formula

By definition, the tangent of the half angle equals:

$tan\frac{x}{2} = \frac{\sqrt{\frac{1-cosx}{2}}}{\sqrt{\frac{1+cosx}{2}}$.

This can be simplified as:

$tan\frac{x}{2} = \sqrt{\frac{\frac{1-cosx}{2}}{\frac{1+cosx}{2}}} = \sqrt{\frac{1-cosx}{1+cosx}}$.

This square root simplifies as both $\frac{sinx}{1+cosx}$ and $\frac{1-cosx}{sinx}$. The proof of this is in the practice problems below, but it involves using the identity $sin^2x+cos^2x=1$. 


This section goes over common examples of problems involving the half-angle formula and their step-by-step solutions.

Example 1

Use the fact that $cos(2x) = 1-2sin^2x$ to derive the half angle formula for sine.


Let $y=2x$. Therefore, $\frac{y}{2} = x$.

Thus, this equation is rewritten as:

$cosy = 1-2sin^2(\frac{y}{2})$.

This equation should be rearranged as:

$2sin^2(\frac{y}{2}) = 1-cosy$.

Then, take the square root of both sides.

$\sqrt{2}sin(\frac{y}{2}) = \sqrt{1 – cosy}$.

Finally, divide both sides by the square root of two to get:

$sin(\frac{y}{2}) = \sqrt{\frac{1-cosy}{2}}$.

Example 2

Find a formula for $cos(\frac{x}{4})$ for any angle $x$.


In this case, $\frac{x}{4} = \frac{1}{2} \times \frac{x}{2}$. Therefore, this is the formula for the cosine of half of the half angle.

Now, let $y=2x$. Then, replace all of the $x$ terms in the equation for the cosine of a half-angle with $\frac{y}{2}$.

The new equation reads:

$cos\left(\frac{\frac{y}{2}}{2}\right) = \sqrt{\frac{cos{\frac{y}{2}+1}}{2}}$.

Then, use the original half angle formula and the fact that $\frac{\frac{y}{2}}{2} = \frac{x}{4}$ to get:

$cos(\frac{x}{4}) = \sqrt{\frac{\sqrt{\frac{cosx+1}{2}}+1}{2}}$.

Though this looks complicated, this can be simplified. Begin by combining the terms in the numerator of the first square root to get:

$cos(\frac{x}{4}) = \sqrt{\frac{\sqrt{cosx+1}+\sqrt{2}}{2\sqrt{2}}}$.

This is equal to:

$\sqrt{\frac{\sqrt{cosx+1}+\sqrt{2}}{2\sqrt{2}}} = \frac{1}{2}\sqrt{\sqrt{2cosx+2}+2}$.

Example 3

Given that $cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, find $sin(\frac{\pi}{12})$.


Recall that only the cosine value of an angle is needed to find the sine of half the angle. Using the formula, $sin(\frac{\pi}{12}) =$


This simplifies to:


Example 4

Given that $cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}$, find $cos(\frac{5\pi}{12})$.


Using the formula given above, the cosine of $\frac{5\pi}{12}$ is:


This is the same equation as example 3. Therefore, this simplifies to:


Example 5

Find $tan(\frac{7\pi}{8})$ given that $tan(\frac{7\pi}{4}) = -1$.


There are multiple ways to do this problem because there are multiple formulas for the tangent of a half-angle.

Notably, however, all of these formulas require the cosine of the original angle. In this case, however, only the tangent is given.

Finding cosine for the angle $\frac{7\pi}{4}$ radians is not difficult. Since the tangent is equal to $-1$, the absolute values of sine and cosine are equal at $\frac{7\pi}{4}$ radians. This only happens when their absolute values are equal to $\frac{\sqrt{2}}{2}$.

Additionally, this angle is in the fourth quadrant, where sine is negative and cosine is positive. 

Now any of the half-angle formulas for tangent are usable. Here is one way:

$tan(\frac{7\pi}{8}) = \frac{1-cos(\frac{7\pi}{4})}{sin(\frac{7\pi}{4})} = \frac{1-\frac{\sqrt{2}}{2}}{\frac{-\sqrt{2}}{2}} = \frac{\frac{2-\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -\frac{2-\sqrt{2}}{\sqrt{2}} =1- \sqrt{2}$.

Practice Questions

1. Given that $\cos \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$, what is the value of $\cos \dfrac{\pi}{12}$?

2. Given that $\cos \dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}$, what is the value of $\sin \dfrac{5\pi}{12}$?

3. Which of the following shows the exact value of $\tan \dfrac{3\pi}{8}$?

4. Which of the following shows the exact value of $\cos \dfrac{7\pi}{12}$?

5. Which of the following shows the exact value of $\sin\dfrac{5\pi}{8}$?


Open Problems

1. Derive the formula for the cosine of half of an angle using the fact that $\cos 2x = 2\cos^2x – 1$
2. Prove tangent half angle formulas $tan(\frac{x}{2}) = \frac{sinx}{1+cosx}$ and $tan(\frac{x}{2}) = \frac{1-cosx}{sinx}$.

Open Problem Solutions

1. Let $y=2x$. Then, $cosy = 2cos^2(\frac{y}{2})-1$. This rearranges as $\cos y+1 = 2\cos^2(\frac{y}{2})$. Taking the square root of both sides yields $\sqrt{\cos y+1} = \sqrt{2}\cos(\frac{y}{2})$. Then, dividing both sides by $\sqrt{2}$ yields $\sqrt{\frac{\cos y+1}{2}} = \cos(\frac{y}{2})$.

2. $\sin^2 x+ \cos^2x=1$. Therefore, $1- \cos^2x=\sin^2 x$. Additionally, $1- \cos^2 x=(1-\cos x)(1+\cos x)$.
Multiply $\frac{1-\cos x}{1+\cos x}$ by $1=\frac{1-\cos x}{1- \cos x}$ to get $\frac{(1-\cos x)^2}{1- \cos^2 x}$. Since $1- \cos^2x=\sin^2 x$, this becomes $\frac{(1-\cos x)^2}{\sin^2x}$. The square root of this is $\frac{1- \cos x}{\sin x}$.
Similarly, $\frac{1-\cos x}{1+\cos x} \times \frac{1+\cos x}{1+ \cos x} = \frac{1-\cos^2 x}{(1+ \cos x)^2} = \frac{\sin^2 x}{(1+ \cos x)^2}$. The square root of this is $\frac{\sin x}{1+\cos x}$.

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