# 3.4.5: Half Angle Formulas


Derivation of sine and cosine formulas for half a given angle

After all of your experience with trig functions, you are feeling pretty good. You know the values of trig functions for a lot of common angles, such as $$30^{\circ}$$, $$60^{\circ}$$ etc. And for other angles, you regularly use your calculator. Suppose someone gave you an equation like this:

$$\cos 75^{\circ}$$

Could you solve it without the calculator? You might notice that this is half of $$150^{\circ}$$. This might give you a hint!

### Half Angle Formulas

Here we'll attempt to derive and use formulas for trig functions of angles that are half of some particular value.

To do this, we'll start with the double angle formula for cosine: $$\cos 2\theta =1−2\sin ^2\theta$$. Set $$\theta =\dfrac{\alpha}{2}$$, so the equation above becomes $$\cos 2\dfrac{\alpha}{2}=1−2\sin ^2\dfrac{\alpha}{2}$$.

Solving this for $$\sin \dfrac{\alpha}{2}$$, we get:

\begin{aligned} \cos 2\dfrac{\alpha}{2}&=1−2\sin 2\dfrac{\alpha}{2} \\ \cos \alpha&=1−2\sin 2\dfrac{\alpha}{2} \\ 2 \sin ^2\dfrac{\alpha}{2}&=1−\cos \alpha \\ \sin ^2\dfrac{\alpha}{2}&=\dfrac{1−\cos \alpha }{ 2} \\ \sin \dfrac{\alpha}{2}&=\pm \sqrt{\dfrac{1−\cos \alpha }{2}} \end{aligned}

$$\sin \dfrac{\alpha}{2}=\sqrt{\dfrac{1−\cos \alpha }{2}}$$ if $$\dfrac{\alpha}{2}$$ is located in either the first or second quadrant.

$$\sin \dfrac{\alpha}{2}=−\sqrt{\dfrac{1−\cos \alpha }{2}}$$ if $$\dfrac{\alpha}{2}$$ is located in the third or fourth quadrant.

This formula shows how to find the sine of half of some particular angle.

One of the other formulas that was derived for the cosine of a double angle is:

$$\cos 2\theta =2\cos ^2\theta −1$$. Set $$\theta =\dfrac{\alpha}{2}$$, so the equation becomes $$\cos 2\dfrac{\alpha}{2}=−1+2\cos ^2\dfrac{\alpha}{2}$$. Solving this for $$\cos \dfrac{\alpha}{2}$$, we get:

\begin{aligned} \cos 2\dfrac{\alpha}{2}&=2\cos ^2\dfrac{\alpha}{2}−1 \\ \cos \alpha&=2\cos ^2\dfrac{\alpha}{2}−1 \\ 2\cos ^2\dfrac{\alpha}{2}&=1+\cos \alpha \\ \cos ^2\dfrac{\alpha}{2}&=\dfrac{1+\cos \alpha }{2} \\ \cos \dfrac{\alpha}{2} &=\pm \sqrt{\dfrac{1+\cos \alpha }{2}}\end{aligned}

$$\cos \dfrac{\alpha}{2}=\sqrt{\dfrac{1+\cos \alpha }{2}}$$ if $$\dfrac{\alpha}{2}$$ is located in either the first or fourth quadrant.

$$\cos \dfrac{\alpha}{2}=−\sqrt{\dfrac{1+\cos \alpha }{2}}$$ if $$\dfrac{\alpha}{2}$$ is located in either the second or fourth quadrant.

This formula shows how to find the cosine of half of some particular angle.

Let's see some examples of these two formulas (sine and cosine of half angles) in action.

1. Determine the exact value of $$\sin 15^{\circ}$$.

Using the half angle identity, $$\alpha =30^{\circ}$$, and $$15^{\circ}$$ is located in the first quadrant. Therefore, $$\sin \dfrac{\alpha}{2}=\sqrt{\dfrac{1−\cos \alpha }{2}}$$.

\begin{aligned} \sin 15^{\circ} &=\sqrt{\dfrac{1−\cos 30^{\circ} }{2}}\\&=\sqrt{\dfrac{1−\dfrac{\sqrt{3}}{2}}{2}}=\sqrt{\dfrac{\dfrac{2−\sqrt{3}}{2}}{2}}=\sqrt{\dfrac{2−\sqrt{3}}{4}} \end{aligned}

Plugging this into a calculator, $$\sqrt{\dfrac{2−\sqrt{3}}{4}}\approx 0.2588$$. Using the sine function on your calculator will validate that this answer is correct.

2. Use the half angle identity to find exact value of $$\sin 112.5^{\circ}$$

Since $$\sin \dfrac{225^{\circ} }{2}=\sin 112.5^{\circ}$$, use the half angle formula for sine, where $$\alpha =225^{\circ}$$. In this example, the angle $$112.5^{\circ}$$ is a second quadrant angle, and the SIN of a second quadrant angle is positive.

\begin{aligned} \sin 112.5^{\circ} &=\sin \dfrac{225^{\circ} }{2} \\&=\pm \sqrt{\dfrac{1−\cos 225^{\circ} }{2}}\\ &=+\sqrt{\dfrac{1−\left(−\dfrac{\sqrt{2}}{2}\right)}{2}} \\ &=\sqrt{\dfrac{\dfrac{2}{2}+\dfrac{\sqrt{2}}{2}}{2}}\\ &=\sqrt{\dfrac{2+\sqrt{2}}{4}} \end{aligned}

3. Use the half angle formula for the cosine function to prove that the following expression is an identity: $$2\cos ^2 \dfrac{x}{2}−\cos x=1$$

Use the formula $$\cos \dfrac{\alpha}{2}=\dfrac{1+\cos \alpha}{2}$$ and substitute it on the left-hand side of the expression.

\begin{aligned} 2 \left(\sqrt{\dfrac{1+\cos \theta }{2}}\right)^2−\cos \theta&=1 \\ 2\left(\dfrac{1+\cos \theta }{2}\right)−\cos \theta&=1\\ 1+\cos \theta −\cos \theta&=1 \\ 1&=1 \end{aligned}

##### Example $$\PageIndex{1}$$

Earlier, you were asked you to find $$\cos 75^{\circ}$$. If you use the half angle formula, then $$\alpha =150^{\circ}$$

Substituting this into the half angle formula:

Solution

$$\sin \dfrac{150^{\circ} }{2}=\sqrt{\dfrac{1−\cos \alpha }{2}}=\sqrt{\dfrac{1−\cos 150^{\circ} }{2}}=\sqrt{\dfrac{1+\dfrac{\sqrt{3}}{2}}{2}}=\sqrt{\dfrac{2+\sqrt{3}}{4}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}$$

##### Example $$\PageIndex{2}$$

Prove the identity: $$\tan \dfrac{b}{2} =\dfrac{\sec b}{\sec b \csc b+\csc b}$$

Solution

Step 1: Change right side into sine and cosine.

\begin{aligned} \tan \dfrac{b}{2} &=\dfrac{\sec b}{\sec b \csc b+\csc b} \\ &=\dfrac{1}{\cos b} \div \csc b(\sec b+1) \\ &=\dfrac{1}{\cos b} \div \dfrac{1}{\sin b}\left(\dfrac{1}{\cos b}+1\right) \\ &=\dfrac{1}{\cos b} \div \dfrac{1}{\sin b}\left(\dfrac{1+\cos b}{\cos b}\right) \\ &=\dfrac{1}{\cos b} \div \dfrac{1+\cos b}{\sin b \cos b} \\ &=\dfrac{1}{\cos b} \cdot \dfrac{\sin b \cos b}{1+\cos b} \\ &=\dfrac{\sin b}{1+\cos b} \end{aligned}

Step 2: At the last step above, we have simplified the right side as much as possible, now we simplify the left side, using the half angle formula.

\begin{aligned} \sqrt{\dfrac{1-\cos b}{1+\cos b}} &=\dfrac{\sin b}{1+\cos b} \\ \dfrac{1-\cos b}{1+\cos b} &=\dfrac{\sin ^{2} b}{(1+\cos b)^{2}} \\ (1-\cos b)(1+\cos b)^{2} &=\sin ^{2} b(1+\cos b) \\ (1-\cos b)(1+\cos b) &=\sin ^{2} b \\ 1-\cos ^{2} b &=\sin ^{2} b \end{aligned}

##### Example $$\PageIndex{3}$$

Verify the identity: $$\cot \dfrac{c}{2} =\dfrac{\sin c}{1-\cos c}$$

Solution

Step 1: change cotangent to cosine over sine, then cross-multiply.

\begin{aligned} \cot \dfrac{c}{2} &=\dfrac{\sin c}{1-\cos c} \\ &=\dfrac{\cos \dfrac{c}{2}}{\sin \dfrac{c}{2}}=\sqrt{\dfrac{1+\cos c}{1-\cos c}} \\ \sqrt{\dfrac{1+\cos c}{1-\cos c}} &=\dfrac{\sin c}{1-\cos c} \\ \dfrac{1+\cos c}{1-\cos c} &=\dfrac{\sin ^{2} c}{(1-\cos c)^{2}} \\ (1+\cos c)(1-\cos c)^{2} &=\sin ^{2} c(1-\cos c) \\ (1+\cos c)(1-\cos c) &=\sin ^{2} c \\ 1-\cos ^{2} c &=\sin ^{2} c \end{aligned}

##### Example $$\PageIndex{4}$$

Prove that $$\sin x \tan \dfrac{x}{2}+2\cos x=2\cos ^2 \dfrac{x}{2}$$

Solution

$$\begin{array}{l} \sin x \tan \dfrac{x}{2}+2 \cos x=\sin x\left(\dfrac{1-\cos x}{\sin x}\right)+2 \cos x \\ \sin x \tan \dfrac{x}{2}+2 \cos x=1-\cos x+2 \cos x \\ \sin x \tan \dfrac{x}{2}+2 \cos x=1+\cos x \\ \sin x \tan \dfrac{x}{2}+2 \cos x=2 \cos ^{2} \dfrac{x}{2} \end{array}$$

## Review

Use half angle identities to find the exact values of each expression.

1. $$\sin 22.5^{\circ}$$
2. $$\sin 75^{\circ}$$
3. $$\sin 67.5^{\circ}$$
4. $$\sin 157.5^{\circ}$$
5. $$\cos 22.5^{\circ}$$
6. $$\cos 75^{\circ}$$
7. $$\cos 157.5^{\circ}$$
8. $$\cos 67.5^{\circ}$$
9. Use the two half angle identities presented in this section to prove that $$\tan \left(\dfrac{x}{2}\right)=\pm \sqrt{\dfrac{1−\cos x}{1+\cos x}}$$.
10. Use the result of the previous problem to show that $$\tan \left(\dfrac{x}{2}\right)=\dfrac{1−\cos x}{\sin x}$$.
11. Use the result of the previous problem to show that $$\tan \left(\dfrac{x}{2}\right)=\dfrac{\sin x}{1+\cos x}$$.

Use half angle identities to help you find all solutions to the following equations in the interval $$[0,2\pi)$$.

1. $$\sin ^2x=\cos ^2\left(\dfrac{x}{2}\right)$$
2. $$\tan\left(\dfrac{x}{2}\right)=\dfrac{1−\cos x}{1+\cos x}$$
3. $$\cos ^2x=\sin ^2\left(\dfrac{x}{2}\right)$$
4. $$\sin ^2\left(\dfrac{x}{2}\right)=2\cos ^2x−1$$

## Vocabulary

Term Definition
Half Angle Identity A half angle identity relates a trigonometric function of one half of an argument to a set of trigonometric functions, each containing the original argument.