Lesson Objectives
  • Learn how to use half-angle identities to find an exact value
  • Learn how to find function values of s/2 given information about s
  • Learn how to simplify expressions using the half-angle identities
  • Learn how to verify identities using the half-angle identities

How to Work with Half-Angle Identities


In the last lesson, we learned about the Double-Angle Identities. Here, we will learn about the Half-Angle Identities.

Half-Angle Identities

$$\text{cos}\frac{A}{2}=\pm \sqrt{\frac{1 + \text{cos A}}{2}}$$ $$\text{sin}\frac{A}{2}=\pm \sqrt{\frac{1 - \text{cos A}}{2}}$$ $$\text{tan}\frac{A}{2}=\pm \sqrt{\frac{1 - \text{cos A}}{1 + \text{cos A}}}$$ $$\text{tan}\frac{A}{2}=\frac{\text{sin A}}{1 + \text{cos A}}$$ $$\text{tan}\frac{A}{2}=\frac{1 - \text{cos A}}{\text{sin A}}$$

Using a Half-Angle Identity to Find an Exact Value

In some cases, we may be asked to use a half-angle identity to find an exact value. Let's look at an example.
Example #1: Find the exact value. $$\text{cos}\hspace{.15em}67.5°$$ We can change the form to: $$\text{cos}\hspace{.15em}67.5°=\text{cos}\hspace{.15em}\frac{135°}{2}$$ Let's use our formula from above: $$\text{cos}\frac{A}{2}=\pm \sqrt{\frac{1 + \text{cos A}}{2}}$$ Here, A will be 135°.
Since we are trying to find the cos 67.5°, we will use the principal square root: $$\text{cos}\frac{135°}{2}=\sqrt{\frac{1 + \text{cos 135°}}{2}}$$ $$\text{cos}\hspace{.15em}135°=\frac{-\sqrt{2}}{2}$$ Let's plug in for cos 135°: $$\text{cos}\frac{135°}{2}=\sqrt{\frac{1 + \large{\frac{-\sqrt{2}}{2}}}{2}}$$ Simplify: $$\text{cos}\frac{135°}{2}=\sqrt{\frac{1 + \large{\frac{-\sqrt{2}}{2}}}{2}\cdot \frac{2}{2}}$$ $$\text{cos}\frac{135°}{2}=\sqrt{\frac{2 - \sqrt{2}}{4}}$$ $$\text{cos}\frac{135°}{2}=\frac{\sqrt{2 - \sqrt{2}}}{2}$$ $$\text{cos}\hspace{.1em}67.5°=\frac{\sqrt{2 - \sqrt{2}}}{2}$$ Let's look at another example.
Example #2: Find the exact value. $$\text{tan}\hspace{.15em}\frac{7π}{12}$$ For these types of problems, it is usually easier to work with degrees. $$\frac{7π}{12}\cdot \frac{180 °}{π}=105°$$ $$\text{tan}\hspace{.15em}105°$$ We can change the form to: $$\text{tan}\hspace{.15em}105°=\text{tan}\hspace{.15em}\frac{210°}{2}$$ Let's use our formula from above: $$\text{tan}\frac{A}{2}=\frac{\text{sin A}}{1 + \text{cos A}}$$ Here, A will be 210°. $$\text{sin}\hspace{.1em}210°=-\frac{1}{2}$$ $$\text{cos}\hspace{.1em}210°=-\frac{\sqrt{3}}{2}$$ Let's plug in for sin 210° and cos 210°: $$\text{tan}\frac{210°}{2}=\frac{\text{sin 210°}}{1 + \text{cos 210°}}$$ $$\text{tan}\frac{210°}{2}=\frac{-\large{\frac{1}{2}}}{1 - \large{\frac{\sqrt{3}}{2}}}$$ Simplify: $$\text{tan}\frac{210°}{2}=\frac{-\large{\frac{1}{2}}}{1 - \large{\frac{\sqrt{3}}{2}}}\cdot \frac{2}{2}$$ $$\text{tan}\frac{210°}{2}=\frac{-1}{2 - \sqrt{3}}$$ Rationalize the denominator: $$\text{tan}\frac{210°}{2}=\frac{-1}{2 - \sqrt{3}}\cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}}$$ $$\text{tan}\frac{210°}{2}=\frac{-2 - \sqrt{3}}{4 - 3}$$ $$\text{tan}\frac{210°}{2}=\frac{-2 - \sqrt{3}}{1}$$ $$\text{tan}\frac{210°}{2}=-2 - \sqrt{3}$$ $$\text{tan}\frac{7π}{12}=-2 - \sqrt{3}$$

Finding Function Values of s/2 Given information about s

Another type of problem in this section involves finding function values of s/2 given some information about s. Let's look at an example. $$\text{sin}\hspace{.1em}θ=\frac{4}{5}$$ $$90° < θ < 180°$$ Find sin θ/2, cos θ/2, and tan θ/2.
First, let's find the quadrant of θ/2 by dividing each part of our inequality by 2: $$\frac{90°}{2}< \frac{θ}{2}< \frac{180°}{2}$$ $$45° < \frac{θ}{2}< 90°$$ This tells us that θ/2 will be in quadrant I, where all trigonometric functions are positive.
Since we are given the sine of θ, we can use the Pythagorean identity to find the cosine θ. $$\left(\frac{4}{5}\right)^2 + \text{cos}^2 θ=1$$ $$\text{cos}^2 θ=1 - \frac{16}{25}$$ $$\text{cos}^2 θ=\frac{25}{25}- \frac{16}{25}$$ $$\text{cos}^2 θ=\frac{9}{25}$$ Since we are dealing with θ, we are in quadrant II. Note, this is not θ/2 but θ. This tells us that we want the negative square root. $$\text{cos}\hspace{.15em}θ=-\sqrt{\frac{9}{25}}$$ $$\text{cos}\hspace{.15em}θ=-\frac{3}{5}$$ Now, let's plug into our formulas, since θ/2 is in quadrant I, we will use the principal square root only: $$\text{cos}\frac{A}{2}=\pm \sqrt{\frac{1 + \text{cos A}}{2}}$$ $$\text{sin}\frac{A}{2}=\pm \sqrt{\frac{1 - \text{cos A}}{2}}$$ Let's begin with cos θ/2 $$\text{cos}\frac{θ}{2}=\sqrt{\frac{1 - \large{\frac{3}{5}}}{2}}$$ $$\text{cos}\frac{θ}{2}=\sqrt{\frac{1 - \large{\frac{3}{5}}}{2}\cdot \frac{10}{10}}$$ $$\text{cos}\frac{θ}{2}=\sqrt{\frac{10 - 6}{20}}$$ $$\text{cos}\frac{θ}{2}=\sqrt{\frac{4}{20}}$$ $$\text{cos}\frac{θ}{2}=\sqrt{\frac{1}{5}}$$ $$\text{cos}\frac{θ}{2}=\frac{1}{\sqrt{5}}$$ $$\text{cos}\frac{θ}{2}=\frac{\sqrt{5}}{5}$$ Now, let's find sin θ/2: $$\text{sin}\frac{θ}{2}=\sqrt{\frac{1 - \left(-\frac{3}{5}\right)}{2}}$$ $$\text{sin}\frac{θ}{2}=\sqrt{\frac{\frac{8}{5}}{2}}$$ $$\text{sin}\frac{θ}{2}=\sqrt{\frac{8}{10}}$$ $$\text{sin}\frac{θ}{2}=\sqrt{\frac{4}{5}}$$ $$\text{sin}\frac{θ}{2}=\frac{2}{\sqrt{5}}$$ $$\text{sin}\frac{θ}{2}=\frac{2\sqrt{5}}{5}$$ To find the tan θ/2, we can use the definition of tangent: $$\text{tan}\frac{θ}{2}=\frac{\text{sin}\frac{θ}{2}}{\text{cos}\frac{θ}{2}}$$ $$\text{tan}\frac{θ}{2}=\frac{2 \sqrt{5}}{5}\cdot \frac{5}{\sqrt{5}}=2$$

Verifying an Identity with Half-Angle Identities

Lastly, we may need to verify an identity using half-angle identities. Let's look at an example.
Example #4: Verify each identity. $$1 + \text{sin}\hspace{.15em}β=\left(\text{sin}\frac{β}{2}+ \text{cos}\frac{β}{2}\right)^2$$ $$=\text{sin}^2 \frac{β}{2}+ 2\text{sin}\frac{β}{2}\text{cos}\frac{β}{2}+ \text{cos}^2 \frac{β}{2}$$ Let's use our Pythagorean Identity: $$\text{sin}^2 \frac{β}{2}+ \text{cos}^2 \frac{β}{2}=1$$ We will replace this: $$=1 + 2\text{sin}\frac{β}{2}\text{cos}\frac{β}{2}$$ Now, we can use our double-angle identity from the last section: $$\text{sin 2A}=2\text{sinAcosA}$$ $$=1 + \text{sin}\hspace{.1em}2\frac{β}{2}$$ $$=1 + \text{sin}\hspace{.1em}β ✓$$

Skills Check:

Example #1

Find the exact value. $$\text{tan}\hspace{.1em}337.5°$$

Please choose the best answer.

A
$$\sqrt{3}- 1$$
B
$$\sqrt{2}- 1$$
C
$$\frac{\sqrt{2 + \sqrt{2}}}{2}$$
D
$$1 - \sqrt{2}$$
E
$$-\frac{\sqrt{2 - \sqrt{2}}}{2}$$

Example #2

Find the exact value. $$\text{tan}\hspace{.1em}θ=-\frac{4\sqrt{33}}{33}$$ $$270° < θ < 360°$$ Find: $$\text{cos}\hspace{.1em}\frac{θ}{2}$$

Please choose the best answer.

A
$$-\frac{\sqrt{98 + 14\sqrt{33}}}{14}$$
B
$$\frac{\sqrt{98 - 14\sqrt{33}}}{14}$$
C
$$\frac{33}{19}$$
D
$$-\frac{\sqrt{5}}{2}$$
E
$$\frac{\sqrt{19 + \sqrt{33}}}{5}$$

Example #3

Find the exact value. $$\text{tan}\hspace{.1em}θ=-\frac{4}{3}$$ $$\frac{5π}{2}< θ < 3π$$ Find: $$\text{tan}\hspace{.1em}\frac{θ}{2}$$

Please choose the best answer.

A
$$-1$$
B
$$-\frac{1}{2}$$
C
$$2$$
D
$$\sqrt{5}$$
E
$$\frac{2\sqrt{5}}{5}$$
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