Lesson Objectives
• Learn how to work with double-angle identities
• Learn how to work with product-to-sum identities
• Learn how to work with sum-to-product identities

## Double-Angle, Product-to-Sum, and Sum-to-Product Identities

At this point, we have learned about the fundamental identities and the sum and difference identities. In this lesson, we will focus on the double-angle identities, along with the product-to-sum identities, and the sum-to-product identities.

### Double-Angle Identities

$$\text{cos 2A}=\text{cos}^2 A - \text{sin}^2 A$$ $$\text{cos 2A}=1 - 2 \text{sin}^2 A$$ $$\text{cos 2A}=2 \text{cos}^2 A - 1$$ $$\text{sin 2A}=2 \text{sin A}\text{cos A}$$ $$\text{tan 2A}=\frac{\text{2 tan A}}{1 - \text{tan}^2 \text{A}}$$ Our first example will involve finding the function values of 2θ when given information about θ. Let's look at an example.
Example #1 Find the exact value of each. $$\text{cos θ}=\frac{2\sqrt{2}}{3}$$ $$\text{sin θ}> 0$$ Find cos 2θ, sin 2θ, and tan 2θ. Since we are given cos θ, we need to find sin θ and tan θ in order to perform all of our calculations. Let's begin with our Pythagorean Identity to find sine: $$\text{cos}^2 θ + \text{sin}^2 θ=1$$ $$\left(\frac{2 \sqrt{2}}{3}\right)^2 + \text{sin}^2 θ=1$$ $$\frac{8}{9}+ \text{sin}^2 θ=1$$ $$\text{sin}^2 θ=\frac{9}{9}- \frac{8}{9}$$ $$\text{sin}^2 θ=\frac{1}{9}$$ $$\text{sin}\hspace{.1em}θ=\pm \sqrt{\frac{1}{9}}$$ Since we know that θ is in quadrant I, we will use the principal square root. $$\text{sin}\hspace{.1em}θ=\frac{1}{3}$$ Now, let's find the tan θ: $$\text{tan}\hspace{.1em}θ=\frac{\text{sin}\hspace{.1em}θ}{\text{cos}\hspace{.1em}θ}$$ $$\text{tan}\hspace{.1em}θ=\frac{1}{3}\cdot \frac{3}{2\sqrt{2}}=\frac{1}{2\sqrt{2}}$$ $$=\frac{1}{2\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{4}$$ Now that we have the values for sin θ, cos θ, and tan θ, let's find cos 2θ, sin 2θ, and tan 2θ: $$\text{cos 2θ}=\text{cos}^2 θ - \text{sin}^2 θ$$ $$\text{cos 2θ}=\left(\frac{2 \sqrt{2}}{3}\right)^2 - \left(\frac{1}{3}\right)^2$$ $$\text{cos 2θ}=\frac{8}{9}- \frac{1}{9}$$ $$\text{cos 2θ}=\frac{7}{9}$$ $$\text{sin 2θ}=2\text{sin θ}\cdot \text{cos θ}$$ $$\text{sin 2θ}=2 \cdot \frac{1}{3}\cdot \frac{2\sqrt{2}}{3}=\frac{4\sqrt{2}}{9}$$ $$\text{tan 2θ}=\frac{2 \text{tan θ}}{1 - \text{tan}^2 θ}$$ $$\text{tan 2θ}=\frac{2 \cdot \large{\frac{\sqrt{2}}{4}}}{1 - \left(\large{\frac{\sqrt{2}}{{4}}}\right)^2}$$ $$=\frac{\large{\frac{2\sqrt{2}}{4}}}{1 - \large{\frac{2}{{16}}}}$$ $$=\frac{\large{\frac{2\sqrt{2}}{4}}}{\large{\frac{16}{16}}- \large{\frac{2}{{16}}}}$$ $$=\frac{\large{\frac{2\sqrt{2}}{4}}}{\large{\frac{14}{16}}}$$ $$=\frac{2\sqrt{2}}{4}\cdot \frac{16}{14}=\frac{4 \sqrt{2}}{7}$$ Let's now look at a similar example.
Example #2: Find the exact value of all six trigonometric functions. $$\text{cos 2θ}=-\frac{7}{25}$$ θ is in quadrant IV.
Let's begin by finding sin θ. We can use our double-angle identity: $$\text{cos 2θ}=1 - 2\text{sin}^2 θ$$ $$-\frac{7}{25}=1 - 2\text{sin}^2 θ$$ Swap Sides: $$2\text{sin}^2 θ=\frac{7}{25}+ 1$$ $$2\text{sin}^2 θ=\frac{7}{25}+ \frac{25}{25}$$ $$2\text{sin}^2 θ=\frac{32}{25}$$ Multiply both sides by 1/2: $$\frac{1}{2}\cdot 2\text{sin}^2 θ=\frac{1}{2}\cdot \frac{32}{25}$$ $$\text{sin}^2 θ=\frac{16}{25}$$ $$\text{sin θ}=\pm \sqrt{\frac{16}{25}}$$ Since sin θ is negative in quadrant IV, we will use the negative square root: $$\text{sin θ}=- \frac{4}{5}$$ To find cos θ, let's use our Pythagorean identity: $$\text{sin}^2 θ + \text{cos}^2 θ=1$$ $$\left(-\frac{4}{5}\right)^2 + \text{cos}^2 θ=1$$ $$\frac{16}{25}+ \text{cos}^2 θ=1$$ $$\text{cos}^2 θ=\frac{25}{25}- \frac{16}{25}$$ $$\text{cos}^2 θ=\frac{9}{25}$$ $$\text{cos θ}=\pm \sqrt{\frac{9}{25}}$$ Since cos θ is positive in quadrant IV, we will use the principal square root: $$\text{cos θ}=\frac{3}{5}$$ Now, we can use sin θ and cos θ to find the other four trigonometric function values. $$\text{tan θ}=\frac{\text{sin θ}}{\text{cos θ}}$$ $$\text{tan θ}=-\frac{4}{5}\cdot \frac{5}{3}=-\frac{4}{3}$$ $$\text{cot θ}=\frac{1}{\text{tan θ}}$$ $$\text{cot θ}=\frac{1}{\large{-\frac{4}{3}}}=-\frac{3}{4}$$ $$\text{csc θ}=\frac{1}{\text{sin θ}}$$ $$\text{csc θ}=\frac{1}{\large{-\frac{4}{5}}}=-\frac{5}{4}$$ $$\text{sec θ}=\frac{1}{\text{cos θ}}$$ $$\text{sec θ}=\frac{1}{\large{\frac{3}{5}}}=\frac{5}{3}$$

### Verifying a Double-Angle Identity

We will also see problems that ask us to verify an identity using the double-angle identity formulas. Let's look at a couple of examples.
Example #3: Verify each identity. $$\text{cot}\hspace{.1em}x (1 - \text{cos}\hspace{.1em}2x)=\text{sin}\hspace{.1em}2x$$ Since the right-hand side is less complex, let's flip for formating: $$\text{sin}\hspace{.1em}2x=\text{cot}\hspace{.1em}x (1 - \text{cos}\hspace{.1em}2x)$$ Let's write cotangent in terms of sine and cosine: $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}(1 - \text{cos}\hspace{.1em}2x)$$ Let's use our double-angle identity for cosine: $$\text{cos 2}\hspace{.1em}x=1 - 2 \text{sin}^2 x$$ $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}(1 - (1 - 2\text{sin}^2 x))$$ Simplify: $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}(1 - 1 + 2\text{sin}^2 x)$$ $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}(2\text{sin}^2 x)$$ $$=2\text{cos}\hspace{.1em}x \hspace{.1em}\text{sin}\hspace{.1em}x$$ Let's use our double-angle identity for sine: $$\text{sin}\hspace{.1em}2x=2 \text{sin}\hspace{.1em}x \text{cos}\hspace{.1em}x$$ $$=\text{sin}\hspace{.1em}2x ✓$$ Example #4: Verify each identity. $$1 + \text{cos}\hspace{.1em}2x=\text{cot}\hspace{.1em}x \hspace{.1em}\text{sin}\hspace{.1em}2x$$ Let's write the right-hand side in terms of sine and cosine: $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}\cdot \text{sin}\hspace{.1em}2x$$ Use the double-angle identity for sine: $$\text{sin}\hspace{.1em}2x=2 \text{sin}\hspace{.1em}x \hspace{.1em}\text{cos}\hspace{.1em}x$$ $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}\cdot 2 \text{sin}\hspace{.1em}x \hspace{.1em}\text{cos}\hspace{.1em}x$$ Simplify: $$=2\text{cos}^2 x$$ Use the double-angle identity for cosine: $$\text{cos}\hspace{.1em}2x=2 \text{cos}^2 x - 1$$ Alternative Form: $$1 + \text{cos}\hspace{.1em}2x=2 \text{cos}^2 x$$ Replace: $$=2\text{cos}^2 x$$ $$=1 + \text{cos}\hspace{.1em}2x ✓$$

### Product-to-Sum Identities

Additionally, we will be asked to simplify or find the exact value using the product-to-sum identities. $$\text{cos A}\hspace{.1em}\text{cos B}=\frac{1}{2}\left[\text{cos}(A + B) + \text{cos}(A - B)\right]$$ $$\text{sin A}\hspace{.1em}\text{sin B}=\frac{1}{2}\left[\text{cos}(A - B) - \text{cos}(A + B)\right]$$ $$\text{sin A}\hspace{.1em}\text{cos B}=\frac{1}{2}\left[\text{sin}(A + B) + \text{sin}(A - B)\right]$$ $$\text{cos A}\hspace{.1em}\text{sin B}=\frac{1}{2}\left[\text{sin}(A + B) - \text{sin}(A - B)\right]$$ Let's look at an example.
Example #5: Find the exact value of each expression. $$2 \text{sin}\hspace{.1em}105° \hspace{.1em}\text{cos}\hspace{.1em}75°$$ We will use our identity from above: $$\text{sin A}\hspace{.1em}\text{cos B}=\frac{1}{2}\left[\text{sin}(A + B) + \text{sin}(A - B)\right]$$ $$2 \text{sin}\hspace{.1em}105° \hspace{.1em}\text{cos}\hspace{.1em}75°$$ $$=2\left(\frac{1}{2}\left[\text{sin}(105° + 75°) + \text{sin}(105° - 75°)\right]\right)$$ $$=2\left(\frac{1}{2}\left[\text{sin}(180°) + \text{sin}(30°)\right]\right)$$ $$=\text{sin}(180°) + \text{sin}(30°)$$ $$=0 + \frac{1}{2}$$ $$=\frac{1}{2}$$

### Sum-to-Product Identities

Lastly, we will be asked to simplify or find the exact value using the sum-to-product identities. $$\text{sin A}+ \text{sin B}=2 \text{sin}\left(\frac{A + B}{2}\right) \hspace{.1em}\text{cos}\left(\frac{A - B}{2}\right)$$ $$\text{sin A}- \text{sin B}=2 \text{cos}\left(\frac{A + B}{2}\right) \hspace{.1em}\text{sin}\left(\frac{A - B}{2}\right)$$ $$\text{cos A}+ \text{cos B}=2 \text{cos}\left(\frac{A + B}{2}\right) \hspace{.1em}\text{cos}\left(\frac{A - B}{2}\right)$$ $$\text{cos A}- \text{cos B}=-2 \text{sin}\left(\frac{A + B}{2}\right) \hspace{.1em}\text{sin}\left(\frac{A - B}{2}\right)$$ Let's look at an example.
Example #6: Write as a product. $$\text{cos}\hspace{.1em}13θ - \text{cos}\hspace{.1em}5θ$$ We will use our identity from above: $$\text{cos A}- \text{cos B}=-2 \text{sin}\left(\frac{A + B}{2}\right) \hspace{.1em}\text{sin}\left(\frac{A - B}{2}\right)$$ $$\text{cos}\hspace{.1em}13θ - \text{cos}\hspace{.1em}5θ$$ $$=-2 \text{sin}\left(\frac{13 θ + 5 θ}{2}\right) \text{sin}\left(\frac{13 θ - 5 θ}{2}\right)$$ $$=-2 \text{sin}\left(\frac{18 θ}{2}\right) \text{sin}\left(\frac{8 θ}{2}\right)$$ $$=-2 \text{sin}\hspace{.1em}9θ \hspace{.1em}\text{sin}\hspace{.1em}4θ$$

#### Skills Check:

Example #1

Find the exact value. $$\text{cos}\hspace{.1em}θ=-\frac{8}{17}$$ θ is in quadrant II.
Find sin 2θ.

A
$$\frac{161}{289}$$
B
$$\frac{161}{240}$$
C
$$-\frac{289}{161}$$
D
$$-\frac{289}{240}$$
E
$$-\frac{240}{289}$$

Example #2

Find the exact value. $$4\text{sin}\frac{11π}{12}- 4\text{sin}\frac{5π}{12}$$

A
$$2\sqrt{2}$$
B
$$-2\sqrt{6}$$
C
$$2\sqrt{6}$$
D
$$-2\sqrt{2}$$
E
$$-\sqrt{3}$$

Example #3

Find the exact value. $$-3\text{cos}\hspace{.1em}105° \hspace{.1em}\text{sin}\hspace{.1em}45°$$

A
$$\frac{-3\sqrt{3}- 3}{4}$$
B
$$\frac{-3 + 3\sqrt{3}}{4}$$
C
$$-\frac{3}{2}$$
D
$$\frac{2}{3}$$
E
$$\frac{\sqrt{5}+ 1}{3}$$

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