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, the sum and difference identities for cosine, and the sum and difference identities for sine and tangent. In this lesson, we will focus on the double-angle identities, along with the product-to-sum identities, and the sum-to-product identities.
Example #1: Find the exact value of each. $$\text{cos}\hspace{.1em}θ=\frac{2\sqrt{2}}{3}$$ $$\text{sin}\hspace{.1em}θ > 0$$ 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 sin θ: $$\text{cos}^2 θ + \text{sin}^2 θ=1$$ $$\left(\frac{2 \sqrt{2}}{3}\right)^2 + \text{sin}^2 θ=1$$ $$\frac{8}{9}+ \text{sin}^2 θ=1$$ $$\text{sin}^2 θ=\frac{9}{9}- \frac{8}{9}$$ $$\text{sin}^2 θ=\frac{1}{9}$$ $$\text{sin}\hspace{.1em}θ=\pm \sqrt{\frac{1}{9}}$$ Since we know that θ is in quadrant I (sine and cosine are both positive), 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: $$\text{cos}\hspace{.1em}2θ=\text{cos}^2 θ - \text{sin}^2 θ$$ $$=\left(\frac{2 \sqrt{2}}{3}\right)^2 - \left(\frac{1}{3}\right)^2$$ $$=\frac{8}{9}- \frac{1}{9}$$ $$=\frac{7}{9}$$ $$\text{sin}2θ=2 \hspace{.1em}\text{sin}θ \cdot \text{cos}θ$$ $$=2 \cdot \frac{1}{3}\cdot \frac{2\sqrt{2}}{3}$$ $$=\frac{4\sqrt{2}}{9}$$ $$\text{tan}\hspace{.1em}2θ=\frac{2 \text{tan}\hspace{.1em}θ}{1 - \text{tan}^2 θ}$$ $$=\frac{2 \cdot \Large{\frac{\sqrt{2}}{4}}}{1 - \left(\large{\frac{\sqrt{2}}{{4}}}\right)^2}$$ $$=\frac{\Large{\frac{2\sqrt{2}}{4}}}{1 - \large{\frac{2}{{16}}}}$$ $$=\frac{\Large{\frac{2\sqrt{2}}{4}}}{\large{\frac{16}{16}}- \large{\frac{2}{{16}}}}$$ $$=\frac{\Large{\frac{2\sqrt{2}}{4}}}{\large{\frac{14}{16}}}$$ $$=\frac{2\sqrt{2}}{4}\cdot \frac{16}{14}$$ $$=\frac{4 \sqrt{2}}{7}$$ Let's now look at a similar example.
Example #2: Find the exact value of all six trigonometric functions. $$\text{cos}\hspace{.1em}2θ=-\frac{7}{25}$$ θ is in quadrant IV.
Let's begin by finding sin θ. We can use our double-angle identity: $$\text{cos}\hspace{.1em}2θ=1 - 2\text{sin}^2 θ$$ $$-\frac{7}{25}=1 - 2\text{sin}^2 θ$$ 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: $$\frac{1}{2}\cdot 2\text{sin}^2 θ=\frac{1}{2}\cdot \frac{32}{25}$$ $$\text{sin}^2 θ=\frac{16}{25}$$ $$\text{sin}\hspace{.1em}θ=\pm \sqrt{\frac{16}{25}}$$ Since sin θ is negative in quadrant IV, we will use the negative square root: $$\text{sin}\hspace{.1em}θ=- \frac{4}{5}$$ To find cos θ, let's use our Pythagorean identity: $$\text{sin}^2 θ + \text{cos}^2 θ=1$$ $$\left(-\frac{4}{5}\right)^2 + \text{cos}^2 θ=1$$ $$\frac{16}{25}+ \text{cos}^2 θ=1$$ $$\text{cos}^2 θ=\frac{25}{25}- \frac{16}{25}$$ $$\text{cos}^2 θ=\frac{9}{25}$$ $$\text{cos}\hspace{.1em}θ=\pm \sqrt{\frac{9}{25}}$$ Since cos θ is positive in quadrant IV, we will use the principal square root: $$\text{cos}\hspace{.1em}θ=\frac{3}{5}$$ Now, we can use sin θ and cos θ to find the other four trigonometric function values. $$\text{tan}\hspace{.1em}θ=\frac{\text{sin}\hspace{.1em}θ}{\text{cos}\hspace{.1em}θ}$$ $$\text{tan}\hspace{.1em}θ=-\frac{4}{5}\cdot \frac{5}{3}=-\frac{4}{3}$$ $$\text{cot}\hspace{.1em}θ=\frac{1}{\text{tan}\hspace{.1em}θ}$$ $$\text{cot}\hspace{.1em}θ=\frac{1}{\large{-\frac{4}{3}}}=-\frac{3}{4}$$ $$\text{csc}\hspace{.1em}θ=\frac{1}{\text{sin}\hspace{.1em}θ}$$ $$\text{csc}\hspace{.1em}θ=\frac{1}{\large{-\frac{4}{5}}}=-\frac{5}{4}$$ $$\text{sec}\hspace{.1em}θ=\frac{1}{\text{cos}\hspace{.1em}θ}$$ $$\text{sec}\hspace{.1em}θ=\frac{1}{\large{\frac{3}{5}}}=\frac{5}{3}$$
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: $$\text{cos}\hspace{.1em}2x=1 - 2 \hspace{.1em}\text{sin}^2 x$$ $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}(1 - (1 - 2 \hspace{.1em}\text{sin}^2 x))$$ Simplify: $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}(1 - 1 + 2 \hspace{.1em}\text{sin}^2 x)$$ $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}(2 \hspace{.1em}\text{sin}^2 x)$$ $$=2\text{cos}\hspace{.1em}x \hspace{.1em}\text{sin}\hspace{.1em}x$$ Let's use our double-angle identity for sine: $$\text{sin}\hspace{.1em}2x=2 \hspace{.1em}\text{sin}\hspace{.1em}x \text{cos}\hspace{.1em}x$$ $$=\text{sin}\hspace{.1em}2x ✓$$ 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: $$=1 + \text{cos}\hspace{.1em}2x ✓$$
Example #5: Find the exact value of each expression. $$2 \hspace{.1em}\text{sin}\hspace{.1em}105° \hspace{.1em}\text{cos}\hspace{.1em}75°$$ We will use our identity from above: $$\text{sin}\hspace{.1em}A\hspace{.1em}\text{cos}\hspace{.1em}B=\frac{1}{2}\left[\text{sin}(A + B) \hspace{.1em}+ \hspace{.1em}\text{sin}(A - B)\right]$$ $$2 \hspace{.1em}\text{sin}\hspace{.1em}105° \hspace{.1em}\text{cos}\hspace{.1em}75°$$ $$=2\left(\frac{1}{2}\left[\text{sin}(105° + 75°) \hspace{.1em}+\hspace{.1em}\text{sin}(105° - 75°)\right]\right)$$ $$=2\left(\frac{1}{2}\left[\text{sin}(180°) \hspace{.1em}+ \hspace{.1em}\text{sin}(30°)\right]\right)$$ $$=\text{sin}(180°) \hspace{.1em}+ \hspace{.1em}\text{sin}(30°)$$ $$=0 + \frac{1}{2}$$ $$=\frac{1}{2}$$
Example #6: Write as a product. $$\text{cos}\hspace{.1em}13θ - \text{cos}\hspace{.1em}5θ$$ We will use our identity from above: $$\text{cos}\hspace{.1em}A- \text{cos}\hspace{.1em}B=-2 \hspace{.1em}\text{sin}\left(\frac{A + B}{2}\right) \hspace{.1em}\text{sin}\left(\frac{A - B}{2}\right)$$ $$\text{cos}\hspace{.1em}13θ - \text{cos}\hspace{.1em}5θ$$ $$=-2 \hspace{.1em}\text{sin}\left(\frac{13 θ + 5 θ}{2}\right) \text{sin}\left(\frac{13 θ - 5 θ}{2}\right)$$ $$=-2 \hspace{.1em}\text{sin}\left(\frac{18 θ}{2}\right) \text{sin}\left(\frac{8 θ}{2}\right)$$ $$=-2 \hspace{.1em}\text{sin}\hspace{.1em}9θ \hspace{.1em}\text{sin}\hspace{.1em}4θ$$
Double-Angle Identities
We will derive these formulas in the practice test section. A full step-by-step is provided in the practice test solution video. $$\text{cos}\hspace{.1em}2A=\text{cos}^2 A - \text{sin}^2 A$$ $$\text{cos}\hspace{.1em}2A=1 - 2 \hspace{.1em}\text{sin}^2 A$$ $$\text{cos}\hspace{.1em}2A=2\hspace{.1em}\text{cos}^2 A - 1$$ $$\text{sin}\hspace{.1em}2A=2 \hspace{.1em}\text{sin}\hspace{.1em}A\hspace{.1em}\text{cos}\hspace{.1em}A$$ $$\text{tan}\hspace{.1em}2A=\frac{2 \hspace{.1em}\text{tan}\hspace{.1em}A}{1 - \text{tan}^2 \text{A}}$$ 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. $$\text{cos}\hspace{.1em}θ=\frac{2\sqrt{2}}{3}$$ $$\text{sin}\hspace{.1em}θ > 0$$ 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 sin θ: $$\text{cos}^2 θ + \text{sin}^2 θ=1$$ $$\left(\frac{2 \sqrt{2}}{3}\right)^2 + \text{sin}^2 θ=1$$ $$\frac{8}{9}+ \text{sin}^2 θ=1$$ $$\text{sin}^2 θ=\frac{9}{9}- \frac{8}{9}$$ $$\text{sin}^2 θ=\frac{1}{9}$$ $$\text{sin}\hspace{.1em}θ=\pm \sqrt{\frac{1}{9}}$$ Since we know that θ is in quadrant I (sine and cosine are both positive), 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: $$\text{cos}\hspace{.1em}2θ=\text{cos}^2 θ - \text{sin}^2 θ$$ $$=\left(\frac{2 \sqrt{2}}{3}\right)^2 - \left(\frac{1}{3}\right)^2$$ $$=\frac{8}{9}- \frac{1}{9}$$ $$=\frac{7}{9}$$ $$\text{sin}2θ=2 \hspace{.1em}\text{sin}θ \cdot \text{cos}θ$$ $$=2 \cdot \frac{1}{3}\cdot \frac{2\sqrt{2}}{3}$$ $$=\frac{4\sqrt{2}}{9}$$ $$\text{tan}\hspace{.1em}2θ=\frac{2 \text{tan}\hspace{.1em}θ}{1 - \text{tan}^2 θ}$$ $$=\frac{2 \cdot \Large{\frac{\sqrt{2}}{4}}}{1 - \left(\large{\frac{\sqrt{2}}{{4}}}\right)^2}$$ $$=\frac{\Large{\frac{2\sqrt{2}}{4}}}{1 - \large{\frac{2}{{16}}}}$$ $$=\frac{\Large{\frac{2\sqrt{2}}{4}}}{\large{\frac{16}{16}}- \large{\frac{2}{{16}}}}$$ $$=\frac{\Large{\frac{2\sqrt{2}}{4}}}{\large{\frac{14}{16}}}$$ $$=\frac{2\sqrt{2}}{4}\cdot \frac{16}{14}$$ $$=\frac{4 \sqrt{2}}{7}$$ Let's now look at a similar example.
Example #2: Find the exact value of all six trigonometric functions. $$\text{cos}\hspace{.1em}2θ=-\frac{7}{25}$$ θ is in quadrant IV.
Let's begin by finding sin θ. We can use our double-angle identity: $$\text{cos}\hspace{.1em}2θ=1 - 2\text{sin}^2 θ$$ $$-\frac{7}{25}=1 - 2\text{sin}^2 θ$$ 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: $$\frac{1}{2}\cdot 2\text{sin}^2 θ=\frac{1}{2}\cdot \frac{32}{25}$$ $$\text{sin}^2 θ=\frac{16}{25}$$ $$\text{sin}\hspace{.1em}θ=\pm \sqrt{\frac{16}{25}}$$ Since sin θ is negative in quadrant IV, we will use the negative square root: $$\text{sin}\hspace{.1em}θ=- \frac{4}{5}$$ To find cos θ, let's use our Pythagorean identity: $$\text{sin}^2 θ + \text{cos}^2 θ=1$$ $$\left(-\frac{4}{5}\right)^2 + \text{cos}^2 θ=1$$ $$\frac{16}{25}+ \text{cos}^2 θ=1$$ $$\text{cos}^2 θ=\frac{25}{25}- \frac{16}{25}$$ $$\text{cos}^2 θ=\frac{9}{25}$$ $$\text{cos}\hspace{.1em}θ=\pm \sqrt{\frac{9}{25}}$$ Since cos θ is positive in quadrant IV, we will use the principal square root: $$\text{cos}\hspace{.1em}θ=\frac{3}{5}$$ Now, we can use sin θ and cos θ to find the other four trigonometric function values. $$\text{tan}\hspace{.1em}θ=\frac{\text{sin}\hspace{.1em}θ}{\text{cos}\hspace{.1em}θ}$$ $$\text{tan}\hspace{.1em}θ=-\frac{4}{5}\cdot \frac{5}{3}=-\frac{4}{3}$$ $$\text{cot}\hspace{.1em}θ=\frac{1}{\text{tan}\hspace{.1em}θ}$$ $$\text{cot}\hspace{.1em}θ=\frac{1}{\large{-\frac{4}{3}}}=-\frac{3}{4}$$ $$\text{csc}\hspace{.1em}θ=\frac{1}{\text{sin}\hspace{.1em}θ}$$ $$\text{csc}\hspace{.1em}θ=\frac{1}{\large{-\frac{4}{5}}}=-\frac{5}{4}$$ $$\text{sec}\hspace{.1em}θ=\frac{1}{\text{cos}\hspace{.1em}θ}$$ $$\text{sec}\hspace{.1em}θ=\frac{1}{\large{\frac{3}{5}}}=\frac{5}{3}$$
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: $$\text{cos}\hspace{.1em}2x=1 - 2 \hspace{.1em}\text{sin}^2 x$$ $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}(1 - (1 - 2 \hspace{.1em}\text{sin}^2 x))$$ Simplify: $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}(1 - 1 + 2 \hspace{.1em}\text{sin}^2 x)$$ $$=\frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}(2 \hspace{.1em}\text{sin}^2 x)$$ $$=2\text{cos}\hspace{.1em}x \hspace{.1em}\text{sin}\hspace{.1em}x$$ Let's use our double-angle identity for sine: $$\text{sin}\hspace{.1em}2x=2 \hspace{.1em}\text{sin}\hspace{.1em}x \text{cos}\hspace{.1em}x$$ $$=\text{sin}\hspace{.1em}2x ✓$$ 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: $$=1 + \text{cos}\hspace{.1em}2x ✓$$
Product-to-Sum Identities
Additionally, we will be asked to simplify or find the exact value using the product-to-sum identities. $$\text{cos}\hspace{.1em}A\hspace{.1em}\text{cos}\hspace{.1em}B=\frac{1}{2}\left[\text{cos}(A + B)\hspace{.1em}+ \hspace{.1em}\text{cos}(A - B)\right]$$ $$\text{sin}\hspace{.1em}A\hspace{.1em}\text{sin}\hspace{.1em}B=\frac{1}{2}\left[\text{cos}(A - B)\hspace{.1em}- \hspace{.1em}\text{cos}(A + B)\right]$$ $$\text{sin}\hspace{.1em}A\hspace{.1em}\text{cos}\hspace{.1em}B=\frac{1}{2}\left[\text{sin}(A + B)\hspace{.1em}+\hspace{.1em}\text{sin}(A - B)\right]$$ $$\text{cos}\hspace{.1em}A\hspace{.1em}\text{sin}\hspace{.1em}B=\frac{1}{2}\left[\text{sin}(A + B)\hspace{.1em}- \hspace{.1em}\text{sin}(A - B)\right]$$ Let's look at an example.Example #5: Find the exact value of each expression. $$2 \hspace{.1em}\text{sin}\hspace{.1em}105° \hspace{.1em}\text{cos}\hspace{.1em}75°$$ We will use our identity from above: $$\text{sin}\hspace{.1em}A\hspace{.1em}\text{cos}\hspace{.1em}B=\frac{1}{2}\left[\text{sin}(A + B) \hspace{.1em}+ \hspace{.1em}\text{sin}(A - B)\right]$$ $$2 \hspace{.1em}\text{sin}\hspace{.1em}105° \hspace{.1em}\text{cos}\hspace{.1em}75°$$ $$=2\left(\frac{1}{2}\left[\text{sin}(105° + 75°) \hspace{.1em}+\hspace{.1em}\text{sin}(105° - 75°)\right]\right)$$ $$=2\left(\frac{1}{2}\left[\text{sin}(180°) \hspace{.1em}+ \hspace{.1em}\text{sin}(30°)\right]\right)$$ $$=\text{sin}(180°) \hspace{.1em}+ \hspace{.1em}\text{sin}(30°)$$ $$=0 + \frac{1}{2}$$ $$=\frac{1}{2}$$
Sum-to-Product Identities
Lastly, we will be asked to simplify or find the exact value using the sum-to-product identities. $$\text{sin}\hspace{.1em}A \hspace{.1em}+ \hspace{.1em}\text{sin}\hspace{.1em}B=2 \hspace{.1em}\text{sin}\left(\frac{A + B}{2}\right) \hspace{.1em}\text{cos}\left(\frac{A - B}{2}\right)$$ $$\text{sin}\hspace{.1em}A \hspace{.1em}- \hspace{.1em}\text{sin}\hspace{.1em}B=2 \hspace{.1em}\text{cos}\left(\frac{A + B}{2}\right) \hspace{.1em}\text{sin}\left(\frac{A - B}{2}\right)$$ $$\text{cos}\hspace{.1em}A \hspace{.1em}+ \hspace{.1em}\text{cos}\hspace{.1em}B=2 \hspace{.1em}\text{cos}\left(\frac{A + B}{2}\right) \hspace{.1em}\text{cos}\left(\frac{A - B}{2}\right)$$ $$\text{cos}\hspace{.1em}A \hspace{.1em}- \hspace{.1em}\text{cos}\hspace{.1em}B=-2\hspace{.1em}\text{sin}\left(\frac{A + B}{2}\right) \hspace{.1em}\text{sin}\left(\frac{A - B}{2}\right)$$ 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: $$\text{cos}\hspace{.1em}A- \text{cos}\hspace{.1em}B=-2 \hspace{.1em}\text{sin}\left(\frac{A + B}{2}\right) \hspace{.1em}\text{sin}\left(\frac{A - B}{2}\right)$$ $$\text{cos}\hspace{.1em}13θ - \text{cos}\hspace{.1em}5θ$$ $$=-2 \hspace{.1em}\text{sin}\left(\frac{13 θ + 5 θ}{2}\right) \text{sin}\left(\frac{13 θ - 5 θ}{2}\right)$$ $$=-2 \hspace{.1em}\text{sin}\left(\frac{18 θ}{2}\right) \text{sin}\left(\frac{8 θ}{2}\right)$$ $$=-2 \hspace{.1em}\text{sin}\hspace{.1em}9θ \hspace{.1em}\text{sin}\hspace{.1em}4θ$$
Skills Check:
Example #1
Find the exact value. $$\text{cos}\hspace{.1em}θ=-\frac{8}{17}$$ θ is in quadrant II.
Find sin 2θ.
Please choose the best answer.
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. $$4\text{sin}\frac{11π}{12}- 4\text{sin}\frac{5π}{12}$$
Please choose the best answer.
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. $$-3\text{cos}\hspace{.1em}105° \hspace{.1em}\text{sin}\hspace{.1em}45°$$
Please choose the best answer.
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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