Test Objectives
• Demonstrate the ability to find exact trigonometric function values using double-angle identities
• Demonstrate the ability to verify an identity using double-angle identities
• Demonstrate the ability to find exact trigonometric expression values using product-to-sum identities
• Demonstrate the ability to find exact trigonometric expression values using sum-to-product identities
Double-Angle Identities Practice Test:

#1:

Instructions: Derive the given identities.

$$a)\hspace{.1em}\text{cos}\hspace{.1em}2A$$

$$b)\hspace{.1em}\text{sin}\hspace{.1em}2A$$

$$c)\hspace{.1em}\text{tan}\hspace{.1em}2A$$

$$d)\hspace{.1em}\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B$$

#2:

Instructions: Find the exact function value.

$$a)\hspace{.1em}\text{tan}\hspace{.1em}θ=-\frac{12}{5}$$ θ is in quadrant II.
Find sin 2θ.

$$b)\hspace{.1em}\text{sin}\hspace{.1em}θ=\frac{\sqrt{231}}{20}$$ θ is in quadrant II.
Find cos 2θ.

$$c)\hspace{.1em}\text{cos}\hspace{.1em}θ=\frac{5}{13}$$ θ is in quadrant IV.
Find cos 2θ.

$$d)\hspace{.1em}\text{sin}\hspace{.1em}θ=\frac{15}{17}$$ θ is in quadrant I.
Find tan 2θ.

#3:

Instructions: Verify each identity.

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$$a)\hspace{.1em}2\text{sin}^2x(1 + \text{cot}\hspace{.1em}x)=\text{sin}\hspace{.1em}2x - \text{cos}\hspace{.1em}2x + 1$$

$$b)\hspace{.1em}\frac{\text{csc}^2 x}{\text{sin}^2 x + \text{cos}2x}=\frac{\text{tan}^2 x + 1}{\text{sin}^2 x}$$

#4:

Instructions: Find the exact value of each expression.

$$a)\hspace{.1em}3\text{cos}\hspace{.1em}\frac{5π}{12}\hspace{.1em}\text{sin}\hspace{.1em}\frac{13π}{12}$$

$$b)\hspace{.1em}{-}3\left(\text{sin}\hspace{.1em}\frac{π}{12}+ \text{sin}\frac{5π}{12}\right)$$

#5:

Instructions: Write the product as a difference.

$$a)\hspace{.1em}\text{cos}\hspace{.1em}8θ \hspace{.1em}\text{sin}\hspace{.1em}2θ$$

Instructions: Write the sum as a product.

$$b) \hspace{.1em}2(\text{sin}\hspace{.1em}13B + \text{sin}\hspace{.1em}3B)$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.1em}\text{cos}\hspace{.1em}2A$$ $$=\text{cos}(A + A)$$ $$=\text{cos}\hspace{.1em}A \hspace{.1em}\cdot \hspace{.1em}\text{cos}\hspace{.1em}A \hspace{.1em}-\hspace{.1em}\text{sin}\hspace{.1em}A \hspace{.1em}\cdot \hspace{.1em}\text{sin}\hspace{.1em}A$$ $$=\text{cos}^2 A - \text{sin}^2 A$$

$$b)\hspace{.1em}\text{sin}\hspace{.1em}2A$$ $$=\text{sin}(A + A)$$ $$=\text{sin}\hspace{.1em}A \hspace{.1em}\cdot \hspace{.1em}\text{cos}\hspace{.1em}A \hspace{.1em}+ \hspace{.1em}\text{cos}\hspace{.1em}A \hspace{.1em}\cdot \hspace{.1em}\text{sin}\hspace{.1em}A$$ $$=2\hspace{.1em}\text{sin}\hspace{.1em}A\hspace{.1em}\text{cos}\hspace{.1em}A\hspace{.1em}$$

$$c)\hspace{.1em}\text{tan}\hspace{.1em}2A$$ $$=\text{tan}\hspace{.1em}(A + A)$$ $$=\frac{\text{tan}\hspace{.1em}A\hspace{.1em}+ \hspace{.1em}\text{tan}\hspace{.1em}A}{1 \hspace{.1em}-\hspace{.1em}\text{tan}\hspace{.1em}A\hspace{.1em}\cdot \text{tan}\hspace{.1em}A}$$ $$=\frac{2\hspace{.1em}\text{tan}\hspace{.1em}A}{1 \hspace{.1em}- \hspace{.1em}\text{tan}^2 \hspace{.1em}A}$$

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$$d)\hspace{.1em}\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B$$ $$\text{cos}(A + B)=\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B \hspace{.1em}-\hspace{.1em}\text{sin}\hspace{.1em}A\hspace{.1em}\text{sin}\hspace{.1em}B$$ $$\text{cos}(A - B)=\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B \hspace{.1em}+\hspace{.1em}\text{sin}\hspace{.1em}A\hspace{.1em}\text{sin}\hspace{.1em}B$$ Add the left sides and set this equal to the sum of the right sides: $$\text{cos}(A + B) \hspace{.1em}+\hspace{.1em}\text{cos}(A - B)=2\hspace{.1em}\text{cos}\hspace{.1em}A\hspace{.1em}\text{cos}\hspace{.1em}B$$ Multiply both sides by 1/2: $$\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B$$ $$=\frac{1}{2}\left[\text{cos}(A + B) \hspace{.1em}\hspace{.1em}+ \hspace{.1em}\text{cos}(A - B)\right]$$

#2:

Solutions:

$$a)\hspace{.1em}\text{sin}\hspace{.1em}2θ=-\frac{120}{169}$$

$$b)\hspace{.1em}\text{cos}\hspace{.1em}2θ=-\frac{31}{200}$$

$$c)\hspace{.1em}\text{cos}\hspace{.1em}2θ=-\frac{119}{169}$$

$$d)\hspace{.1em}\text{tan}\hspace{.1em}2θ=-\frac{240}{161}$$

#3:

Solutions:

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$$a)\hspace{.1em}2\text{sin}^2x(1 + \text{cot}\hspace{.1em}x)=\text{sin}\hspace{.1em}2x - \text{cos}\hspace{.1em}2x + 1$$ $$\text{sin}\hspace{.1em}2x - \text{cos}\hspace{.1em}2x + 1=2\text{sin}^2x(1 + \text{cot}\hspace{.1em}x)$$ $$=2\text{sin}^2 x\left(1 + \frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}\right)$$ $$=2\text{sin}^2 x\left(\frac{\text{cos}\hspace{.1em}x + \text{sin}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}\right)$$ $$=2\text{sin}\hspace{.1em}x(\text{cos}\hspace{.1em}x + \text{sin}\hspace{.1em}x)$$ $$=2\text{sin}\hspace{.1em}x \hspace{.1em}\text{cos}\hspace{.1em}x + 2\text{sin}^2 x$$ $$=\text{sin}\hspace{.1em}2x + 2\text{sin}^2 x$$ $$=\text{sin}\hspace{.1em}2x + 1 - \text{cos}\hspace{.1em}2x$$ $$=\text{sin}\hspace{.1em}2x - \text{cos}\hspace{.1em}2x + 1 ✓$$

$$b)\hspace{.1em}\frac{\text{csc}^2 x}{\text{sin}^2 x + \text{cos}2x}=\frac{\text{tan}^2 x + 1}{\text{sin}^2 x}$$ $$\frac{\text{tan}^2 x + 1}{\text{sin}^2 x}=\frac{\text{csc}^2 x}{\text{sin}^2 x + \text{cos}2x}$$ $$=\frac{\text{csc}^2 x}{\text{sin}^2 x + \text{cos}^2 x - \text{sin}^2 x}$$ $$=\frac{\text{csc}^2 x}{\text{cos}^2 x}$$ $$=\text{csc}^2 x \hspace{.1em}\text{sec}^2 x$$ $$=\text{csc}^2 x (\text{tan}^2 x + 1)$$ $$=\frac{\text{tan}^2 x + 1}{\text{sin}^2 x}✓$$

#4:

Solutions:

$$a)\hspace{.1em}\frac{-6 + 3\sqrt{3}}{4}$$

$$b)\hspace{.1em}-\frac{3\sqrt{6}}{2}$$

#5:

Solutions:

$$a)\hspace{.1em}\frac{\text{sin}\hspace{.1em}10θ - \text{sin}\hspace{.1em}6θ}{2}$$

$$b)\hspace{.1em}4\hspace{.1em}\text{sin}\hspace{.1em}8B \hspace{.1em}\text{cos}\hspace{.1em}5B$$