- 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
#1:
Instructions: Derive the given identities.
$$a)\hspace{.1em}\text{cos}\hspace{.1em}2A$$
$$b)\hspace{.1em}\text{sin}\hspace{.1em}2A$$
$$c)\hspace{.1em}\text{tan}\hspace{.1em}2A$$
$$d)\hspace{.1em}\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B$$
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#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.
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#3:
Instructions: Verify each identity.
$$a)\hspace{.1em}2\text{sin}^2x(1 + \text{cot}\hspace{.1em}x)=\text{sin}\hspace{.1em}2x - \text{cos}\hspace{.1em}2x + 1$$
$$b)\hspace{.1em}\frac{\text{csc}^2 x}{\text{sin}^2 x + \text{cos}2x}=\frac{\text{tan}^2 x + 1}{\text{sin}^2 x}$$
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#4:
Instructions: Find the exact value of each expression.
$$a)\hspace{.1em}3\text{cos}\hspace{.1em}\frac{5π}{12}\hspace{.1em}\text{sin}\hspace{.1em}\frac{13π}{12}$$
$$b)\hspace{.1em}{-}3\left(\text{sin}\hspace{.1em}\frac{π}{12}+ \text{sin}\frac{5π}{12}\right)$$
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#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.
$$b) \hspace{.1em}2(\text{sin}\hspace{.1em}13B + \text{sin}\hspace{.1em}3B)$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.1em}\text{cos}\hspace{.1em}2A$$ $$=\text{cos}(A + A)$$ $$=\text{cos}\hspace{.1em}A \hspace{.1em}\cdot \hspace{.1em}\text{cos}\hspace{.1em}A \hspace{.1em}-\hspace{.1em}\text{sin}\hspace{.1em}A \hspace{.1em}\cdot \hspace{.1em}\text{sin}\hspace{.1em}A$$ $$=\text{cos}^2 A - \text{sin}^2 A$$
$$b)\hspace{.1em}\text{sin}\hspace{.1em}2A$$ $$=\text{sin}(A + A)$$ $$=\text{sin}\hspace{.1em}A \hspace{.1em}\cdot \hspace{.1em}\text{cos}\hspace{.1em}A \hspace{.1em}+ \hspace{.1em}\text{cos}\hspace{.1em}A \hspace{.1em}\cdot \hspace{.1em}\text{sin}\hspace{.1em}A$$ $$=2\hspace{.1em}\text{sin}\hspace{.1em}A\hspace{.1em}\text{cos}\hspace{.1em}A\hspace{.1em}$$
$$c)\hspace{.1em}\text{tan}\hspace{.1em}2A$$ $$=\text{tan}\hspace{.1em}(A + A)$$ $$=\frac{\text{tan}\hspace{.1em}A\hspace{.1em}+ \hspace{.1em}\text{tan}\hspace{.1em}A}{1 \hspace{.1em}-\hspace{.1em}\text{tan}\hspace{.1em}A\hspace{.1em}\cdot \text{tan}\hspace{.1em}A}$$ $$=\frac{2\hspace{.1em}\text{tan}\hspace{.1em}A}{1 \hspace{.1em}- \hspace{.1em}\text{tan}^2 \hspace{.1em}A}$$
$$d)\hspace{.1em}\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B$$ $$\text{cos}(A + B)=\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B \hspace{.1em}-\hspace{.1em}\text{sin}\hspace{.1em}A\hspace{.1em}\text{sin}\hspace{.1em}B$$ $$\text{cos}(A - B)=\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B \hspace{.1em}+\hspace{.1em}\text{sin}\hspace{.1em}A\hspace{.1em}\text{sin}\hspace{.1em}B$$ 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: $$\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B$$ $$=\frac{1}{2}\left[\text{cos}(A + B) \hspace{.1em}\hspace{.1em}+ \hspace{.1em}\text{cos}(A - B)\right]$$
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#2:
Solutions:
$$a)\hspace{.1em}\text{sin}\hspace{.1em}2θ=-\frac{120}{169}$$
$$b)\hspace{.1em}\text{cos}\hspace{.1em}2θ=-\frac{31}{200}$$
$$c)\hspace{.1em}\text{cos}\hspace{.1em}2θ=-\frac{119}{169}$$
$$d)\hspace{.1em}\text{tan}\hspace{.1em}2θ=-\frac{240}{161}$$
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#3:
Solutions:
$$a)\hspace{.1em}2\text{sin}^2x(1 + \text{cot}\hspace{.1em}x)=\text{sin}\hspace{.1em}2x - \text{cos}\hspace{.1em}2x + 1$$ $$\text{sin}\hspace{.1em}2x - \text{cos}\hspace{.1em}2x + 1=2\text{sin}^2x(1 + \text{cot}\hspace{.1em}x)$$ $$=2\text{sin}^2 x\left(1 + \frac{\text{cos}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}\right)$$ $$=2\text{sin}^2 x\left(\frac{\text{cos}\hspace{.1em}x + \text{sin}\hspace{.1em}x}{\text{sin}\hspace{.1em}x}\right)$$ $$=2\text{sin}\hspace{.1em}x(\text{cos}\hspace{.1em}x + \text{sin}\hspace{.1em}x)$$ $$=2\text{sin}\hspace{.1em}x \hspace{.1em}\text{cos}\hspace{.1em}x + 2\text{sin}^2 x$$ $$=\text{sin}\hspace{.1em}2x + 2\text{sin}^2 x$$ $$=\text{sin}\hspace{.1em}2x + 1 - \text{cos}\hspace{.1em}2x$$ $$=\text{sin}\hspace{.1em}2x - \text{cos}\hspace{.1em}2x + 1 ✓$$
$$b)\hspace{.1em}\frac{\text{csc}^2 x}{\text{sin}^2 x + \text{cos}2x}=\frac{\text{tan}^2 x + 1}{\text{sin}^2 x}$$ $$\frac{\text{tan}^2 x + 1}{\text{sin}^2 x}=\frac{\text{csc}^2 x}{\text{sin}^2 x + \text{cos}2x}$$ $$=\frac{\text{csc}^2 x}{\text{sin}^2 x + \text{cos}^2 x - \text{sin}^2 x}$$ $$=\frac{\text{csc}^2 x}{\text{cos}^2 x}$$ $$=\text{csc}^2 x \hspace{.1em}\text{sec}^2 x$$ $$=\text{csc}^2 x (\text{tan}^2 x + 1)$$ $$=\frac{\text{tan}^2 x + 1}{\text{sin}^2 x}✓$$
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#4:
Solutions:
$$a)\hspace{.1em}\frac{-6 + 3\sqrt{3}}{4}$$
$$b)\hspace{.1em}-\frac{3\sqrt{6}}{2}$$
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#5:
Solutions:
$$a)\hspace{.1em}\frac{\text{sin}\hspace{.1em}10θ - \text{sin}\hspace{.1em}6θ}{2}$$
$$b)\hspace{.1em}4\hspace{.1em}\text{sin}\hspace{.1em}8B \hspace{.1em}\text{cos}\hspace{.1em}5B$$
