- Demonstrate the ability to use a half-angle identity to find an exact value
- Demonstrate the ability to find function values of s/2 given information about s
- Demonstrate the ability to verify a trigonometric identity using half-angle identities
#1:
Instructions: Derive the given identities.
$$a)\hspace{.1em}\text{cos}\hspace{.1em}\frac{A}{2}$$
$$b)\hspace{.1em}\text{sin}\hspace{.1em}\frac{A}{2}$$
$$c)\hspace{.1em}\text{tan}\hspace{.1em}\frac{A}{2}$$
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#2:
Instructions: Find the exact value.
$$a)\hspace{.1em}\text{sin}\hspace{.1em}345°$$
$$b)\hspace{.1em}\text{cos}\hspace{.1em}112.5°$$
$$c)\hspace{.1em}\text{sin}\hspace{.1em}\frac{π}{12}$$
$$d)\hspace{.1em}\text{cos}\hspace{.1em}\frac{9π}{8}$$
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#3:
Instructions: Find the exact value.
$$a)\hspace{.1em}\text{tan}\hspace{.1em}θ=\frac{4}{3}$$ $$180° < θ < 270°$$ Find: $$\text{cos}\hspace{.1em}\frac{θ}{2}$$
$$b)\hspace{.1em}\text{tan}\hspace{.1em}θ=\frac{1}{4}$$ $$180° < θ < 270°$$ Find: $$\text{tan}\hspace{.1em}\frac{θ}{2}$$
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#4:
Instructions: Find the exact value.
$$a)\hspace{.1em}\text{sin}\hspace{.1em}θ=-\frac{4}{5}$$ $$180° < θ < 270°$$ Find: $$\text{sin}\hspace{.1em}\frac{θ}{2}$$
$$b)\hspace{.1em}\text{cos}\hspace{.1em}θ=-\frac{11}{17}$$ $$90° < θ < 180°$$ Find: $$\text{sin}\hspace{.1em}\frac{θ}{2}$$
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#5:
Instructions: Verify each identity.
$$a)\hspace{.1em}\text{sec}^2 \frac{θ}{2}=\frac{2}{1 + \text{cos}\hspace{.1em}θ}$$
$$b) \hspace{.1em}\text{sin}^2 \frac{θ}{2}=\frac{\text{tan}\hspace{.1em}θ - \text{sin}\hspace{.1em}θ}{2 \hspace{.1em}\text{tan}\hspace{.1em}θ}$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.1em}2\hspace{.1em}\text{cos}^2 x - 1=\text{cos}\hspace{.1em}2x$$ Add 1 to each side: $$2\hspace{.1em}\text{cos}^2 x - 1 + 1=\text{cos}\hspace{.1em}2x + 1$$ $$2\hspace{.1em}\text{cos}^2 x=1 + \text{cos}\hspace{.1em}2x$$ Divide both sides by 2: $$\require{cancel}\frac{\cancel{2}\hspace{.1em}\text{cos}^2 x}{\cancel{2}}=\frac{1 + \text{cos}\hspace{.1em}2x}{2}$$ $$\hspace{.1em}\text{cos}^2 x=\frac{1 + \text{cos}\hspace{.1em}2x}{2}$$ Solve for cos x (note: we need +/- since we don't know the quadrant): $$\hspace{.1em}\text{cos}\hspace{.1em}x=\pm\sqrt{\frac{1 + \text{cos}\hspace{.1em}2x}{2}}$$ Replace each x with A/2: $$\text{cos}\frac{A}{2}=\pm\sqrt{\frac{1 + \text{cos}\cancel{2}\cdot \frac{A}{\cancel{2}}}{2}}$$ $$\text{cos}\frac{A}{2}=\pm\sqrt{\frac{1 + \text{cos}\hspace{.1em}A}{2}}$$
$$b)\hspace{.1em}1 - 2\hspace{.1em}\text{sin}^2 x=\text{cos}\hspace{.1em}2x$$ Subtract 1 from each side: $$1 - 1 - 2\hspace{.1em}\text{sin}^2 x=\text{cos}\hspace{.1em}2x - 1$$ $$-2\hspace{.1em}\text{sin}^2 x=-1 + \text{cos}\hspace{.1em}2x$$ Divide both sides by -2: $$\frac{\cancel{-2}\hspace{.1em}\text{sin}^2 x}{\cancel{-2}}=\frac{-1 + \text{cos}\hspace{.1em}2x}{-2}$$ $$\text{sin}^2 x=\frac{1 - \text{cos}\hspace{.1em}2x}{2}$$ Solve for sin x (note: we need +/- since we don't know the quadrant): $$\text{sin}\hspace{.1em}x=\pm \sqrt{\frac{1 - \text{cos}\hspace{.1em}2x}{2}}$$ Replace each x with A/2: $$\text{sin}\frac{A}{2}=\pm \sqrt{\frac{1 - \text{cos}\cancel{2}\cdot \frac{A}{\cancel{2}}}{2}}$$ $$\text{sin}\frac{A}{2}=\pm \sqrt{\frac{1 - \text{cos}A}{2}}$$
$$c)\hspace{.1em}\text{tan}\frac{A}{2}$$ $$=\large{\frac{\text{sin}\frac{A}{2}}{\text{cos}\frac{A}{2}}}$$ $$=\large{\frac{\pm \sqrt{\frac{1 \hspace{.1em}- \hspace{.1em}\text{cos}A}{2}}}{\pm\sqrt{\frac{1 \hspace{.1em}+ \hspace{.1em}\text{cos}\hspace{.1em}A}{2}}}}$$ $$=\pm \sqrt{\frac{1 - \text{cos}A}{\cancel{2}}\cdot \frac{\cancel{2}}{1 + \text{cos}A}}$$ $$=\pm \sqrt{\frac{1 - \text{cos}A}{1 + \text{cos}A}}$$
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#2:
Solutions:
$$a)\hspace{.1em}\text{sin}\hspace{.1em}345°=-\frac{\sqrt{2 - \sqrt{3}}}{2}$$
$$b)\hspace{.1em}\text{cos}\hspace{.1em}112.5°=-\frac{\sqrt{2 - \sqrt{2}}}{2}$$
$$c)\hspace{.1em}\text{sin}\hspace{.1em}\frac{π}{12}=\frac{\sqrt{2 - \sqrt{3}}}{2}$$
$$d)\hspace{.1em}\text{cos}\hspace{.1em}\frac{9π}{8}=-\frac{\sqrt{2 + \sqrt{2}}}{2}$$
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#3:
Solutions:
$$a)\hspace{.1em}\text{cos}\hspace{.1em}\frac{θ}{2}=-\frac{\sqrt{5}}{5}$$
$$b)\hspace{.1em}\text{tan}\hspace{.1em}\frac{θ}{2}=-\sqrt{33 + 8\sqrt{17}}$$
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#4:
Solutions:
$$a)\hspace{.1em}\text{sin}\hspace{.1em}\frac{θ}{2}=\frac{2\sqrt{5}}{5}$$
$$b)\hspace{.1em}\text{sin}\hspace{.1em}\frac{θ}{2}=\frac{\sqrt{238}}{17}$$
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#5:
Solutions:
$$a)\hspace{.1em}\text{sec}^2 \frac{θ}{2}=\frac{2}{1 + \text{cos}\hspace{.1em}θ}$$ $$\frac{2}{1 + \text{cos}\hspace{.1em}θ}=\text{sec}^2 \frac{θ}{2}$$ $$=\frac{1}{\text{cos}^2 \large{\frac{θ}{2}}}$$ $$=\frac{1}{\left(\text{cos}\hspace{.1em}\large{\frac{θ}{2}}\right)^2}$$ $$=\frac{1}{\left(\pm \sqrt{\large{\frac{1 + \text{cos}\hspace{.1em}θ}{2}}}\right)^2}$$ $$=\large{\frac{1}{\large{\frac{1 + \text{cos}\hspace{.1em}θ}{2}}}}$$ $$=\frac{2}{1 + \text{cos}\hspace{.1em}θ}✓$$
$$b)\hspace{.1em}\hspace{.1em}\text{sin}^2 \frac{θ}{2}=\frac{\text{tan}\hspace{.1em}θ - \text{sin}\hspace{.1em}θ}{2 \hspace{.1em}\text{tan}\hspace{.1em}θ}$$ $$=\frac{\large{\frac{\text{sin}\hspace{.1em}θ}{\text{cos}\hspace{.1em}θ}- \text{sin}\hspace{.1em}θ}}{2\large{\frac{\text{sin}\hspace{.1em}θ}{\text{cos}\hspace{.1em}θ}}}$$ $$=\frac{\large{\frac{\text{sin}\hspace{.1em}θ}{\text{cos}\hspace{.1em}θ}- \text{sin}\hspace{.1em}θ}}{2\large{\frac{\text{sin}\hspace{.1em}θ}{\text{cos}\hspace{.1em}θ}}}\cdot \frac{\text{cos}\hspace{.1em}θ}{\text{cos}\hspace{.1em}θ}$$ $$=\frac{\text{sin}\hspace{.1em}θ - \text{sin}\hspace{.1em}θ \hspace{.1em}\text{cos}\hspace{.1em}θ}{2 \hspace{.1em}\text{sin}\hspace{.1em}θ}$$ $$=\frac{1 - \text{cos}\hspace{.1em}θ}{2}$$ $$=\left(\pm \sqrt{\frac{1 - \text{cos}\hspace{.1em}θ}{2}}\right)^2$$ $$=\left(\text{sin}\frac{θ}{2}\right)^2$$ $$=\text{sin}^2 \frac{θ}{2}✓$$
