Trigonometry / Unit 6: Trigonometric Identities
6.6 Half-Angle Formulas
Notes
Solving the double-angle formula for the half angle gives:
- $\displaystyle \sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}$ and $\displaystyle \cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}$.
- $\displaystyle \tan\frac{\theta}{2} = \frac{\sin\theta}{1 + \cos\theta} = \frac{1 - \cos\theta}{\sin\theta}$.
- Choose the sign from the quadrant of $\displaystyle \frac{\theta}{2}$, not of $\theta$.
- Example: $\displaystyle \cos 15^\circ = \sqrt{\frac{1 + \cos 30^\circ}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}$.
Practice
What is the exact value of $\cos 15^\circ$?
Use $\displaystyle \cos\frac{30^\circ}{2} = \sqrt{\frac{1 + \cos 30^\circ}{2}}$, positive because $15^\circ$ is in Quadrant I.
What is the exact value of $\sin 22.5^\circ$?
$\displaystyle \sin\frac{45^\circ}{2} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{2}}{4}}$.
To find $\displaystyle \cos\frac{\theta}{2}$ when $\theta = 300^\circ$, which sign does the half-angle formula take?
The half angle is $150^\circ$, in Quadrant II, where cosine is negative.
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