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Chapter 5: Problem 24

Find the exact values of the sine, cosine, and tangent of the angle. $$\frac{5 \pi}{12}$$

Short Answer

Expert verified
The trigonometric ratios for the angle \( \frac{5 \pi}{12} \) are as follows: \( \sin(\frac{5 \pi}{12}) = \frac{\sqrt{6} + \sqrt{2}}{4} \), \( \cos(\frac{5 \pi}{12}) = \frac{\sqrt{6} - \sqrt{2}}{4} \), and \( \tan(\frac{5 \pi}{12}) = \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} \)

Step by step solution

01

Express the Given Angle in Terms of Known Angles

The given angle is \( \frac{5 \pi}{12} \). This can be expressed as a sum of \( \frac{\pi}{4} \) and \( \frac{\pi}{6} \), both of which are familiar angles. Thus, \( \frac{5 \pi}{12} = \frac{\pi}{4} + \frac{\pi}{6} \)
02

Calculate the Sine of the Angle

Use the formula for sine of sum of angles: \( \sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta \). Substituting \( \alpha = \frac{\pi}{4} \) and \( \beta = \frac{\pi}{6} \), we get \( \sin(\frac{5 \pi}{12}) = \sin(\frac{\pi}{4}) \cos(\frac{\pi}{6}) + \cos(\frac{\pi}{4}) \sin(\frac{\pi}{6}) = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4} \)
03

Calculate the Cosine of the Angle

Use the formula for cosine of sum of angles: \( \cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta \). Substituting \( \alpha = \frac{\pi}{4} \) and \( \beta = \frac{\pi}{6} \), we get \( \cos(\frac{5 \pi}{12}) = \cos(\frac{\pi}{4}) \cos(\frac{\pi}{6}) - \sin(\frac{\pi}{4}) \sin(\frac{\pi}{6}) = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4} \)
04

Calculate the Tangent of the Angle

The tangent of an angle is the ratio of the sine to the cosine. Thus, \( \tan(\frac{5 \pi}{12}) = \frac{\sin(\frac{5 \pi}{12})}{\cos(\frac{5 \pi}{12})} = \frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{\sqrt{6} - \sqrt{2}}{4}} = \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} \)

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