Question
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Select the correct option. Find the exact value of \( \cos \frac{13 \pi}{6}+\cos \frac{7 \pi}{6} \), \( \begin{array}{l}0 \\ -\frac{\sqrt{3}}{2} \\ 0\end{array} \)

Ask by Elliott Schwartz. in the United Arab Emirates
Feb 26,2025

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Answer

The exact value of \( \cos \frac{13 \pi}{6} + \cos \frac{7 \pi}{6} \) is 0.

Solution

To solve for cos(13π/6) + cos(7π/6), start by reducing the angles when necessary. 1. Note that 13π/6 is larger than 2π. We can subtract 2π (which is 12π/6) to get:   13π/6 − 12π/6 = π/6.   Thus, cos(13π/6) = cos(π/6) = √3/2. 2. For cos(7π/6), notice this angle is in the third quadrant. Its reference angle is:   7π/6 − π = π/6.   Since cosine is negative in the third quadrant,   cos(7π/6) = −cos(π/6) = −√3/2. 3. Adding the two values:   cos(13π/6) + cos(7π/6) = (√3/2) + (−√3/2) = 0. Thus, the exact value is 0.

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To find \( \cos \frac{13 \pi}{6} + \cos \frac{7 \pi}{6} \), we first find the cosine values of the individual angles. The angle \( \frac{13 \pi}{6} \) is equivalent to \( \frac{13 \pi}{6} - 2\pi = \frac{13 \pi}{6} - \frac{12 \pi}{6} = \frac{\pi}{6} \). Thus, \( \cos \frac{13 \pi}{6} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \). Next, for \( \frac{7 \pi}{6} \), this angle is in the third quadrant where cosine values are negative. Specifically, \( \cos \frac{7 \pi}{6} = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2} \). Now we sum these results: \[ \cos \frac{13 \pi}{6} + \cos \frac{7 \pi}{6} = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = 0. \] So the correct option is: \( 0 \)

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