1. The problem states: Given $\cos a = \frac{5}{7}$ and the terminal side of angle $a$ lies in a certain quadrant, find the quadrant and evaluate $\cos\left(-\frac{a}{2}\right)$ using the half-angle formula. 2. First, determine the quadrant of angle $a$. Since $\cos a = \frac{5}{7}$ is positive, $a$ lies either in Quadrant I or Quadrant IV. 3. The problem states the terminal side of $a$ lies in a quadrant where cosine is positive. So $a$ is in Quadrant I or IV. 4. Now, consider the angle $-\frac{a}{2}$. The negative sign means the angle is measured clockwise from the positive x-axis. 5. The half-angle formula for cosine is: $$\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}$$ 6. For $\cos\left(-\frac{a}{2}\right)$, use the even property of cosine: $$\cos\left(-\frac{a}{2}\right) = \cos\left(\frac{a}{2}\right)$$ 7. Since $a$ is in Quadrant I or IV, $\frac{a}{2}$ lies in Quadrant I or II (half the angle). 8. Because $\frac{a}{2}$ is between $0$ and $\pi$, $\cos\left(\frac{a}{2}\right)$ is positive if $\frac{a}{2}$ is in Quadrant I and negative if in Quadrant II. 9. Since $a$ is in Quadrant IV (because cosine is positive and angle is negative), $\frac{a}{2}$ lies in Quadrant IV or II depending on $a$. 10. To be precise, let's assume $a$ is in Quadrant IV (angle between $270^\circ$ and $360^\circ$ or $\frac{3\pi}{2}$ and $2\pi$). 11. Then $\frac{a}{2}$ lies between $\frac{3\pi}{4}$ and $\pi$, which is Quadrant II, where cosine is negative. 12. Therefore, for $\cos\left(\frac{a}{2}\right)$, we take the negative root: $$\cos\left(\frac{a}{2}\right) = -\sqrt{\frac{1 + \cos a}{2}}$$ 13. Substitute $\cos a = \frac{5}{7}$: $$\cos\left(\frac{a}{2}\right) = -\sqrt{\frac{1 + \frac{5}{7}}{2}} = -\sqrt{\frac{\frac{7}{7} + \frac{5}{7}}{2}} = -\sqrt{\frac{\frac{12}{7}}{2}} = -\sqrt{\frac{12}{14}} = -\sqrt{\frac{6}{7}}$$ 14. Thus, $$\cos\left(-\frac{a}{2}\right) = \cos\left(\frac{a}{2}\right) = -\sqrt{\frac{6}{7}}$$ 15. Final answers: - The terminal side of angle $a$ lies in Quadrant IV. - $\cos\left(-\frac{a}{2}\right) = -\sqrt{\frac{6}{7}}$