" Hañ AHTe 39cos((7pi )/(2)+alpha ) , ecjin cosalpha =-(5)/(13)H alpha in (0,5pi ;pi )

To solve the given expression, we need to find the value of $\alpha$ that satisfies the equation $39\cos(\frac{7\pi}{2}+\alpha) = \cos\alpha = -\frac{5}{13}$, where $\alpha \in (0, 5\pi; \pi)$.<br /><br />First, let's simplify the expression $39\cos(\frac{7\pi}{2}+\alpha)$:<br /><br />$\cos(\frac{7\pi}{2}+\alpha) = \cos(\frac{7\pi}{2}+\alpha - 2\pi) = \cos(\frac{3\pi}{2}+\alpha) = -\sin(\alpha)$<br /><br />So, the expression becomes $39(-\sin(\alpha)) = -39\sin(\alpha)$.<br /><br />Now, we have the equation $-39\sin(\alpha) = \cos\alpha = -\frac{5}{13}$.<br /><br />We can use the Pythagorean identity $\sin^2(\alpha) + \cos^2(\alpha) = 1$ to find the value of $\sin(\alpha)$:<br /><br />$\sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \left(-\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}$<br /><br />Taking the square root of both sides, we get:<br /><br />$\sin(\alpha) = \pm\frac{12}{13}$<br /><br />Since $\alpha \in (0, 5\pi; \pi)$, we know that $\sin(\alpha) > 0$. Therefore, $\sin(\alpha) = \frac{12}{13}$.<br /><br />Now, we can find the value of $\alpha$ by taking the inverse sine of $\frac{12}{13}$:<br /><br />$\alpha = \arcsin\left(\frac{12}{13}\right)$<br /><br />Therefore, the value of $\alpha$ that satisfies the given equation is $\alpha = \arcsin\left(\frac{12}{13}\right)$.