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 $\alpha$ in the interval $(0, 5\pi; \pi)$ such that $cos(\frac{7\pi}{2} + \alpha) = -\frac{5}{13}$.<br /><br />First, let's simplify the expression $cos(\frac{7\pi}{2} + \alpha)$:<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 />Now, we have $-sin(\alpha) = -\frac{5}{13}$, which simplifies to $sin(\alpha) = \frac{5}{13}$.<br /><br />Since $\alpha$ is in the interval $(0, 5\pi; \pi)$, we know that $\alpha$ is in the third quadrant where sine is negative. Therefore, $sin(\alpha) = -\frac{5}{13}$.<br /><br />Now, we can use the Pythagorean identity $sin^2(\alpha) + cos^2(\alpha) = 1$ to find the value of $cos(\alpha)$:<br /><br />$sin^2(\alpha) + cos^2(\alpha) = 1$<br /><br />$(\frac{5}{13})^2 + cos^2(\alpha) = 1$<br /><br />$\frac{25}{169} + cos^2(\alpha) = 1$<br /><br />$cos^2(\alpha) = 1 - \frac{25}{169}$<br /><br />$cos^2(\alpha) = \frac{144}{169}$<br /><br />$cos(\alpha) = \pm \frac{12}{13}$<br /><br />Since $\alpha$ is in the third quadrant where cosine is negative, we have $cos(\alpha) = -\frac{12}{13}$.<br /><br />Therefore, the value of $\alpha$ in the interval $(0, 5\pi; \pi)$ such that $cos(\frac{7\pi}{2} + \alpha) = -\frac{5}{13}$ is $\alpha = \pi + \arccos(-\frac{12}{13})$.