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Suppose that $\sin \alpha=\frac{3}{\sqrt{13}}$ and $0^{\circ} < \alpha < 90^{\circ}$ . Find the exact values of $\cos \left(\frac{\alpha}{2}\right)$ and $\tan \left(\frac{\alpha}{2}\right)$ . $\cos \left(\frac{\alpha}{2}\right)=$ $\tan \left(\frac{\alpha}{2}\right)=$

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Evaluate an exponential function

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Q. Suppose that

\sin \alpha=\frac{3}{\sqrt{13}}

and

0^{\circ} < \alpha < 90^{\circ}

. Find the exact values of

\cos \left(\frac{\alpha}{2}\right)

and

\tan \left(\frac{\alpha}{2}\right)

.

\cos \left(\frac{\alpha}{2}\right)=

\tan \left(\frac{\alpha}{2}\right)=

Given information: We are given that $\sin(\alpha) = \frac{3}{\sqrt{13}}$ and $\alpha$ is in the first quadrant $0^\circ < \alpha < 90^\circ$ . To find $\cos(\alpha/2)$ and $\tan(\alpha/2)$ , we can use the half-angle formulas for sine and cosine. The half-angle formula for cosine is: $\newline$ $\cos(\alpha/2) = \pm\sqrt{\left(\frac{1 + \cos(\alpha)}{2}\right)}$ $\newline$ Since $\alpha$ is in the first quadrant, $\cos(\alpha)$ will be positive, and so will $\cos(\alpha/2)$ . We need to find $\cos(\alpha)$ first.
Find $\cos(\alpha)$ : To find $\cos(\alpha)$ , we use the Pythagorean identity: $\newline$ $\sin^2(\alpha) + \cos^2(\alpha) = 1$ $\newline$ $\left(\frac{3}{\sqrt{13}}\right)^2 + \cos^2(\alpha) = 1$ $\newline$ $\frac{9}{13} + \cos^2(\alpha) = 1$ $\newline$ $\cos^2(\alpha) = 1 - \frac{9}{13}$ $\newline$ $\cos^2(\alpha) = \frac{4}{13}$ $\newline$ $\cos(\alpha) = \pm\sqrt{\frac{4}{13}}$ $\newline$ Since $\alpha$ is in the first quadrant, $\cos(\alpha)$ is positive, so $\cos(\alpha)$ $0$ .
Find $\cos(\alpha/2)$ : Now we can find $\cos(\alpha/2)$ using the half-angle formula: $\newline$ $\cos(\alpha/2) = \sqrt{(1 + \cos(\alpha))/2}$ $\newline$ $\cos(\alpha/2) = \sqrt{(1 + 2/\sqrt{13})/2}$ $\newline$ $\cos(\alpha/2) = \sqrt{(\sqrt{13} + 2)/2\sqrt{13}}$ $\newline$ $\cos(\alpha/2) = \sqrt{(\sqrt{13} + 2)/(2\sqrt{13})}$ $\newline$ $\cos(\alpha/2) = \sqrt{((\sqrt{13} + 2)\sqrt{13})/(2\sqrt{13}\sqrt{13})}$ $\newline$ $\cos(\alpha/2) = \sqrt{((\sqrt{13} + 2)\sqrt{13}/26)}$ $\newline$ $\cos(\alpha/2) = \sqrt{(13 + 2\sqrt{13})/26}$ $\newline$ $\cos(\alpha/2) = \sqrt{(13 + 2\sqrt{13})/26}$
Find $\tan(\alpha/2)$ : Next, we find $\tan(\alpha/2)$ using the half-angle formula for tangent, which is: $\newline$ $\tan(\alpha/2) = \pm\sqrt{(1 - \cos(\alpha))/(1 + \cos(\alpha))}$ $\newline$ Since $\alpha$ is in the first quadrant, $\tan(\alpha/2)$ will also be positive. We already found $\cos(\alpha) = \frac{2}{\sqrt{13}}$ , so we can substitute it into the formula: $\newline$ $\tan(\alpha/2) = \sqrt{(1 - \frac{2}{\sqrt{13}})/(1 + \frac{2}{\sqrt{13}})}$ $\newline$ $\tan(\alpha/2) = \sqrt{(\sqrt{13} - 2)/(\sqrt{13} + 2)}$ $\newline$ $\tan(\alpha/2) = \sqrt{(\sqrt{13} - 2)\sqrt{13}/(\sqrt{13} + 2)\sqrt{13}}$ $\newline$ $\tan(\alpha/2) = \sqrt{(13 - 2\sqrt{13})/(13 + 2\sqrt{13})}$ $\newline$ $\tan(\alpha/2) = \sqrt{(13 - 2\sqrt{13})/(13 + 2\sqrt{13})}$

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