the value of one trigonometric function of an acute angle 3 , find the volues of the ramaining five trigonometric functions of . Gue the eract answar as a olfied fraction, if necessary, Rationalice denominators if necescary. \[ \cos 5=\frac{\sqrt{5}}{5} \]

To find the values of the remaining five trigonometric functions, we start with the given cosine value, \( \cos 5 = \frac{\sqrt{5}}{5} \). From the Pythagorean identity, we know that \( \sin^2 \theta + \cos^2 \theta = 1 \). Plugging in the cosine value: \[ \sin^2 5 + \left(\frac{\sqrt{5}}{5}\right)^2 = 1 \Rightarrow \sin^2 5 + \frac{5}{25} = 1 \Rightarrow \sin^2 5 + \frac{1}{5} = 1 \] Solving for \( \sin^2 5 \): \[ \sin^2 5 = 1 - \frac{1}{5} = \frac{4}{5} \Rightarrow \sin 5 = \frac{2\sqrt{5}}{5} \] Now, for the remaining functions, we can find: - \( \tan 5 = \frac{\sin 5}{\cos 5} = \frac{\frac{2\sqrt{5}}{5}}{\frac{\sqrt{5}}{5}} = 2 \) - \( \csc 5 = \frac{1}{\sin 5} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2} \) (after rationalizing the denominator) - \( \sec 5 = \frac{1}{\cos 5} = \frac{5}{\sqrt{5}} = \sqrt{5} \) (after rationalizing the denominator) - \( \cot 5 = \frac{1}{\tan 5} = \frac{1}{2} \) Thus, the remaining values are: \[ \sin 5 = \frac{2\sqrt{5}}{5}, \quad \tan 5 = 2, \quad \csc 5 = \frac{\sqrt{5}}{2}, \quad \sec 5 = \sqrt{5}, \quad \cot 5 = \frac{1}{2}. \]