Evaluate each expression without using a calculator. [Hint: Make a sketch of a right triangle.] (a) \( \sin \left(\arctan \left(\frac{8}{15}\right)\right) \) (b) \( \sec \left(\arcsin \left(\frac{15}{17}\right)\right) \)

To evaluate the expressions, we'll use right triangles and the definitions of trigonometric functions. For (a), let's consider \( \theta = \arctan\left(\frac{8}{15}\right) \). Here, the opposite side is 8, and the adjacent side is 15. Using the Pythagorean theorem, we can find the hypotenuse: \[ h = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17. \] Now, we can find \( \sin(\theta) \): \[ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{8}{17}. \] So, \[ \sin \left(\arctan \left(\frac{8}{15}\right)\right) = \frac{8}{17}. \] For (b), we set \( \phi = \arcsin\left(\frac{15}{17}\right) \). This means \( \sin(\phi) = \frac{15}{17} \), so in our right triangle, the opposite side is 15, and the hypotenuse is 17. Again, we use the Pythagorean theorem to find the adjacent side \( b \): \[ b = \sqrt{17^2 - 15^2} = \sqrt{289 - 225} = \sqrt{64} = 8. \] Now we can find \( \sec(\phi) \): \[ \sec(\phi) = \frac{1}{\cos(\phi)} = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{17}{8}. \] So, \[ \sec \left(\arcsin \left(\frac{15}{17}\right)\right) = \frac{17}{8}. \] Final answers: (a) \( \frac{8}{17} \) (b) \( \frac{17}{8} \)