Answer
(a) $sin~2u=\frac{15}{17}$
(b) $cos~2u=\frac{8}{17}$
(c) $tan~2u=\frac{15}{8}$
Work Step by Step
$sec^2u=tan^2u+1$
$sec^2u=(\frac{3}{5})^2+1=\frac{34}{25}$
$sec~u=\frac{\sqrt {34}}{5}~~~~$ ($0\lt u\lt\frac{\pi}{2}$)
$cos~u=\frac{1}{sec~u}=\frac{5}{\sqrt {34}}=\frac{5\sqrt {34}}{34}$
$tan~u=\frac{sin~u}{cos~u}$
$\frac{3}{5}=\frac{sin~u}{\frac{5\sqrt {34}}{34}}$
$sin~u=\frac{3}{5}\frac{5\sqrt {34}}{34}=\frac{3\sqrt {34}}{34}$
(a) $sin~2u=2~sin~u~cos~u=2(\frac{3\sqrt {34}}{34})(\frac{5\sqrt {34}}{34})=\frac{15}{17}$
(b) $cos~2u=cos^2u-sin^2u=\frac{25}{34}-\frac{9}{34}=\frac{8}{17}$
(c) $tan~2u=\frac{2~tan~u}{1-tan^2u}=\frac{2(\frac{3}{5})}{1-(\frac{3}{5})^2}=\frac{\frac{6}{5}}{\frac{16}{25}}=\frac{15}{8}$
