Calculate, without using tables: $\;\;\;$ $\left(\sin 4 \alpha + 2 \sin 2 \alpha\right) \cos \alpha$ $\;\;$ if $\;\;$ $\sin \alpha = \dfrac{1}{4}$, $\;$ $\alpha$ is in the first quadrant.
Given $\;$ $\alpha$ $\;$ is in the first quadrant $\implies$ both $\sin \alpha$, $\cos \alpha \;$ are positive.
Given: $\;$ $\sin \alpha = \dfrac{1}{4}$
$\therefore \;$ $\cos \alpha = \sqrt{1 - \sin^2 \alpha}= \sqrt{1 - \dfrac{1}{16}} = \dfrac{\sqrt{15}}{4}$
Now, $\;$ $\sin 2 \alpha = 2 \sin \alpha \cos \alpha = 2 \times \dfrac{1}{4} \times \dfrac{\sqrt{15}}{4} = \dfrac{\sqrt{15}}{8}$
$\cos 2 \alpha = \cos^2 \alpha - \sin^2 \alpha = \dfrac{15}{16} - \dfrac{1}{16} = \dfrac{14}{16} = \dfrac{7}{8}$
and, $\;$ $\sin 4 \alpha = \sin \left[2 \left(2 \alpha\right)\right] = 2 \sin \left(2 \alpha\right) \cos \left(2 \alpha\right) = 2 \times \dfrac{\sqrt{15}}{8} \times \dfrac{7}{8} = \dfrac{7 \sqrt{15}}{32}$
Given expression: $\;$ $\left(\sin 4 \alpha + 2 \sin 2 \alpha\right) \cos \alpha$
$= \left(\dfrac{7 \sqrt{15}}{32} + 2 \times \dfrac{\sqrt{15}}{8}\right) \times \dfrac{\sqrt{15}}{4}$
$= \dfrac{15 \sqrt{15}}{32} \times \dfrac{\sqrt{15}}{4} = \dfrac{225}{128}$