If sinθ + cosθ = (1)/(2), then 16 (sin (2θ) + (4θ) + (6θ)) is equal...

$sin\theta + cos\theta = \dfrac{1}{2}$

$\sin^{2}\theta + \cos^{2}\theta + 2sin\theta cos\theta = \dfrac{1}{4} ⇒ \sin 2\theta = - \dfrac{3}{4}$

Now:

$\cos 4\theta = 1 - 2\sin^{2}2\theta$

$= 1 - 2{\left(- \dfrac{3}{4}\right)}^{2}$

$= 1 - 2 \times \dfrac{9}{16} = - \dfrac{1}{8}$

And $\sin 6\theta = 3\sin 2\theta - 4\sin^{3}2\theta$

$= \left(3 - 4\sin^{2}2\theta\right) \cdot \sin 2\theta$

$= \left[3 - 4\left(\dfrac{9}{16}\right)\right] \cdot \left(- \dfrac{3}{4}\right)$

$= \left[\dfrac{3}{4}\right] \times \left(- \dfrac{3}{4}\right) = - \dfrac{9}{16}$

So, $16\left[\sin 2\theta + \cos 4\theta + \sin 6\theta\right]$

$= 16\left(- \dfrac{3}{4} - \dfrac{1}{8} - \dfrac{9}{16}\right) = - 23$