We decompose the fraction into fractionReduce(3+2*N,12)\pi = fraction(1,4)\pi+ fractionReduce(N,6)\pi

and apply the addition theorem with \sin(\alpha + \beta)=\sin (\alpha) \cdot \cos (\beta) + \sin (\beta) \cdot \cos (\alpha).

We look into the table of known values of the sine and cosine functions

\begin{array}{c|c c c c c} x & 0 & \frac{\pi}{6} & \frac{\pi}{4} & \frac{\pi}{3} & \frac{\pi}{2} \\ \hline \sin(x) & 0 & \frac{1}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{3}}{2} & 1 \\ \cos (x) & 1 & \frac{\sqrt{3}}{2} & \frac{\sqrt{2}}{2} & \frac{1}{2} & 0 \end{array}

and substitute these values:

Thus, {\color{orange}w} = \sin \left(fractionReduce(3+2*N,12)\pi\right) = \sin \left(fraction(1,4)\pi+ fractionReduce(N,6)\pi\right) = \sin \left(fraction(1,4)\pi\right) \cdot \cos \left(fractionReduce(N,6)\pi\right) + \cos \left(fraction(1,4)\pi\right) \cdot \sin \left(fractionReduce(N,6)\pi\right).

Evaluating then gives {\color{orange}w} = \left( \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{-2*N+5}}{2}+ \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{2*N-1}}{2}\right), which simplifies to {\color{orange}w} =\dfrac{\sqrt{2*(-2*N+5)} +\sqrt{2*(2*N-1)}}{4} .