Is it complicated or complex?

Call the expression $\mathscr O$ .

$\color{#0C6AC7}{\mathcal{METHOD~1}}$ The value of $\small{\sin \left( \dfrac{\pi n}{2} + \dfrac{\pi}{4}\right) }$ for starting values of $n$ is $\small{\dfrac1{\sqrt 2},\dfrac1{\sqrt 2},-\dfrac1{\sqrt 2},-\dfrac1{\sqrt 2}\color{#3D99F6}{;}\dfrac1{\sqrt 2},\dfrac1{\sqrt 2},-\dfrac1{\sqrt 2},-\dfrac1{\sqrt 2}\color{#3D99F6}{;}\cdots}$ .

We can form four infinite GP's by grouping $1,5,9,13,\cdots$ terms $;;$ $2,6,10,14,\cdots$ terms $;;$ $3,7,11,15,\cdots$ terms $;;$ $4,8,12,16,\cdots$ terms $;;$ as follows $:$

$\large \mathscr O=\small{\dfrac1{\sqrt 2}\left[\color{#D61F06}{\displaystyle\sum_{n=0}^{\infty}\dfrac1{16^n}}+\color{#D61F06}{\displaystyle\sum_{n=0}^{\infty}\dfrac1{\color{#333333}{2}\cdot16^n}}-\color{#D61F06}{\displaystyle\sum_{n=0}^{\infty}\dfrac1{ \color{#333333}{4}\cdot16^n}}-\color{#D61F06}{\displaystyle\sum_{n=0}^{\infty}\dfrac1{\color{#333333}{8} \cdot16^n}}\right] }$

$\large =\dfrac1{\sqrt 2} \cdot \underbrace{\color{#D61F06}{\displaystyle\sum_{n=0}^{\infty}\dfrac1{16^n}}}_{\text{Infinite GP}}\left(1+\dfrac{1}2-\dfrac 14-\dfrac{1}8\right)$

$\large =\dfrac{3\sqrt 2}{5}$

$\Large 3+2+5=\boxed{\color{#0C6AC7}{10}}$

$\color{#0C6AC7}{\mathcal{METHOD~2}}$

Use $\sin x =\dfrac{e^{ix}-e^{-ix}}{2i}$

$\dfrac{1}{2i}\left(\displaystyle \sum_{n=0}^{\infty}\dfrac{e^{ i\left(\frac{\pi n}{2} +\frac{\pi}{4}\right)}}{2^n}-\displaystyle \sum_{n=0}^{\infty} \dfrac{ e^{-i\left( \frac{\pi n}{2} + \frac{\pi}{4}\right)}}{2^n}\right)$

Writing 5-6 terms you will get 2 infinite GP's

2^(-0.5) -2^(2.5) + 2^(4.5).............+ 2^(-1.5)-2^(3.5)..........

Applying formula of infinite Gp we get required result