Let $x,y$ denote hereafter variables in the interval $I:=[0,\pi]$. Denote $$S:=\{-\sin(y)<\alpha\sin(x)+\sin(x+y)<\sin(y)\}\subset I^2$$ the set to be measured, and $\Delta:=\{x+y<\pi\}\subset I^2$.

Note that intersecting $S$ with $\Delta$ one inequality that defines $S$ is automatically satisfied, namely $$S\cap\Delta= \{-\sin(y)<\alpha\sin(y)+\sin(x+y)<\sin(y),\;x+y<\pi \}=$$$$ =\{ \alpha\sin(y)+\sin(x+y)<\sin(y),\;x+y<\pi\}.$$ 

Similarly, the elementary inequality $\sin(x+y)\le\sin(x)-\sin(y)$ for $x+y\ge\pi$ gives $$S^c\cap\Delta^c= \{-\sin(y)\ge\alpha\sin(y)+\sin(x+y),\;x+y\ge\pi \}.$$

It is then straightforward to check that the area preserving affine transformation $(x,y)\mapsto (\pi-x,x+y)$ maps (a.e.) $S\cap\Delta$ onto $S^c\cap\Delta^c$, so they have the same area. Then $|S|=|S\cap\Delta|+|S\cap\Delta^c|=|S^c\cap\Delta^c|+|S\cap\Delta^c|=|\Delta^c|=\frac12|I^2|,$ ending the computation.

Let $x,y$ denote hereafter variables in the interval $I:=[0,\pi]$. Denote $$S:=\{-\sin(y)<\alpha\sin(x)+\sin(x+y)<\sin(y)\}\subset I^2$$ the set to be measured, and $\Delta:=\{x+y<\pi\}\subset I^2$.

Note that intersecting $S$ with $\Delta$ one inequality that defines $S$ is automatically satisfied, namely $$S\cap\Delta= \{-\sin(y)<\alpha\sin(x)+\sin(x+y)<\sin(y),\;x+y<\pi \}=$$ $$ =\{ \alpha\sin(x)+\sin(x+y)<\sin(y),\;x+y<\pi\}.$$

Similarly, the elementary inequality $\sin(x+y)\le\sin(y)-\sin(x)$ for $x+y\ge\pi$ gives $$S^c\cap\Delta^c= \{-\sin(y)\ge\alpha\sin(x)+\sin(x+y),\;x+y\ge\pi \}.$$

It is then straightforward to check that the area preserving affine transformation $(x,y)\mapsto (\pi-x,x+y)$ maps (a.e.) $S\cap\Delta$ onto $S^c\cap\Delta^c$, so they have the same area. Then $|S|=|S\cap\Delta|+|S\cap\Delta^c|=|S^c\cap\Delta^c|+|S\cap\Delta^c|=|\Delta^c|=\frac12|I^2|,$ ending the computation.

Let $x,y$ denote hereafter variables in the interval $I:=[0,\pi]$. Denote $$S:=\{-\sin(y)<\alpha\sin(x)+\sin(x+y)<\sin(y)\}\subset I^2$$ the set to be measured, and $\Delta:=\{x+y<\pi\}\subset I^2$.

Note that intersecting $S$ with $\Delta$ one inequality that defines $S$ is automatically satisfied, namely $$S\cap\Delta= \{-\sin(y)<\alpha\sin(y)+\sin(x+y)<\sin(y),\;x+y<\pi \}=$$$$ =\{ \alpha\sin(y)+\sin(x+y)<\sin(y),\;x+y<\pi\}.$$

Similarly, the elementary inequality $\sin(x+y)\le\sin(x)-\sin(y)$ for $x+y\ge\pi$ gives $$S^c\cap\Delta^c= \{-\sin(y)\ge\alpha\sin(y)+\sin(x+y),\;x+y\ge\pi \}.$$

It is then straightforward to check that the area preserving affine transformation $(x,y)\mapsto (\pi-x,x+y)$ maps (a.e.) $S\cap\Delta$ onto $S^c\cap\Delta^c$, so they have the same area. Then $|S|=|S\cap\Delta|+|S\cap\Delta^c|=|S^c\cap\Delta^c|+|S\cap\Delta^c|=|\Delta|=\frac12|I^2|,$ ending the computation.

Let $x,y$ denote hereafter variables in the interval $I:=[0,\pi]$. Denote $$S:=\{-\sin(y)<\alpha\sin(x)+\sin(x+y)<\sin(y)\}\subset I^2$$ the set to be measured, and $\Delta:=\{x+y<\pi\}\subset I^2$.

Note that intersecting $S$ with $\Delta$ one inequality that defines $S$ is automatically satisfied, namely $$S\cap\Delta= \{-\sin(y)<\alpha\sin(y)+\sin(x+y)<\sin(y),\;x+y<\pi \}=$$$$ =\{ \alpha\sin(y)+\sin(x+y)<\sin(y),\;x+y<\pi\}.$$

Similarly, the elementary inequality $\sin(x+y)\le\sin(x)-\sin(y)$ for $x+y\ge\pi$ gives $$S^c\cap\Delta^c= \{-\sin(y)\ge\alpha\sin(y)+\sin(x+y),\;x+y\ge\pi \}.$$

It is then straightforward to check that the area preserving affine transformation $(x,y)\mapsto (\pi-x,x+y)$ maps (a.e.) $S\cap\Delta$ onto $S^c\cap\Delta^c$, so they have the same area. Then $|S|=|S\cap\Delta|+|S\cap\Delta^c|=|S^c\cap\Delta^c|+|S\cap\Delta^c|=|\Delta^c|=\frac12|I^2|,$ ending the computation.