The trick is finding values for $x$ at which the sin and cos are nice numbers. Here's how it goes:
1) Use $x = -\pi/2$, so that $\sin (x) = -1$, and $\cos(x) = 0$, then the inequality reads $\sin (y) -1 \geq 0$, so $\sin(y)=1$.
2) We've reduced the problem to finding a number $A = \cos (\alpha)$ in the interval $[-1,1]$, such that $A \times \cos(x) \leq \sin(x) + 1$ for all x. Let us just assume that $x$ is such that $\cos (x) > 0$, then you can divide by $\cos(x)$ to get $A \leq \tan (x) + \sec (x)$, but the right hand side can get arbitrarily close to $0$, so $A \leq 0$. Similarly, you can look at $x$ where $\cos (x) < 0$ and you'll find that $A \geq 0$, hence you can conclude $A = \cos(\alpha) =0$.
In total, $\sin(y) + \cos(\alpha) = 1 + 0 = 1$.