Find the value of this trigonometric expression 
If $$\operatorname{sin}(x)+\operatorname{sin}(y)\ge\operatorname{cos}(x)\cdot\operatorname{cos}(\alpha)$$ $\forall x\in\mathbb{R}$ then find the value of $$\operatorname{sin}(y)+\operatorname{cos}(\alpha)$$

I have only been able to prove the fact that $$\operatorname{sin}(y)\ge\operatorname{cos}(\alpha)$$ by plugging $x=\pm\frac{\pi}{2}$
In other words we can say that $$2\ge\operatorname{sin}(y)+\operatorname{cos}(\alpha)\ge2\operatorname{cos}(\alpha)\ge-2$$
How to proceed further$?$
 A: Write $Y:=\sin y$ and $A:=\cos \alpha$. The claim is that
$$A\cos x-\sin x\le Y$$ for every $x$. The maximum value of $A\cos x-\sin x$ is $\sqrt{A^2+1}$; this can be seen using calculus, or via the identity
$$A\cos x-\sin x = \sqrt{A^2+1}\left(\frac A{\sqrt{A^2+1}}\cos x-\frac1{\sqrt{A^2+1}}\sin x\right) = \sqrt{A^2+1}\sin(\theta-x)$$ for $\theta:=\arctan A$. Therefore we require
$$\sqrt{A^2+1}\le Y.$$ But $1\le\sqrt{A^2+1}$ and $Y\le1$, hence $A=0$ and $Y=1$ and $\sin  y+ \cos \alpha = Y+A=1$.
A: When $x$ is a multiple of $\frac{\pi}{2}$, the inequality simplifies to:
$$x = 0 \implies \sin(y) \ge \cos(\alpha)$$
$$x = \frac{\pi}{2} \implies 1 + \sin(y) \ge 0$$
$$x = \pi \implies \sin(y) \ge -\cos(\alpha)$$
$$x = \frac{\pi}{2} \implies -1 + \sin(y) \ge 0$$
The last of these is equivalent to $\sin(y) \ge 1$.  But since the sine of a real number is never greater than 1, we must have $\boxed{\sin(y) = 1}$.  Plugging this back into the original inequality gives:
$$\sin(x) + 1 \ge \cos(x)\cos(\alpha)$$
If $\cos(x) = 0$, then we just get the tautology $\pm 1 + 1 \ge 0$.  Otherwise, let's divide by $\cos(x)$, but remember that dividing by a negative number flips the order.
$$\cos(x) > 0 \implies \frac{\sin(x) + 1}{\cos(x)} \ge \cos(\alpha)$$
$$\cos(x) < 0 \implies \frac{\sin(x) + 1}{\cos(x)} \le \cos(\alpha)$$
Now, $\frac{d}{dx} \frac{\sin(x) + 1}{\cos(x)} = \frac{1 + \sin(x)}{\cos^2(x)}$, so the left-hand expression has a critical point when $\sin(x) = -1$, or $x = \frac{3\pi}{2}$.  But this
means $\cos(x) = 0$, making the expression an undefined $\frac{0}{0}$.  But we can use L'Hôpital's Rule.
$$\lim_{x \rightarrow \frac{\pi}{2}} \frac{\sin(x) + 1}{\cos(x)} = \lim_{x \rightarrow \frac{\pi}{2}} \frac{\cos(x)}{-\sin(x)} = \frac{\cos(\frac{3\pi}{2})}{-\sin(\frac{3\pi}{2})} = \frac{0}{-1} = 0$$
Plugging in this limit gives:
$$\cos(x) \rightarrow 0^+ \implies 0 \ge \cos(\alpha)$$
$$\cos(x) \rightarrow 0^- \implies 0 \le \cos(\alpha)$$
Which can only be simultaneously true for both positive and negative values of $\cos(x)$ if $\boxed{\cos(\alpha) = 0}$.
Therefore, $\boxed{\sin(y) + \cos(\alpha) = 1}$.
