# If $\sin(x)+\sin(y)\ge \cos(\alpha) \times \cos(x)$ $\forall x\in \mathbb R$, then $\sin(y)+\cos(\alpha)$ is equal to? [closed]

If $$\sin(x)+\sin(y)\ge \cos(\alpha) \times \cos(x)$$, $$\forall x\in \mathbb R$$, then $$\sin(y)+\cos(\alpha)$$ is equal to ?

My thinking:- I have break the left hand side on $$sinC + sinD$$ and right hand side with $$2cosAcosB$$ but doing this , I can't get my answer ?

## closed as off-topic by Saad, Alexander Gruber♦Apr 29 at 1:55

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• I did not down vote the question, but I, for one, can't sort out what you are asking. Surely you don't believe the initial inequality determines the value of $\sin y +\cos \alpha$. – lulu Apr 17 at 18:48
• I don't think so. If you have any approach then tell me – Abhishek Kumar Apr 17 at 18:51
• I don't understand, Since the value is undetermined by the assumptions, there is no approach. Please edit your post to ask a clear question. – lulu Apr 17 at 18:53
• Well, I'm not sure. It seems difficult to make that inequality hold for all $x$. I'd search for an example..it's likely to be an extreme one and, possibly, you can then show that you've got the only case. – lulu Apr 17 at 19:14
• To be precise: the inequality certainly holds for all $x$ if $\sin y =1$ and $\cos \alpha =0$. It's not clear to me that this is the only possibility, but perhaps it is. Anyway, I'd start from there. – lulu Apr 17 at 19:18

## 1 Answer

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, $$\sin(y) + \cos(\alpha) = 1 + 0 = 1$$.