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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 ?

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

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    $\begingroup$ 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 $\sin y +\cos \alpha$. $\endgroup$ – lulu Apr 17 at 18:48
  • $\begingroup$ I don't think so. If you have any approach then tell me $\endgroup$ – Abhishek Kumar Apr 17 at 18:51
  • $\begingroup$ 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. $\endgroup$ – lulu Apr 17 at 18:53
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    $\begingroup$ 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. $\endgroup$ – lulu Apr 17 at 19:14
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    $\begingroup$ 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. $\endgroup$ – lulu Apr 17 at 19:18
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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, $\sin(y) + \cos(\alpha) = 1 + 0 = 1$.

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