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The value of $A + B$ should be less than $\frac{-\pi}{2}$ for the cosine to be negative. Then relation should be $-\pi \leq A + B < -\frac{\pi}{2}$. But the question's relation is not this; that contains right hand closed, which should be open. Now, which relation is correct?

Actually I was reading the proof of $$\arcsin x + \arcsin y = -\pi - \arcsin(x\sqrt{1 - y^2} + y\sqrt{1 - x^2})$$.

Let $\arcsin x = A$ & $\arcsin y = B$. $$A + B \in [\frac{-\pi}{2} ,0)$$ Since $$x^2 + y^2 > 1$$. . . .$$\cos(A + B) < 0$$ $A + B$ either lies in II or III quadrant. $$\therefore -\pi \leq A + B \leq -\frac{\pi}{2}$$. . . .

Now, how could the book write $\leq - \frac{\pi}{2}$ since $\cos(A + B) < 0$? [But if it doesn't hold good, then the above relation can't be proved; really in a dilemma]

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    The correct conclusion is that $\frac{4k+1}{2}\pi < A + B < \frac{4k+3}{2}\pi$ for some integer $k$. You need to know more about $A$ and $B$ to restrict this. – Alex Zorn Mar 04 '15 at 18:40
  • If $A = \arcsin{x}$, then $A \in [-\pi/2,\pi/2]$, and similarly for $B$. So $A + B \in [-\pi,\pi]$. If you also know that $cos(A + B) < 0$, then that means $A + B \in [-\pi,-\pi/2) \cup (\pi/2,\pi]$. – Alex Zorn Mar 04 '15 at 21:01
  • I can't really follow the proof, and I expect you may have left out something important. I don't know where $A + B \in [\pi/2,0)$ comes from, nor where $x^2 + y^2 > 1$ comes from. – Alex Zorn Mar 04 '15 at 21:03
  • @Alex Zorn: The prequisite condition for the relation was $$-1 \leq x , y < 0$$, which means $A + B \in [- 90 , 0)$. Now, $\cos(A + B) < 0$ which means, to me , $-180 \leq A + B < -90$ . . . –  Mar 05 '15 at 00:54
  • . . . But the domain of $\arcsin$ is $[- 180, -90]$. But that can't be inferred from $\cos(A + B) <0$ since it is open inequality, not closed. But the book seems to have ignored it; just wrote $A + B$ lies either in 2nd or 3rd quadrant & hence $-180\leq A + B \leq -90$. But if it is $\leq -90$, it'll not satisfy that cosine of $A + B$ is negative. That's my problem, sir. –  Mar 05 '15 at 01:00
  • Yeah, then you're correct. If $cos(A + B) < 0$, and $x,y$ both negative, then you should get $-\pi \leq A + B < -\pi/2$, and the book is wrong. If instead it says $cos(A + B) \leq 0$, then it's $-\pi \leq A + B \leq -\pi/2$. – Alex Zorn Mar 05 '15 at 02:11
  • @lab bhattacharjee: Anything can I do for getting an answer! Can you help? –  Mar 10 '15 at 13:35
  • @user36790, Just focus the first comment made by Alex Zorn based on http://www.mathymind.com/all-sin-tan-cos-rule-in-trigonometry/ OR http://www.math-only-math.com/all-sin-tan-cos-rule.html – lab bhattacharjee Mar 10 '15 at 15:08

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