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$\cos^{-1}x+\cos^{-1}y=\cos^{-1}\Big(xy-\sqrt{1-x^2}\sqrt{1-y^2}\Big)$ true for all $x$ ?

My Attempt

Let $a=\cos^{-1}x$, $b=\cos^{-1}y\implies$$\cos a=x$, $\cos b=y$ and $a,b\in[0,\pi]\implies a+b\in[0,2\pi]$ $$ \cos(a+b)=\cos a\cos b-\sin a\sin b=xy-\sqrt{1-x^2}\sqrt{1-y^2}\\=\cos\bigg[\cos^{-1}\Big(xy-\sqrt{1-x^2}\sqrt{1-y^2}\Big)\bigg]\\ a+b=\color{red}{\cos^{-1}x+\cos^{-1}y=2n\pi\pm\cos^{-1}\Big(xy-\sqrt{1-x^2}\sqrt{1-y^2}\Big)} $$ Case 1: $a+b\in[0,\pi)$ $$ \cos^{-1}x+\cos^{-1}y=\cos^{-1}\Big(xy-\sqrt{1-x^2}\sqrt{1-y^2}\Big) $$

Case 2: $a+b\in[\pi,2\pi]$ $$ \cos^{-1}x+\cos^{-1}y=2\pi-\cos^{-1}\Big(xy-\sqrt{1-x^2}\sqrt{1-y^2}\Big) $$ Case 1-: $a+b\in[0,\pi)$

Note I have checked a possible solution to this in link, but it gives a different expression differ from my attempt.

Sooraj S
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  • See https://math.stackexchange.com/questions/3027787/expression-for-cos-1x-pm-cos-1y – lab bhattacharjee Dec 06 '18 at 06:46
  • @labbhattacharjee thanx. but the given post does not say anything about all cases for the expression nor the domain of $x,y$ for which the given expression is valid !. I hardly find any post on the complete expression for $\cos^{-1}x+\cos^{-1}y$ on internet. – Sooraj S Dec 06 '18 at 06:47
  • For $x=y=-1$, the LHS is $2\pi$, the RHS is $0$. – Andrei Dec 06 '18 at 06:53
  • @Andrei For $x=y=-1\implies a=b=\pi$. RHS=$2\pi-\cos^{-1}1=2\pi=LHS$, it is case 2. srry abt that, just edited the domain. – Sooraj S Dec 06 '18 at 07:12
  • @labbhattacharjee could you please comment on my attempt ? – Sooraj S Dec 06 '18 at 07:16

1 Answers1

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Using Principal values and Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$,

$\cos^{-1}x+\cos^{-1}y\le2\pi$

$$\cos^{-1}x+\cos^{-1}y=\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)})$$

will hold true if $\cos^{-1}x+\cos^{-1}y\le\pi$

$\iff\sin^{-1}x+\sin^{-1}y\ge0\iff\sin^{-1}x\ge-\sin^{-1}y=\sin^{-1}(-y)$

$\iff x\ge -y\iff x+y\ge0$

$$\cos^{-1}x+\cos^{-1}y=2\pi-\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)})$$

will hold true if $\cos^{-1}x+\cos^{-1}y>\pi$

$\iff\sin^{-1}x+\sin^{-1}y<0\iff x+y<0$

lab bhattacharjee
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  • Thanks. I just have one more query. Could u please confirm that, $\cos^{-1}x-\cos^{-1}y=\begin{cases}\cos^{-1}\bigg(xy+\sqrt{1-x^2}\sqrt{1-y^2}\bigg),;x-y\leq 0\ -2\pi+\cos^{-1}\bigg(xy+\sqrt{1-x^2}\sqrt{1-y^2}\bigg),;x-y> 0\end{cases}$ – Sooraj S Dec 06 '18 at 17:48
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    @ss1729, First replace $y$ with $-z$ in the identity derived in the answer. Then use https://math.stackexchange.com/questions/1224415/how-do-i-prove-that-arccosx-arccos-x-pi-when-x-in-1-1 – lab bhattacharjee Dec 06 '18 at 18:24