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As mentioned in Proof for the formula of sum of arcsine functions $\arcsin x+\arcsin y$ for $\sin^{-1}x+\sin^{-1}y$

$$ \sin^{-1}x+\sin^{-1}y= \begin{cases} \sin^{-1}( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 \le 1 \\\ \pi - \sin^{-1}( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 > 1, 0< x,y \le 1\\ -\pi - \sin^{-1}( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 > 1, -1< x,y \le 0 \end{cases} $$ Can we have similar expression for $\cos^{-1}x\pm\cos^{-1}y$ ?

Sooraj S
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1 Answers1

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Alternatively, to get rid of all those cases, we can write: $$\sin^{-1}x\pm\sin^{-1}y=\text{atan2}(x\sqrt{1-y^2}\pm y\sqrt{1-x^2},\ \sqrt{1-x^2}\sqrt{1-y^2}\mp xy)$$ and: $$\cos^{-1}x\pm\cos^{-1}y=\text{atan2}(y\sqrt{1-x^2}\pm x\sqrt{1-y^2},\ xy\mp\sqrt{1-x^2}\sqrt{1-y^2})$$

See atan2.

Klaas van Aarsen
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  • thanx. but the cases are really important and my doubt is there ! – Sooraj S Dec 06 '18 at 00:55
  • How are the cases important, and which doubt do you have @ss1729? Note that we have a formula that transforms an angle uniquely into x,y. Arcsin and arccos don't do the reverse but atan2 does. – Klaas van Aarsen Dec 06 '18 at 08:29