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$$ \cos ^{-1} x-\cos ^{-1} y=\left\{\begin{array}{l} \cos ^{-1}\left(x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right) ; \text { if }-1 \leq x, y \leq 1 \quad \text{and} \quad x \leq y \\ -\cos ^{-1}\left(x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right) ; \text { if }-1 \leq y \leq 0,0<x \leq 1 \quad \text{and} \quad x \geqslant 1 \end{array}\right. $$

I'm having some issues proving for different cases, this is what I tried so far:

Let $\cos ^{-1} x=\alpha, \quad \cos ^{-1} y=\beta \quad \Longrightarrow \quad x=\cos \alpha, y=\cos \beta$

$$ \begin{aligned} \cos (\alpha-\beta) &=\cos \alpha \cos \beta+\sin \alpha \sin \beta \\ &=\cos \alpha \cos \beta+\sqrt{1-\cos ^{2} \alpha} \sqrt{1-\cos ^{2} \beta} \\ &=\left(x y+\sqrt{1-\cos ^{2} \alpha} \sqrt{1-\cos ^{2} \beta}\right) \end{aligned} $$

$$ \begin{aligned} \therefore \alpha-\beta &=\cos ^{-1} x-\cos ^{-1} y \\ &=\cos ^{-1}\left(x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right) \end{aligned} $$

Dan
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1 Answers1

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$\cos^{-1}z$ will be real $\iff-1\le z\le1$

Now as $0\le\cos^{-1}x,\cos^{-1}y\le\pi,$

$$-\pi<\cos^{-1}x-\cos^{-1}y\le\pi$$

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

$$\iff\cos^{-1}x-\cos^{-1}y\ge0$$

$$\iff\dfrac\pi2-\sin^{-1}x\ge\dfrac\pi2-\sin^{-1}y\iff\sin^{-1}x\le\sin^{-1}y\iff x\le y$$ as $\sin^{-1}x$ is an increasing function

Similarly, $$\cos^{-1}(xy+\sqrt{(1-x^2)(1-y^2)})\text{ will be }=\cos^{-1}y-\cos^{-1}x\iff y\le x$$

See also: Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $

lab bhattacharjee
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