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I was looking at this strange and obscure identity an answer of the user @sirous Calculating $\cos^{-1}{\frac{3}{\sqrt10}} + \cos^{-1}{\frac{2}{\sqrt5}}$. But where did it come from outside?

$$\cos^{-1} x +\cos^{-1}y=\cos^{-1}\left[xy-\sqrt{(1-x^2)(1-y^2)}\,\right]$$

or equivalentely being $\cos^{-1}=\arccos$

$$ \arccos x+\arccos y= \begin{cases} \arccos\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right)& x+y\ge0\\ 2\pi-\arccos\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right)& x+y<0 \end{cases}$$

Can it probably be proved with two different graphs?

Blue
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Sebastiano
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1 Answers1

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If one sets $\alpha=\cos^{-1}x$ and $\beta=\cos^{-1}y$ takes cosine of both sides, this becomes $$ \cos(\alpha+\beta) = xy - \sqrt{(1-x^2)(1-y^2)}; $$ but since $\cos\alpha=x$ and $\sin\alpha=\sqrt{1-x^2}$, and $\cos\beta=y$ and $\sin\beta=\sqrt{1-y^2}$, this is just a disguised version of the usual angle-addition formula $$ \cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta. $$ (assuming $\alpha,\beta$ are in the first quadrant for example)

Greg Martin
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