Proving a trigonometric identity (involving arcos(x))

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Not very familiar with how to do this problem. Not sure on the process of how to get the answer. Thanks for the help to those who help answer it! :)

Assuming all angles are acute, show that: $\cos ^{-1}x+\cos ^{-1}y=\lbrack xy-\sqrt{\lbrace (1-x^{2})(1-y^{2}\rbrace }\rbrack $

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2 Answers

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Hint:

Compute the cosine of both sides. You'll need to calculate $\sin(\arccos x)$, which shouldn't be too hard, since you know all angles are acute.

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let $X =\arccos(x)\implies x=\cos(X)$

$ Y =\arccos(y)\implies y=\cos(Y)$

now;

$\cos(X+Y) = \cos(X)\cos(Y)-\sin(X)\sin(Y)$

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

$X+Y=\arccos\bigg(xy- \sqrt{1-x^2}\sqrt{1-y^2}\bigg) $

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

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