Proving a trigonometric identity (involving arcos(x))

Question

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 $

do you mean $\arccos(x)+\arccos(y) = \arccos(xy-\sqrt{(1-x^2)(1-y^2)})$ — The Integrator, Jun 27 '18 at 13:49
Use https://math.stackexchange.com/questions/1116974/why-its-true-arcsinx-arccosx-frac-pi2. and https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y — lab bhattacharjee, Jun 27 '18 at 14:29

score 0 · Answer 1 · answered Jun 27 '18 at 13:52

0

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.

answered Jun 27 '18 at 13:52

Bernard

score 0 · Answer 2 · answered Jun 27 '18 at 13:54

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) $

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

2 Answers2