• En

please help! prove that! arccos (-x)= π-arccos x

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  • En

please help! prove that! arccos (-x)= π-arccos x

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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  • En
please help :)
Remember the definition of arccos x. It's the _angle_ whose cosine is x. So this identity says that two angles, one of whose cosine is x, and another whose cosine is (-x) add up to pi (that is 180 degrees).
A figure of the unit circle might illustrate it better: |dw:1441508969839:dw|

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  • En
okay i get it..after that?
  • En
@beginnersmind after that ? how am i going to start to prove it?
I would probably write it up in words. Maybe there's a way to do it by manipulating the identity, but I don't see it.
Ummmm this is what I would do :) \[\large\rm \arccos (-x)=\color{orangered}{π-\arccos x}\]Let's call this right side something... like... y.\[\large\rm y=\pi-\arccos x\]Let's subtract pi from each side, then multiply by -1, giving us,\[\large\rm \pi-y=\arccos x\]Take the inverse,\[\large\rm \cos(\pi-y)=x\]So if you subtract an angle from pi within cosine, that's the same as flipping it over the x-axis. Therefore:\[\large\rm \cos(\pi-y)=-\cos(y)=x\]Inverse again,\[\large\rm y=\arccos(-x)\]Relate this back to the original equation. Since the right side is arccos(-x) and the left side is arccos(-x), proofed! :O Maybe not the most straight forward approach :\ but whatev
From this step:\[\large\rm \cos(\pi-y)=x\]If you would like a better justification, you can apply your cosine angle difference identity:\[\large\rm \cos(\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\]\[\large\rm \cos(\pi-y)=\cos \pi \cos y+\sin \pi \sin y\]Which will simplify down to -cos y
I think you should be able to use cos(x) = -cos(pi-x). It's more basic than the subtraction formula.
ya i was hoping so :) wasn't sure how basic it was though hehe
  • En
thanks a lot! I'm gonna go and try to understand it :)))
  • En
@zepdrix and @beginnersmind i understood it now :)
yay \c:/

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