En
  • En
please help! prove that! arccos (-x)= π-arccos x
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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En
  • En
please help :)
beginnersmind
  • beginnersmind
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).
beginnersmind
  • beginnersmind
A figure of the unit circle might illustrate it better: |dw:1441508969839:dw|

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En
  • En
okay i get it..after that?
En
  • En
@beginnersmind after that ? how am i going to start to prove it?
beginnersmind
  • beginnersmind
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.
zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
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
beginnersmind
  • beginnersmind
I think you should be able to use cos(x) = -cos(pi-x). It's more basic than the subtraction formula.
zepdrix
  • zepdrix
ya i was hoping so :) wasn't sure how basic it was though hehe
En
  • En
thanks a lot! I'm gonna go and try to understand it :)))
En
  • En
@zepdrix and @beginnersmind i understood it now :)
zepdrix
  • zepdrix
yay \c:/

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