1. En

2. 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).

3. beginnersmind

A figure of the unit circle might illustrate it better: |dw:1441508969839:dw|

4. En

okay i get it..after that?

5. En

@beginnersmind after that ? how am i going to start to prove it?

6. 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.

7. 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

8. 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

9. beginnersmind

I think you should be able to use cos(x) = -cos(pi-x). It's more basic than the subtraction formula.

10. zepdrix

ya i was hoping so :) wasn't sure how basic it was though hehe

11. En

thanks a lot! I'm gonna go and try to understand it :)))

12. En

@zepdrix and @beginnersmind i understood it now :)

13. zepdrix

yay \c:/