Cosine of (1 Viewer)

[studynerd]

I just want to clarify my concept. We started doing 3d trig recently and I came across the fact that inverse cos of any number less then 1 plus the invese cos of the same number but negative equals to 180degrees. Why is this?

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[Shadowdude]

I just want to clarify my concept. We started doing 3d trig recently and I came across the fact that inverse cos of any number less then 1 plus the invese cos of the same number but negative equals to 180degrees. Why is this?

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I'm going to guess that it's some property of the arccos graph. In fact, I think you'll find that it's always true.

 

[braintic]

For starters, the inverse cosine function is defined in radians. So the sum is pi.

Inverse cos is defined in the 1st and second quadrants only.

Let y be a 1st quadrant angle and let x=cosy, so y=arccos x (1)

cos(pi-y)=-cosy=-x (cos is -ve in 2nd quadrant)
pi-y = arccos(-x) (2)

Adding (1) & (2): pi = arccos(x) + arccos(-x)


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If you want a more graphical approach:

Picture the arccos graph, y=arccos(x)
If you translate the graph down by pi/2, you get f(x)=arccos(x) - pi/2.
You should see that this new graph is an odd function.
So f(-x) = -f(x)
arccos(-x) - pi/2 = - [arccos(x) - pi/2]

Rearranging gives arccos(x) + arccos(-x) = pi

 

[integral95]

Yeah Shadowdude is right that it's always true.

Here the general proof.

 

[RealiseNothing]

Yeah Shadowdude is right that it's always true.

Here the general proof.

#cyclicargument

 

[braintic]

Yeah Shadowdude is right that it's always true.

Here the general proof.

Do you realise that you are quoting the rule that this guy is trying to prove?

As RealiseNothing says - it is a cyclical argument.