Thread: How the heck do you simplify this expression?

Can anyone please, please show me how arccos(cos2x) = (- or +)2x +2(pi)(k)


k = integer

I'm bamboozled.

Hello,

Originally Posted by Kaitosan

Can anyone please, please show me how arccos(cos2x) = (- or +)2x +2(pi)(k)


k = integer

I'm bamboozled.

Because cos(2x)=cos(-2x)=cos(2x+2kpi)=cos(-2x+2kpi) ^^

cos(-2x)=cos(2x+2kpi)

What? You didn't even explain how you came about that answer. What is the reasoning behind this? I don't get it. Please repeat.

Originally Posted by Moo

Hello,

Because cos(2x)=cos(-2x)=cos(2x+2kpi)=cos(-2x+2kpi) ^^

Originally Posted by Kaitosan

cos(-2x)=cos(2x+2kpi)

What? You didn't even explain how you came about that answer. What is the reasoning behind this? I don't get it. Please repeat.

arccos and cos are inverse operators so they cancel each other out giving the solution of 2x. The 2kpi is because cos(2x) repeats itself every pi radians

Originally Posted by Kaitosan

cos(-2x)=cos(2x+2kpi)

What? You didn't even explain how you came about that answer. What is the reasoning behind this? I don't get it. Please repeat.

*Ahem* You were given an explanation.

Originally Posted by Moo

Hello,

Because cos(2x)=cos(-2x)=cos(2x+2kpi)=cos(-2x+2kpi) ^^

So you were told:

1. cos(-2x) = cos(2x) (a symmetry property should be familiar with).

2. cos(2x) = cos(2x + 2kpi) (another symmetry property you should be familiar with).

Thanks guys. I didn't know that if cosx is within its inverse, they cancel each other. Is this also the case with arcsin(sinx) = x and arctan(tanx) = x? What about tan(arctanx) and cos(arccosx) or something?

Originally Posted by Kaitosan

Thanks guys. I didn't know that if cosx is within its inverse, they cancel each other. Is this also the case with arcsin(sinx) = x and arctan(tanx) = x? What about tan(arctanx) and cos(arccosx) or something?

Yes, that is the case