Prove Trig Identity

[Oxymoron]

How would you prove using the cancellation laws 2arccos(x) = arccos(2x² - 1). I am stumped. Any guidance is appreciated.

[uart]

Start by taking cos of both sides to get :

cos ( acos(x) + acos(x) ) = 2x^2 -1

It's pretty easy from there, so I'll leave itfor you to have a go. :)

[Oxymoron]

What if I let θ = acos(x) which implies that x = cosθ

Then the expression becomes...

2θ = acos(2(cosθ )² - 1)
2θ = acos(2cos²θ - 1)

Then take cos of both sides and use the cancellation law...

cos2θ = 2cos²θ - 1

This is a well known trigonometric identity. So if we start from this and work backwards the identity in the question can easily be shown.

[Oxymoron]

Start with the double angle identity

cos(2θ) = 2cos²θ -1

Then take the square outside the brackets

cos(2θ) = 2(cosθ )² -1

Since cos(acos(x)) = x we may write it as

cos(2θ) = cos(acos(2(cosθ )² -1))

Now take inverse cos of both sides

acos(cos(2θ)) = acos(cos(acos(2(cosθ )² -1)))

Cancel out the acos(cos(x)) terms

2θ = acos(2(cosθ )² -1)

Let θ = acos(α) implies α = cosθ

2acos(α ) = acos(2α² - 1)

QED

[uart]

I just used the usual cos(A+B) expansion on the LHS to give :

cos^2( acos(x) ) - sin^2 (acos(x))

and then added zero in the form of cos^2(.) + sin^2(.) - 1 to the LHS to give,

2 cos^2( acos(x) ) - 1

[Oxymoron]

Another question.

For what values of A is the trigonometric identity cos2A = 2cos²A - 1 valid? I thought it valid for all real numbers. But there must be a trick??

Any hints would be appreciated. Thanks.

[jtolliver]

For what values of A is the trigonometric identity cos2A = 2cos²A - 1 valid? I thought it valid for all real numbers. But there must be a trick??
It is valid for all A. It follows from the pythagorean identity and the identity cos (A+B) = cos A cos B - sin A sin B. That identity gives cos 2A = cos (A+A) = cos² A - sin² A. Adding on cos² A + sin² A - 1 (which is 0 by the pythagorean identity) to the right side gives the identity cos 2A = 2cos² A - 1