Double identities involving inverse

Probably help to post the original question/example, but I guess they're just using a variable A to represent the angle.
Are you expected to draw a unit circle / triangle? If so cos(A) will be the base (1/2) as the hypotenuse is 1
So
tan(A) = sqrt(1-A^2)/A = sqrt(3)
for instance.

Edit - as above, theyre just using it to say the expression arccos(x) represents some angle A.

Yes

Damn book likes to make things confusing

Its a bit verbose and you/they could simplify it as
cos(2A) = cos^2(A)-1 = x^2-1
I guess they've done it like this as x is frequently used as a variable name for an angle (rather than a dimensionless side ratio in this case) so you could be thinking about cos(2x) = cos^2(x)-1... which is obviously wrong for this example.

Also, you should know that
cos^2(A) = (cos(A))^2
The left hand side is just a common way of writing the right hand side. You have to be careful with
cos^(-1)(x) = 1/cos(x)
and is not the inverse arccos(x).

Can you explain part b for me. I see the identity for sin 2A=2(SinaCosa)

I’m not sure how they’ve introduced the trig Pythagoras identity sin²x+cos²x=1, due to the 2 seemingly disappearing from the original function.

Its just a combination of the double angle sin and pythagoras, as you say, so
sin(2A) = 2 sin(A) cos(A) = 2 sin(A) x
using pythagoras
sin(A) = sqrt(1 - cos^2(A)) = sqrt(1-x^2)
so
sin(2A) = 2 x sqrt(1-x^2)

Mmm, let’s see if I follow correctly.

So, we derive 2SinAx from the double identity.

I take it we take 2 out of the equation by dividing by 2 first, so we’re left with sin A.

Taking sin A, we have to introduce cos by using Pythagoras identity. Hence,

So, sin²a=1-cos²a
Sin²a=1-x² due to cos a=x

Hence, sin a=√1-x²


I believe I made sense of that now lol