Question

Hello :) I need a little help with
arccos((x-2)/2) = arctan(___) for 2 < x < 4

Answer 1

Let arccos((x-2)/2) = y
cos(y) = (x-2)/2
Find tan(y)

Answer 2

I"m not sure how I could do that?

my teacher was trying to get us to do a triangle idea like

Answer 3

\[
y^2+(x-2)^2 = 2^2
\]

Answer 4

Find the length of the side opposite to the angle theta.
tan(theta) = opposite / adjacent

Answer 5

\[
\tan(\theta) = \frac{y}{x-2}
\]

Answer 6

mmm I got y^2 = -x^2 + 4x ? :)

Answer 7

oh wait hmmm

Answer 8

Yeah, you would have to take the square root.

Answer 9

ok :)
so I get (sqrt(4x - x^2)) / (x-2) as the arctan?

Answer 10

Yes

Answer 11

Thank you! :D <3

Answer 12

the positive root might also be a solution.

Answer 13

I mean negative root^

Answer 14

because of the given domain?

Answer 15

because \(y\) was squared. Suppose \(y\) was negative.

Answer 16

ohhhhh okay ^_^