Question

Assume x is greater than 0. Simplify sin(2arccos(.5x)). The answer should have no trig functions in it, but it will have an x.

Answer 1

is that 0.5x?

Answer 2

yes! sorry for the confusion

Answer 3

what is is the equivalent form of sin(2u) ?

Answer 4

2cosxsinx ?

Answer 5

correct. If you apply the identity to the original expression, what do you get?

Answer 6

im not sure how to do that

Answer 7

you get 2 sin(arccos(x/2)) cos(arccos(x/2)), yes?

Answer 8

Ah ok! got that! Can I then cancel out the cos(arccos(0.5))?

Answer 9

no, you can not

Answer 10

you can not do so without stating the domian first

Answer 11

Oh I just saw the mistake in that. That was dumb. Should I use half angle identities then?

Answer 12

no you don't have too. What is the domain?

Answer 13

The domain of arccos and arcsin is -1 to 1

Answer 14

ok, if the domain of arccos(x) is [-1,1], what is the domain of arccos(x/2) ?

Answer 15

[-2, 2] ?

Answer 16

correct. Now, you're allowed to do cos(arccos(x/2)) = x/2.
Now you see why you need to state the domain first. If you didn't,
cos(arccos(x/2)) = x/2 is false if x is not in [-2,2]

Answer 17

say x = 10. cos(arccos(10/2)) is undefined, but 10/2 = 5. Undefined is not the same as 5 (which is defined)

Answer 18

now what about sin(arccos(x/2)) ?

Answer 19

well i can't cancel them out... Im not sure

Answer 20

maybe the half-angle identities?

Answer 21

no, half angle identity would probably make it worse.

Let's let u = arccos(x/2), then cos(u) = x/2, yes?

Answer 22

so am i solving for u?

Answer 23

sin(u) = sqrt(4-x^2)/2

but since u = arccos(x/2),
sin(u) = sin(arccos(x/2)) = sqrt(4-x^2)/2

Answer 24

so you have,
2 (sqrt(4-x^2)/2) (x/2) = (x/2) sqrt(4-x^2) <-- ans

Answer 25

thank you!!

Answer 26

no prob