Question

Could someone please show me how to go from

5 sin(2 * arcsin[(x-2)/2])

to

5/2 (x – 2) * sqrt(4x – x^2)?


If the above is too ugly for you, the following link shows what I am asking for.:
http://www.wolframalpha.com/input/?i=5+sin(2+*+arcsin%5B(x-2)%2F2%5D)+%3D%3D+5%2F2+(x+%E2%80%93+2)+*+sqrt(4x+%E2%80%93+x%5E2)+%3F

Any help would be greatly appreciated!

Answer 1

\[5 \sin(2 \arcsin(\frac{ x-2 }{ 2 })\]

Answer 2

If you're saying that sin^(-1) (x) = 1/sin(x), then I disagree (and I don't mean to sound rude).

Answer 3

csc(x) = 1 / sin(x)

Answer 4

heheh ahemm one may note that "reciprocal" and "inverse relation" can be easily mistaken

Answer 5

Yes, that is a pretty bad notation. :)

Answer 6

So, @Johnbc and @jdoe0001, if you could at least tell me how to do this in words without typing all the work, I would very much appreciate it!

Answer 7

well...haven't found much yet

Answer 8

Johnbc badly as the notation looks.... is not the reciprocal of sine... but the inverse function

Answer 9

I am aware.

Answer 10

Thank you.

Answer 11

ok

Answer 12

Oh, I think I figure it out!

Answer 13

sin(2x) = 2 sin(x) cos(x)

sin(2 * arcsin[(x-2)/2])
2 sin(arcsin[(x-2)/2]) cos(arcsin[(x-2)/2])

And, I think I can work from there using the triangle I have from the problem I am doing.

Answer 14

hmmm am I wrong by suggesting that 2sin(x) is not the same as sin(2x) ?

Answer 15

Yes.:
http://www.wolframalpha.com/input/?i=sin%282x%29+%3D%3D+2sin%28x%29+%3F

Answer 16

I used the formula from the top-left corner here.:
http://en.wikipedia.org/wiki/Trigonometric_identities#Double-angle.2C_triple-angle.2C_and_half-angle_formulae

Answer 17

I just confirmed that I got the right answer! :)

Thanks!

Answer 18

By the way, I just realized I misread what you said above.

Answer 19

hehe

Answer 20

Suggesting that 2 sin(x) != sin(2x) is correct. :)

Answer 21

Alright, thanks again, but I have to go eat now. :)

Answer 22

Bye.