Question

Tan[(1/2)sin^-1(4/5)]

Answer 1

@terenzreignz

Answer 2

Ok this time, recall the half-angle identities...

Answer 3

Maybe it would be to your advantage to use one of these
\[\huge \tan\left[\frac12 x\right]=\frac{1-\cos(x)}{\sin(x)}=\csc(x) -\cot(x)\]
Whichever of these may be more useful for you.

Answer 4

the second one is familiar

Answer 5

\[\large \tan\left[ \frac12 \sin^{-1}\left(\frac45\right)\right]=\frac{1-\cos\left[ \sin^{-1}\left(\frac45\right)\right]}{\sin\left[ \sin^{-1}\left(\frac45\right)\right]}=\csc\left[ \sin^{-1}\left(\frac45\right)\right] -\cot\left[ \sin^{-1}\left(\frac45\right)\right]\]

Answer 6

Never mind the cut-off part, the first one is easier, anyway.
Now what is
\[\huge \sin\left[ \sin^{-1}\left(\frac45\right)\right]\]?

Answer 7

4/5

Answer 8

That's right. Now what is \[\huge \cos\left[ \sin^{-1}\left(\frac45\right)\right]\]?

Answer 9

3/5

Answer 10

oh... I was about to draw a triangle, but it seems you have it down. Now all you have to do is plug in...
\[\large \tan\left[ \frac12 \sin^{-1}\left(\frac45\right)\right]=\frac{1-\cos\left[ \sin^{-1}\left(\frac45\right)\right]}{\sin\left[ \sin^{-1}\left(\frac45\right)\right]}\]

Answer 11

wow. Ok I understand

Answer 12

very simple, once you get it. Thanks a lot!

Answer 13

No problem :)