Answer:
Step-by-step explanation:
In each case, draw the right triangle which produces the inverse trig value. That is, label the two sides as needed, and calculate the third side.
It should be clear that
[tex]tan(sin^{-1} x/4) = x/\sqrt{16-x^{2} }[/tex]
[tex]sin(tan^{-1} x/4) = x/\sqrt{16+x^{2} }[/tex]
sin(2α) = 2 sinα cosα
So,
[tex]1/2 sin(2sin^{-1} x/4) = (1/2)(x/4) * x/\sqrt{16-x^2} = x^{2} /(8\sqrt{16-x^2})[/tex]
See what you can do with the others.