Question 467040
For x>0, find sin(arccos(4/x))
<pre>
First look at arccos(4/x).  That asks the question:

"What angle has 4/x for its cosine?"

So we draw a right triangle and let the angle be

arccos(4/x)

and since the cosine is the adjacent over the hypotenuse,

we let the adjacent side be 4 and the hypotenuse be x:

Let the opposite side be h.

{{{drawing(200,320,-1,5,-1,7, triangle(0,0,4,0,4,6),
locate(.5,.95,arccos(4/x)), red(arc(0,0,5.5,-5.5,0,56.31)),
locate(2,0,4),locate(1.5,3,x),locate(4.2,3,h)
  )}}}


All we need now is to find the opposite side h by the 
Pythagorean theorem:

{{{c^2=a^2+b^2}}}
{{{x^2=4^2+h^2}}}
{{{x^2-4^2=h^2}}}
{{{x^2-16=h^2}}}
{{{sqrt(x^2-16)=h}}}

So the triangle is now:

{{{drawing(1100/3,320,-1,10,-1,7, triangle(0,0,4,0,4,6),
locate(.5,.95,arccos(4/x)), red(arc(0,0,5.5,-5.5,0,56.31)),
locate(2,0,4),locate(1.5,3,x),locate(4.2,3,sqrt(x^2-16))
  )}}}

So now the sine of arccos(4/x) is found by taking
the opposite side over the hypotenuse:

sin(arccos({{{4/x}}})) = {{{(opposite)/(hypotenuse)= sqrt(x^2-16)/x}}}

Edwin</pre>