Question 247773
I need to write the following as an algebraic expression: sin(2arccos x)

{{{sin(2arccos(x))}}}
<pre><font size = 4 color = "indigo"><b>
let {{{arccos(x) = alpha}}}

then {{{cos(alpha)=x}}}

So draw a right triangle and label one of the acute angles {{{alpha}}}


{{{drawing(200,134,-1,11,-1,7,
locate(1.7,1,alpha), rectangle(9.5,0,10,.5),
triangle(0,0,10,0,10,6) )}}} 

Now we need to put values on the sides of the
triangle to cause {{{cos(alpha)}}} to equal {{{x}}}.

Since the cosine is the adjacent side divided by the
hypotenuse, if we label the adjacent side as x and the
hypotenuse as 1. then {{{cos(alpha)}}} will equal {{{x}}}.

{{{drawing(200,134,-1,11,-1,7, rectangle(9.5,0,10,.5),
locate(1.7,1,alpha), locate(5,0,x), locate(5,4.5,1),
triangle(0,0,10,0,10,6) )}}}

Now we use the Pythagorean theorem to find the opposite
side:

{{{adjacent^2 + opposite^2= hypotenuse^2}}}

{{{x^2+opposite^2=1^2}}}

{{{x^2+opposite^2=1}}}

{{{opposite^2=1-x^2}}}

{{{opposite = sqrt(1-x^2)}}}

So we write this on the opposite side:

{{{drawing(250,134,-1,14,-1,7, locate(10.2,4,sqrt(1-x^2)),rectangle(9.5,0,10,.5),
locate(1.7,1,alpha), locate(5,0,x), locate(5,4.5,1),
triangle(0,0,10,0,10,6) )}}}

Now we go back to the original problem:

{{{sin(2arccos(x))}}}

or

{{{sin(2alpha)}}}

and we use the identity

{{{sin(2alpha)=2sin(alpha)cos(alpha)}}}  

Then using the fact that the sine is the opposite over
the hypotenuse and the cosine is the adjacent over the
hypotenuse, we use the right triangle above:

{{{sin(2alpha)=2sin(alpha)cos(alpha)}}}

{{{sin(2alpha)=2((sqrt(1-x^2))/1)(x/1)}}}

{{{sin(2alpha)=2(sqrt(1-x^2))(x)}}}

{{{sin(2alpha)=2x*sqrt(1-x^2)}}}

Edwin</pre>