Question 1195395
<pre>

{{{y}}}{{{""=""}}}{{{sqrt((1+(sqrt(2)/2)(cos((x*pi)/3)+sin((x*pi)/3)^"")^"")/(1-(sqrt(2)/2)(cos((x*pi)/3)+sin((x*pi)/3)^"")^""))}}}

let {{{sqrt(2)/2}}}{{{""=""}}}{{{a}}}

let {{{cos((x*pi)/3)+sin((x*pi)/3)}}}{{{""=""}}}{{{b}}}

Substituting:

{{{y}}}{{{""=""}}}{{{sqrt((1+ab)/(1-ab))}}}

Under the radical multiply top and bottom by the conjugate
of the denominator, as if you were rationalizing the denominator:

{{{y}}}{{{""=""}}}{{{sqrt(((1+ab)(1+ab))/((1-ab)(1+ab)))}}}

{{{y}}}{{{""=""}}}{{{sqrt(((1+ab)^2)/(1-(ab)^2))}}}

The numerator is a perfect square, so we take the square root of
the numerator and denominator:

{{{y}}}{{{""=""}}}{{{(1+ab^"")/sqrt(1-(ab)^2))}}}

Write the right side as the sum of two fractions:

{{{y}}}{{{""=""}}}{{{(1^"")/sqrt(1-(ab)^2))}}}{{{""+""}}}{{{(ab^"")/sqrt(1-(ab)^2))}}}

Next we draw a right triangle with an angle &theta;, 1 as the hypotenuse, ab
as the adjacent side and the radical as the opposite side:

{{{drawing(200,800/7,-1,9,-1.3,3.7,
line(0,0,5,0),line(5,0,5,3),line(0,0,5,3), 
locate(2.4,-.05,ab),locate(5.1,2.2,sqrt(1-(ab)^2)), 
locate(2.34,2.3,1), locate(1.3,.81,theta)

)}}}

So now the equation is 

{{{y}}}{{{""=""}}}{{{csc(theta)}}}{{{""+""}}}{{{cot(theta)}}} where {{{cos(theta)}}}{{{""=""}}}{{{ab}}}

Since you want y as a function of cot(&theta;), we use an identity for csc(&theta;).

1 + cot<sup>2</sup>(&theta;) = csc<sup>2</sup>(&theta;), solve for csc(&theta;)

    {{{csc(theta)}}}{{{""=""}}}{{{sqrt(1+cot^2(theta))}}}

And we substitute back for ab:

The final equation is 

{{{y}}}{{{""=""}}}{{{sqrt(1+cot^2(theta))}}}{{{""+""}}}{{{cot(theta)}}} where  {{{cos(theta)}}}{{{""=""}}}{{{(sqrt(2)/2)(cos((x*pi)/3)+sin((x*pi)/3)^"")}}}


Edwin</pre>