Question 1143228
<pre>

If {{{t = tan(x)}}} what is {{{1-sin(2x)}}}?

We draw a right triangle. The tangent of x is t, which is the 
same as t/1.  The tangent is the opposite side over tha adjacent side,
so we make the opposite side of angle x be the numerator of t/1
which is t and the adjacent side be the denominator of t/1 which
is 1.  That makes the hypotenuse be {{{sqrt(1^2+t^2)}}} or {{{sqrt(1+t^2)}}}. 

{{{drawing(400,200,-1,9,-1,4,  triangle(0,0,8,0,8,4),
locate(4,0,1), locate(8.2,2.3,t), locate(1,.5,matrix(1,2,Angle,x)),
locate(3,2.8,sqrt(1+t^2))
)}}}

The sine is the opposite over the hypotenuse and the cosine is the
adjacent over the hypotenuse, so

{{{1-sin(2x)=1-2sin(x)cos(x)=1-2(t^""/sqrt(1+t^2))(1^""/sqrt(1+t^2))=1-2t^""/(1+t^2)=(1+t^2)/(1+t^2)-2t^""/(1+t^2)=(1+t^2-2t)/(1+t^2)=(t^2-2t+1)/(1+t^2)=(t-1)^2/(1+t^2)}}}

Edwin</pre>