Question 907799
<pre>
Draw the angle in QII by drawing a right triangle in QII that
has an angle with sine of 7/25.

Since the sine is {{{opposite/hypotenuse=y/r}}} make the opposite
side or y equal to the numerator of 7/25, which is 7, and the 
hypotenuse or r equal to the denominator of 7/25, or 25.  I will
indicate the angle by the red curved line.

{{{drawing(200,200,-30,30,-30,30,triangle(0,0,-24,0,-24,7), 
line(-50,0,50,0),line(0,-50,0,50),locate(-27,6,7),locate(-14,8,25),
red(arc(0,0,12,-12,0,164))
)}}}

Calculate the adjacent side or x by the Pythagorean theorem:

{{{r^2=x^2+y^2}}}

{{{25^2=x^2+7^2}}}

{{{625=x^2+49}}}

{{{576=x^2}}}

{{{"" +- sqrt(576)=x}}}

{{{-24=x}}}

We take x as negative because x goes left in QII.


{{{drawing(200,200,-30,30,-30,30,triangle(0,0,-24,0,-24,7), 
line(-50,0,50,0),line(0,-50,0,50),locate(-27,6,7),locate(-14,8,25),
red(arc(0,0,12,-12,0,164)),locate(-20,0,-24)
)}}}
 
{{{tan(2theta)=2tan(theta)/(1-tan^2(theta))}}}

Since {{{tan(theta)=opposite/(adjacent)=y/x=7/(-24)=-7/24}}},

{{{tan(2theta)=2(-7/24)/(1-(-7/24)^2)}}}

{{{tan(2theta)=(-7/12)/(1-(49/576))}}}

{{{tan(2theta)=(-7/12)/((576/576)-(49/576))}}}

{{{tan(2theta)=(-7/12)/(527/576)}}}

{{{tan(2theta)=(-7/12)*(576/527)}}}

The 12 goes into 576 48 times, so that simplifies to

{{{tan(2theta)=-336/527}}}

Edwin</pre>