Question 1056790
<pre><b>
Draw a right triangle in quadrant IV,
indicate the reference angle (the one inside the
triangle with one side as the x-axis) with an
arc:

{{{drawing(200,200,-8,8,-8,8,
red(arc(0,0,3,-3,310,360)),
line(-9,0,9,0),line(0,-9,0,9), triangle(6,0,6,-7,0,0) )}}}

Since {{{matrix(1,3,tangent,""="",opposite/adjacent)}}},
label the opposite side to the reference angle as the numerator
of -7/6  which is -7 (negative since the opposite side goes
down) and the adjacent side as +6 (positive since it goes
right.

{{{drawing(200,200,-8,8,-8,8,
red(arc(0,0,3,-3,310,360)),
locate(6.2,-3.2,-7), locate(3.2,1.5,"" + 6),
line(-9,0,9,0),line(0,-9,0,9), triangle(6,0,6,-7,0,0) )}}}

Find the hypotenuse by the Pythagorean theorem:

{{{c^2=a^2+b^2}}}
{{{c^2=(6)^2+(-7)^2}}}
{{{c^2=36+49}}}
{{{c^2=85}}}
{{{c="" +- sqrt(85)}}}

We always take the hypotenuse as positive in trigonometry.
Label it {{{"" + sqrt(85)}}}

{{{drawing(200,200,-8,8,-8,8,
red(arc(0,0,3,-3,310,360)),
locate(.05,-3,"" + sqrt(85)),
locate(6.2,-3.2,-7), locate(3.2,1.5,"" + 6),
line(-9,0,9,0),line(0,-9,0,9), triangle(6,0,6,-7,0,0) )}}}

Now we just plug in the formulas for sin(2x), cos(2x), and tan(2x),
using opposite, adjacent, and hypotenuse to find the trig ratios
for x.

{{{matrix(1,3,sin(2x),""="",2sin(x)cos(x))}}}
{{{matrix(1,5,sin(2x),""="",2(-7/sqrt(85))(6/sqrt(85)),""="",-84/85)}}}


{{{matrix(1,3,cos(2x),""="",cos^2(x)-sin^2(x))}}}
{{{matrix(1,7,cos(2x),""="",(6/sqrt(85))^2-((-7)/sqrt(85))^2,""="",36/85-49/85,""="",-13/85)}}}

{{{matrix(1,11,tan(2x),""="",sin(2x)/cos(2x),""="",(-84/85)/(-13/85),""="",(-84/85)*(-85/13),""="",(-84/cross(85))(-cross(85)/36),""="",84/13)}}}

Edwin</pre></b>