Question 985175
{{{tan(A)=2/3}}} A ∈ QIII,  {{{sec(B)=3}}} B ∈ QI find {{{cos(A+B)}}}

<pre>

{{{cos(A+B)}}}{{{""=""}}}{{{cos(A)cos(B)-sin(A)sin(B)}}}.

First we draw the two angles, A, B, in their respective quadrants:

{{{drawing(200,200,-1.2,1.2,-1.2,1.2,
red(locate(-.2,.41,A)),
line(-3,0,3,0), line(0,3,0,-3), line(0,0,cos(3.729595257),sin(3.729595257) ),
red(arc(0,0,.6,-.6,0,214))

 )}}} {{{drawing(200,200,-1.2,1.2,-1.2,1.2,
red(locate(.22,.29,B)),
line(-3,0,3,0), line(0,3,0,-3), line(0,0,cos(1.230959417),sin(1.230959417) ),
red(arc(0,0,.47,-.47,0,71))

 )}}}
Draw perpendiculars to the x-axis from the end of the terminal sides of
the angles, creating right triangles.

{{{drawing(200,200,-1.2,1.2,-1.2,1.2,
red(locate(-.2,.41,A)), green(line(cos(3.729595257),sin(3.729595257),cos(3.729595257),0)),
line(-3,0,3,0), line(0,3,0,-3), line(0,0,cos(3.729595257),sin(3.729595257) ),
red(arc(0,0,.6,-.6,0,214))

 )}}} {{{drawing(200,200,-1.2,1.2,-1.2,1.2,
green(line(cos(1.230959417),sin(1.230959417),cos(1.230959417),0)),
red(locate(.22,.29,B)),
line(-3,0,3,0), line(0,3,0,-3), line(0,0,cos(1.230959417),sin(1.230959417) ),
red(arc(0,0,.47,-.47,0,71))

 )}}}

For angle A:
{{{TANGENT=OPPOSITE/ADJACTENT=y/x}}}. Since the adjacent side, x, goes left
on the x-axis, and the opposite side, y, goes downward from the x-axis, we must
consider the tangent {{{2/3}}} as {{{(-2)/(-3)}}} and put the numerator of {{{(-2)/(-3)}}},
which is -2, on the y=OPPOSITE side and the denominator of {{{(-2)/(-3)}}}, which is
-3, on the x=ADJACENT SIDE.

For angle B:
Since {{{SECANT=HYPOTENUSE/ADJACENT=r/x}}} is in QI we can leave everything positive.
We put the numerator of 3/1, which is 3, on the r=HYPOTENUSE and the denominator
of 3/1, which is 1, on the x=ADJACENT SIDE. 

Then we calculate the third sides of the two right triangles by using the
Pythagorean theorem:  

for angle A:                  for angle B:
{{{r^2=x^2+y^2}}}                   {{{r^2=x^2+y^2}}}                       
{{{r^2=(-3)^2+(-2)^2}}}              {{{3^2=(1)^2+y^2}}}
{{{r^2=9+4}}}                      {{{9=1+y^2}}}
{{{r^2=13}}}                       {{{8 = y^2}}}
{{{r=sqrt(13)}}}                      {{{sqrt(8) = y}}}
                           {{{sqrt(4*2) = y}}} 
                            {{{2sqrt(2) = y}}}

{{{drawing(200,200,-1.2,1.2,-1.2,1.2,
red(locate(-.2,.41,A)), green(line(cos(3.729595257),sin(3.729595257),cos(3.729595257),0)),
line(-3,0,3,0), line(0,3,0,-3), line(0,0,cos(3.729595257),sin(3.729595257) ),



locate(-.8,.23,x=-3), locate(-1.2,-.2,y=-2), locate(-.6,-.3,r=sqrt(13)),  


red(arc(0,0,.6,-.6,0,214))

 )}}} {{{drawing(200,200,-1.2,1.2,-1.2,1.2,
green(line(cos(1.230959417),sin(1.230959417),cos(1.230959417),0)),
red(locate(.22,.29,B)),

locate(0,0,x=1), locate(.34,.5,y=2sqrt(2)), locate(-.2,.65,r=3),

line(-3,0,3,0), line(0,3,0,-3), line(0,0,cos(1.230959417),sin(1.230959417) ),
red(arc(0,0,.47,-.47,0,71))

 )}}}

Now we are able to substitute in

{{{cos(A+B)}}}{{{""=""}}}{{{cos(A)cos(B)-sin(A)sin(B)  remembering that
{{{SINE=OPPOSITE/HYPOTENUSE=y/r}}} and {{{COSINE=ADJACENT/HYPOTENUSE=y/r}}}

{{{cos(A+B)}}}{{{""=""}}}{{{cos(A)cos(B)-sin(A)sin(B)}}}

{{{cos(A+B)}}}{{{""=""}}}{{{((-3)/sqrt(13))(1/3)-((-2)/sqrt(13))(2sqrt(2)/3)}}}

{{{cos(A+B)}}}{{{""=""}}}{{{(-3)/(3sqrt(13))+(4sqrt(2))/(3sqrt(13))}}}

{{{cos(A+B)}}}{{{""=""}}}{{{(-3+4sqrt(2))/(3sqrt(13))}}}

Rationalizing the denominator:

{{{cos(A+B)}}}{{{""=""}}}{{{(-3+4sqrt(2))/(3sqrt(13))}}}{{{""*""}}}{{{sqrt(13)/sqrt(13))}}}

{{{cos(A+B)}}}{{{""=""}}}{{{(-3sqrt(13)+4sqrt(26))/(3*13)}}}

{{{cos(A+B)}}}{{{""=""}}}{{{(4sqrt(26)-3sqrt(13))/39}}}

Edwin</pre>