Question 1159221
If tan α = − 4/3 and cot β = 15/8
 for a second-quadrant angle α and a third-quadrant angle β, find the following.
1. sin (a+b)
2. cos (a+b)
3. tan (a+b)

I got -5 for #1, 20 for #2, and 13/84 for #3. They are all wrong and I do not know why. Maybe because I just substituted the fractions I was given into the equations? 

<pre>
We draw a picture of α in Quadrant II, and a picture of β in Quadrant III

{{{drawing(200,200,-4,4,-4,4,line(-6,0,6,0), line(0,6,0,-6),
line(0,0,-1.5,2),red(arc(0,0,3,-3,0,127),locate(1,1.7,alpha)) )}}}   {{{drawing(200,200,-5,5,-5,5,line(-6,0,6,0), line(0,6,0,-6),
line(0,0,-sqrt(15),-1.6),red(arc(0,0,3,-3,0,200),locate(.2,2.2,beta)) )}}}  

From the ends of the terminal sides we draw vertical lines to the x-axis,
in green, forming right triangles:

{{{drawing(200,200,-4,4,-4,4,line(-6,0,6,0), line(0,6,0,-6),
green(line(-1.5,2,-1.5,0)),
line(0,0,-1.5,2),red(arc(0,0,3,-3,0,127),locate(1,1.7,alpha)) )}}}   {{{drawing(200,200,-5,5,-5,5,line(-6,0,6,0), line(0,6,0,-6),
green(line(-sqrt(15),-1.6,-sqrt(15),0)),
line(0,0,-sqrt(15),-1.6),red(arc(0,0,3,-3,0,200),locate(.2,2.2,beta)) )}}}

We are given that tan α = − 4/3

Since we know that TANGENT = OPPOSITE/ADJACENT = y/x, we label the
opposite side (y) as the numerator of -4/3, which is -4, negative because
it goes left.  We label the adjacent (x) as the denominator of -4/3,
which is +3 because it goes upward from the x-axis. 

We are given that cot(β)= 15/8

Since we know that COTANGENT = ADJACENT/OPPOSITE = x/y, we label the
adjacent side (x) as the numerator of 15/8, thought of as (-15)/(-8) which is
-15, negative because
it goes downward.  We label the opposite (y) as the denominator of (-15)/(-8),
which is -8 because it goes left.


{{{drawing(200,200,-4,4,-4,4,line(-6,0,6,0), line(0,6,0,-6),
locate(-2,0,x=-3),locate(-3,1,y=4),

green(line(-1.5,2,-1.5,0)),
line(0,0,-1.5,2),red(arc(0,0,3,-3,0,127),locate(1,1.7,alpha)) )}}}   {{{drawing(200,200,-5,5,-5,5,line(-6,0,6,0), line(0,6,0,-6),
locate(-4,1,x=-15),locate(-5,-.45,y=-8),
green(line(-sqrt(15),-1.6,-sqrt(15),0)),
line(0,0,-sqrt(15),-1.6),red(arc(0,0,3,-3,0,200),locate(.2,2.2,beta)) )}}}

Next we find the missing sides, the terminal sides, using the Pythagorean theorem:

{{{matrix(7,1,

r^2=x^2+y^2,
r^2=(-3)^2+4^2,
r^2=9+16,
r^2=25,
r="" +- sqrt(25),
r="" +- 5,
r=5)}}}    {{{matrix(6,1,

r^2=x^2+y^2,
r^2=(-15)^2+(-8)^2,
r^2=225+64,
r^2=289,
r="" +- sqrt(289),
r="" +- 17
)}}}

Note that we always take the terminal side r as positive.

{{{drawing(200,200,-4,4,-4,4,line(-6,0,6,0), line(0,6,0,-6),
locate(-2,0,x=-3),locate(-3,1,y=4),
locate(-1.2,2,r=5),
green(line(-1.5,2,-1.5,0)),
line(0,0,-1.5,2),red(arc(0,0,3,-3,0,127),locate(1,1.7,alpha)) )}}}   {{{drawing(200,200,-5,5,-5,5,line(-6,0,6,0), line(0,6,0,-6),
locate(-4,1,x=-15),locate(-5,-.45,y=-8),

locate(-1.5,-.63,r=17),
green(line(-sqrt(15),-1.6,-sqrt(15),0)),
line(0,0,-sqrt(15),-1.6),red(arc(0,0,3,-3,0,200),locate(.2,2.2,beta)) )}}}

Now we're finally ready to calculate

1. sin (a+b)
2. cos (a+b)
3. tan (a+b) 

where

{{{matrix(2,5,
sin(alpha),""="",y/r,""="",4/5,
cos(alpha),""="",x/r,""="",-3/5)}}} 

and

{{{matrix(2,5,
sin(beta),""="",y/r,""="",-8/17,
cos(beta),""="",x/r,""="",-15/17)}}} 

{{{matrix(4,3,
sin(alpha+beta),""="",sin(alpha)cos(beta)+cos(alpha)sin(beta),
sin(alpha+beta),""="",(4/5)(-15/17)+(-3/5)(-8/17),
sin(alpha+beta),""="",(-60/85)+(24/85),
sin(alpha+beta),""="",-36/85)}}}

{{{matrix(4,3,
cos(alpha+beta),""="",cos(alpha)cos(beta)-sin(alpha)sin(beta),
cos(alpha+beta),""="",(-3/5)(-15/17)-(4/5)(-8/17),
cos(alpha+beta),""="",(45/85)+(32/85),
cos(alpha+beta),""="",77/85)}}}

{{{matrix(3,3,
tan(alpha+beta),""="",sin(alpha+beta)/cos(alpha+beta),
tan(alpha+beta),""="",(-36/85)/(77/85),
tan(alpha+beta),""="",-36/77)}}}

Edwin</pre>