Question 625851
<pre>
If tan &#945; = 12/5, &#960; < &#945; < 3 &#960;/2, and sin &#946; = 2/3, 0< &#946; < &#960;/2, find

We draw the graphs of the angles, {{{alpha}}} is in the third
quadrant and {{{beta}}} is in the first:

{{{drawing(300,300,-1.1,1.1,-1.1,1.1,
line(-2,0,2,0), line(0,-2,0,2), red(locate(-.2,.2,alpha)),
green(line(0,0,-cos(1.176),-sin(1.176))), 
red(arc(0,0,.3,-.3,0,247)) )}}}    {{{drawing(300,300,-1.1,1.1,-1.1,1.1,
line(-2,0,2,0), line(0,-2,0,2), red(locate(.25,.2,beta)),
green(line(0,0,cos(.7297),sin(.7297))),
red(arc(0,0,.5,-.5,0,42)) )}}}

Next we draw perpendiculars to the x-axis, which creates right
triangles

{{{drawing(300,300,-1.1,1.1,-1.1,1.1,
line(-2,0,2,0), line(0,-2,0,2), red(locate(-.2,.2,alpha)),
green(line(0,0,-cos(1.176),-sin(1.176))),line(-cos(1.176),-sin(1.176),-cos(1.176),0),
red(arc(0,0,.3,-.3,0,247)) )}}}    {{{drawing(300,300,-1.1,1.1,-1.1,1.1,
line(-2,0,2,0), line(0,-2,0,2), red(locate(.25,.2,beta)),
green(line(0,0,cos(.7297),sin(.7297))),line(cos(.7297),sin(.7297),cos(.7297),0),
red(arc(0,0,.5,-.5,0,42)) )}}}

Since the tangent is opposite/adjacent or y/x, we take the opposite side of 
{{{alpha}}} to be -12 (negative because it goes downward from the x-axis), and
the adjacent to be -5 (negative because it goes left from the origin).  

Since the sine is opposite/hypotenuse or y/r, we take the opposite side of 
{{{beta}}} to be +2 (positive because it goes upward from the x-axis), and
the hypotenuse r to be 3 (the hypotenuse r is ALWAYS taken positive).

{{{drawing(300,300,-1.1,1.1,-1.1,1.1,locate(-.8,-.4,y=-12),locate(-.35,0,x=-5),
line(-2,0,2,0), line(0,-2,0,2), red(locate(-.2,.2,alpha)), locate(-.15,-.4,r),
green(line(0,0,-cos(1.176),-sin(1.176))),line(-cos(1.176),-sin(1.176),-cos(1.176),0),
red(arc(0,0,.3,-.3,0,247)) )}}}    {{{drawing(300,300,-1.1,1.1,-1.1,1.1,
line(-2,0,2,0), line(0,-2,0,2), red(locate(.25,.2,beta)),

locate(.4,0,x),locate(.75,.4,y=2),locate(.1,.4,r=3),


green(line(0,0,cos(.7297),sin(.7297))),line(cos(.7297),sin(.7297),cos(.7297),0),
red(arc(0,0,.5,-.5,0,42)) )}}}

Next we use the Pythagorean theorem to calculate the hypotenuse r for {{{alpha}}}
and the adjacent side x for {{{beta}}}.

r² = x² + y²            r² = x² + y²
r² = (-5)²+(-12)²       3² = x² + 2²
r² = 25 + 144            9 = x² + 4  
r² = 169                 5 = x² 
 r = 13                  {{{sqrt(5)}}} = x

{{{drawing(300,300,-1.1,1.1,-1.1,1.1,locate(-.8,-.4,y=-12),locate(-.35,0,x=-5),
line(-2,0,2,0), line(0,-2,0,2), red(locate(-.2,.2,alpha)), locate(-.15,-.4,r=13),
green(line(0,0,-cos(1.176),-sin(1.176))),line(-cos(1.176),-sin(1.176),-cos(1.176),0),
red(arc(0,0,.3,-.3,0,247)) )}}}    {{{drawing(300,300,-1.1,1.1,-1.1,1.1,
line(-2,0,2,0), line(0,-2,0,2), red(locate(.25,.2,beta)),

locate(.4,0,x=sqrt(5)),locate(.75,.4,y=2),locate(.1,.4,r=3),


green(line(0,0,cos(.7297),sin(.7297))),line(cos(.7297),sin(.7297),cos(.7297),0),
red(arc(0,0,.5,-.5,0,42)) )}}}

Now we can answer the first two questions by merely looking at the triangles,
and knowing sine = opposite/hypotenuse and cosine = adjacent/hypotenuse,
using x for adjacent, y for opposite and r for hypotenuse:

 A) sin &#945; = opp/hyp = y/r = (-12)/13 = -12/13
 B) cos &#946; = adj/hyp = x/r = {{{sqrt(5)}}}/3 

To do the others we have to use identities and then we can
substitute as above.  I will do C):

 C) cos 2&#946; = cos²&#946;-sin²&#946; = ({{{sqrt(5)}}}/3)²-(2/3)² = 5/9-4/9=1/9

Now you can do the others by yourself, using the proper identities and
the above graphs:

 D) sin &#945;/2
 E) cos (&#945;+&#946;)
 F) sin (&#945;+&#946;)/2 

Edwin</pre>