Question 1138000
<pre>
The formula is {{{matrix(1,3,cos(alpha-beta),""="",cos(alpha)cos(beta)+sin(alpha)sin(beta))}}}.

Draw angle α in Quadrant I:

Since sine = {{{y/r}}}, we make y=4 and r=5, so that the sin(α)
will be {{{y/r}}}{{{""=""}}}{{{4/5}}}.

{{{drawing(200,1950/11,-3.9,3.9,-3.9,4.9,grid(1),
red(locate(1.1,1.2,alpha) ),
locate(3.1,2,y=4), locate(.4,2.2,r=5),
line(3,0,3,4),line(0,0,3,0),
line(0,0,3,4), locate(.6,0,x=3),
red(arc(0,0,2,-2,0,54)) )}}}

For the formula, we need sine and cosine, and the cosine is {{{x/r}}}

So we find x by the Pythagorean relation: 

{{{x^2+y^2=r^2}}}
{{{x^2+(4)^2=(5)^2}}}
{{{x^2+16=25}}}
{{{x^2=9}}}
{{{x = "" +- 3}}}

Since x goes to the right, we know to take the positive
value {{{x = ""+3}}}.  So now we know that {{{cos(alpha)=x/r=3/5}}}

Next we draw angle β in Quadrant II:

Since cosine = {{{x/r}}}, we make x=-5 and r=13, so that the cos(β)
will be {{{x/r}}}{{{""=""}}}{{{(-5)/13}}}.

{{{drawing(9800/47,400,-5.9,3.9,-5.9,12.9,graph(9800/47,400,-5.9,3.9,-5.9,12.9),red(locate(1,1.3,beta) ),
locate(-5.8,6,y=12), locate(-2.8,7.3,r=13),
line(-5,0,0,0),line(-5,0,-5,12),
line(0,0,-5,12), locate(-3.5,1,x=-5), red(arc(0,0,3,-3,0,113)) )}}}

For the formula, we need sine and cosine, and the sine is {{{y/r}}}

So we find x by the Pythagorean relation: 

{{{x^2+y^2=r^2}}}
{{{(-5)^2+y^2=(13)^2}}}
{{{25+y^2=169}}}
{{{y^2=144}}}
{{{y = "" +- 12}}}

Since y goes up from the x-axis, we know to take the positive
value {{{y = ""+12}}}.  So now we know that {{{sin(beta)=y/r=12/13}}}

Now we use the formula

{{{matrix(1,3,cos(alpha-beta),""="",cos(alpha)cos(beta)+sin(alpha)sin(beta))}}}.

{{{matrix(1,3,cos(alpha-beta),""="",(3/5)(-5/13)+(4/5)(12/13))}}}

{{{matrix(1,3,cos(alpha-beta),""="",(-15/65)+(48/65))}}}

{{{matrix(1,3,cos(alpha-beta),""="",33/65)}}}

Edwin</pre>