Question 1074693
1) Line {{{y+2x=0}}}
{{{drawing(400,400,-1.1,1.3,-1.2,1.2,grid(0),
red(circle(0,0,1)),locate(0.35,-0.6,blue(y+2x=0)),
blue(line(-1,2,1,-2)),arrow(0,0,1.2,0),
arrow(0,0,0.577,-1.155),circle(0.447,-.894,0.02),
arc(0,0,0.6,0.6,63.4,360),locate(0.23,0.25,theta),
locate(0.465,-0.89,P)
)}}} Maybe your teacher instructor defined trigonometric functions
based on the coordinates of {{{P(x[P],y[P])}}} , with {{{system(cos(theta)=x[P],sin(theta)=y[P],tan(theta)=y[P]/x[P])}}} .
My 9th grade math teacher defined {{{than(theta)=y[Q]}}} {{{drawing(280,420,-1.1,1.3,-2.3,1.3,grid(0),
red(circle(0,0,1)),locate(0.35,-0.6,blue(y+2x=0)),
blue(line(-1,2,1.5,-3)),arrow(0,0,1.2,0),
arrow(0,0,0.577,-1.155),blue(triangle(0,0,1,0,1,-2)),
blue(rectangle(1,0,0.9,-0.1)),
arc(0,0,0.6,0.6,63.4,360),locate(0.23,0.25,theta),
locate(1.05,-1.95,Q),circle(1,-2,0.03)
)}}} .
Either way it is {{{tan(theta)=y[R]/x[R]}}} for some point {{{R(x[R],y[R])}}}
on that blue line that passes through the origin {{{O(0,0)}}} ,
{{{y+2x=0}}} <---> {{{y=-2x}}} .
Of course that {{{tan(theta)=y[R]/x[R]=(y[R]-0)/(x[R]-0)}}} is the slope of the line,
and the slope of {{{y=-2x}}} is {{{highlight(-2))}}} ,
the coefficient of {{{x}}}
in the slope-intercept form of the equation of the line.
 
1) Line {{{-a+3b=0}}}
would have to be graphed on a pait if a-b axes.
I would use the usual x-y axes and say that {{{system (a=x,b=y)}}} .
Then, the line would be {{{-x+3y=0}}} <--> {{{3y=x}}} <--> {{{y=(1/3)x}}} .
{{{tan(theta)=1/3}}}
If {{{sin(theta)>0}}} ,
{{{sin(theta)=sqrt(1-cos^2(theta))}}} and
since {{{tan(theta)=sin(theta)/cos(theta)}}} ,
{{{sqrt(1-cos^2(theta))/cos(theta)=1/3}}}
{{{sqrt(1-cos^2(theta))=cos(theta)/3}}}
Squaring both sides,
{{{1-cos^2(theta)=cos^2(theta)/9}}}
{{{9-9cos^2(theta)=cos^2(theta)}}}
{{{9=10cos^2(theta)}}}
{{{9/10=cos^2(theta)}}}
{{{sin(theta)=sqrt(1-9/10)}}}
{{{sin(theta)=sqrt(1/10)}}}
and
{{{highlight(sin(theta)=sqrt(10)/10)}}}