Question 1175232
<pre>
We have a conflict of notation.  We can't call the x-axis if we call the angle
x also.  So I will call the angle capital X and the x-axis by small x.</pre>
If sec X = 3 and the terminal side x does not lie on the first quadrant, find the
other circular function values of x.<pre>

We will consider 3 as the fraction 3/1.

Since the secant is positive in the first and fourth quadrant and it does not
lie in the first quadrant, then it lies in the fourth quadrant.

We draw a right triangle with an angle in Q4, and remember that the secant
is the hypotenuse over the adjacent (or the r over the x). So we put the
numerator of 3/1, which is 3, on the hypotenuse (or the r) and the denominator
of 3/1, which is 1, on the adjacent side, (or the x). Then we calculate the
opposite side (or the y) by using the Pythagorean theorem:

{{{matrix(7,1,
x^2+y^2=r^2,
1^2+y^2=3^2,
1+y^2=9,
y^2=8,
y="" +- sqrt(8),
y="" +- sqrt(4*2),
y="" +- 2sqrt(2)

)}}}
                                                       _
Since it's Q4 we know that y is negative, so we put -2√2 on the opposite
side (y):

{{{drawing(300,200,-1,1.5,-1,1,line(-2,0,2,0), line(0,-2,0,2),
green(line(3/5,-4/5,3/5,0)),
locate(.2,.2,x=1),locate(.6,-.3,y=-2sqrt(2)),locate(.06,-.4,r=3),
line(0,0,3/5,-4/5)  )}}}

So now we know that

{{{sin(X)=opposite/hypotenuse = y/r=(-2sqrt(2))/3}}}
{{{cos(X)=hypotenuse/opposite = x/r=1/3}}}
{{{tan(X)=hypotenuse/opposite = y/x=(-2sqrt(2))/1=-2sqrt(2)}}}
{{{sec(X)=hypotenuse/opposite = r/x=given=3}}}
{{{csc(X)=hypotenuse/opposite = r/y=3/(-2sqrt(2))=(3/(-2sqrt(2))) 
  *((sqrt(2))/(sqrt(2)))=-3sqrt(2)/(2*2)=-3sqrt(2)/4}}}
{{{cot(X)=adjacent/opposite = x/y=1/(-2sqrt(2))=(1/(-2sqrt(2))) 
  *((sqrt(2))/(sqrt(2)))=-sqrt(2)/(2*2)=-sqrt(2)/4}}}



Edwin</pre>