Question 141788
The point (0,-1) on the unit circle is rotated 315 degrees counter-clockwise.Determine the coordinates of the new point.

<pre><font size = 4 color = "indigo"><b>
{{{drawing(400,400,-1.2,1.2,-1.2,1.2, graph(400,400,-1.2,1.2,-1.2,1.2,0,-(sqrt(-x-.2)/sqrt(-x-.2))*sqrt(.1-x^2)),locate(.05,-1,"(0,-1)"),
circle(0,0,1), line(0,0,-sqrt(2)/2,-sqrt(2)/2), locate(-1,-.7,"(?,?)") )}}}

Now we subtract 315°-270° and find that the angle between the slanted
radius and the left side of the x-axis is 45° (indicated by the green
curved line, so we label it 45°:

{{{drawing(400,400,-1.2,1.2,-1.2,1.2, graph(400,400,-1.2,1.2,-1.2,1.2,0,-(sqrt(-x-.2)/sqrt(-x-.2))*sqrt(.1-x^2)),locate(.05,-1,"(0,-1)"), locate(-.25,-.04,"45°"), locate(-.4,-.4,1),
circle(0,0,1), line(0,0,-sqrt(2)/2,-sqrt(2)/2), locate(-1,-.7,"(?,?)") )}}}

Now from the point in question we draw a perpendicular up to the
x-axis:

{{{drawing(400,400,-1.2,1.2,-1.2,1.2, graph(400,400,-1.2,1.2,-1.2,1.2,0,-(sqrt(-x-.2)/sqrt(-x-.2))*sqrt(.1-x^2)),locate(.05,-1,"(0,-1)"), locate(-.25,-.04,"45°"), locate(-.4,-.4,1), line(-sqrt(2)/2,-sqrt(2)/2,-sqrt(2)/2,0),
circle(0,0,1), line(0,0,-sqrt(2)/2,-sqrt(2)/2), locate(-1,-.7,"(?,?)") )}}}

This forms a special right triangle.  

We are supposed to know that if a right triangle has an acute angle 
of 45°, then it is isosceles and the sides of the 45° angle are equal.
The hypotenuse is the radius 1.

We can calculate the two equal sides of that right triangle by
the Pythagorean theorem:

{{{a^2+b^2=c^2}}}

Since we know that a = b we can substitute a for b: 

{{{a^2+a^2=1^2}}}

{{{2a^2=1}}}

{{{a^2=1/2}}}

{{{a=sqrt(1/2)}}}

{{{a=sqrt((1*2)/(2*2))}}}

{{{a=sqrt(2)/2}}} 

So the vertical and horizontal sides of that right triangle 
are both equal to {{{sqrt(2)/2}}}.

{{{drawing(400,400,-1.2,1.2,-1.2,1.2, graph(400,400,-1.2,1.2,-1.2,1.2,0,-(sqrt(-x-.2)/sqrt(-x-.2))*sqrt(.1-x^2)),locate(.05,-1,"(0,-1)"), locate(-.25,-.04,"45°"), locate(-.4,-.4,1), line(-sqrt(2)/2,-sqrt(2)/2,-sqrt(2)/2,0),
circle(0,0,1), line(0,0,-sqrt(2)/2,-sqrt(2)/2), locate(-1,-.7,"(?,?)"),
locate(-.67,-.25,sqrt(2)/2),locate(-.5,.25,sqrt(2)/2)

 )}}}

The point is in the 3rd quadrant, so both its coordinates are
negative, so the coordinates of the point is ({{{-sqrt(2)/2}}},{{{-sqrt(2)/2}}}).

{{{drawing(400,400,-1.2,1.2,-1.2,1.2, graph(400,400,-1.2,1.2,-1.2,1.2,0,-(sqrt(-x-.2)/sqrt(-x-.2))*sqrt(.1-x^2)),locate(.05,-1,"(0,-1)"), locate(-.25,-.04,"45°"), locate(-.4,-.4,1), line(-sqrt(2)/2,-sqrt(2)/2,-sqrt(2)/2,0),
circle(0,0,1), line(0,0,-sqrt(2)/2,-sqrt(2)/2), 
locate(-.67,-.25,sqrt(2)/2),locate(-.5,.25,sqrt(2)/2),

locate(-1,-.88,"("),locate(-.95,-.8,-sqrt(2)/2),locate(-.77,-.9,","),
locate(-.73,-.8,-sqrt(2)/2), locate(-.56,-.88,")") 

 )}}}

Edwin</pre>