Trigonometry 1

(a)
the coordinates of P is
(cos θ, sin θ)
(b)
Let the equations of the tangents to the circle from the point (2,2) are
y-2=m(x-2)
y=mx+2(1-m)

sub into x^2+y^2=1
x^2+[mx+2(1-m)]^2=1
x^2+m^2x^2+4mx(1-m)+4(1-m)^2=1
(1+m^2)x^2+4m(1-m)x+4(1-m)^2-1=0
Since discriminant = 0
16m^2(1-m)^2-4(1+m^2)[4(1-m)^2-1]=0
4m^2(1-m)^2-(1+m^2)[4(1-m)^2-1]=0
4m^2(1-m)^2-4(1+m^2)(1-m)^2=-(1+m^2)

-4(1-m)^2=-1-m^2
3m^2-8m+3=0
m=1/6(8+ √28) or 1/6(8- √28)

m=(4+ √7)/3 or (4- √7)/3
the slopes of AB is (4+ √7)/3
the slopes of AC is (4- √7)/3
(c)

Notice that 2-sin θ is the vertical distance between A and P
Also 2-cos θ is the horizontal distance between A and P
So 2-sin θ/2-cos θ is the slope of the line AP
But the line AP is between the lines AB and AC
So the maximum value of 2-sin θ/2-cos θ is (4+ √7)/3

the minimum value of 2-sin θ/2-cos θ is (4- √7)/3