Andrei Lazanu proved the double angle formulae
illustrated in the diagram but there were no
solutions to the second half of this question so
it becomes a Tough Nut.
The diagram starts from a right angled triangle,
of sides 2t and 2 and where consequently
tanq = t. In this triangle, a line making
an angle q with the hypotenuse is drawn.
This way, an isosceles triangle is formed, and
2q is the angle exterior to this isosceles
triangle. Let the sides DA and DB of this
isosceles triangle be x units. Then the length
of DC must be 2-x units. Using Pythagoras
Theorem for triangle ADC we find x.
x2=(2t)2+(2-x)2.
Hence x=1+t2 and so the length of side DC is
2-(1+t2)=1-t2.
The formulae for the sine, cosine and tangent of
2q in terms of t, where t=tanq,
follow directly from the ratios of the sides of
the right angled triangle ADC and we get
tan2q =
2t1-t2
, sin2q =
2t1+t2
, cos2q =
1-t21+t2
For the rest of the question there is a lot of
information leading you step by step to a
solution.
The conditions suggested in the question give you
two equations which you can solve to find p '
and q ' . Use the same formulation to give
(p ' ' ,q ' ' ).