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

2 t

and 2 and where consequently

\tan θ = t

. In this triangle, a line making an angle

θ

with the hypotenuse is drawn. This way, an isosceles triangle is formed, and

2 θ

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

x^{2} = (2 t)^{2} + (2 - x)^{2} .

Hence

x = 1 + t^{2}

and so the length of side DC is

2 - (1 + t^{2}) = 1 - t^{2}

.

The formulae for the sine, cosine and tangent of

2 θ

in terms of

t

, where

t = \tan θ

, follow directly from the ratios of the sides of the right angled triangle ADC and we get

\tan 2 θ = \frac{2 t}{1 - t^{2}}, \sin 2 θ = \frac{2 t}{1 + t^{2}}, \cos 2 θ = \frac{1 - t^{2}}{1 + t^{2}}

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'')