Hi MHF, I need some help with 2 problems that I have been stuck with please

1) If $\displaystyle \displaystyle sin(\dfrac{\pi}{4} - \theta) = \dfrac {5}{13}$ and

0 <= theta <= pi/4 , find the value of $\displaystyle \displaystyle \dfrac {cos2\theta}{cos(\dfrac{\pi}{4} + \theta)} $.

2) In triangle ABC,the angle B is twice the size as compared to that of angle A.

Show that

a) $\displaystyle \displaystyle b = 2a cos A $

b) the area of triangle ABC = $\displaystyle \displaystyle a^2 cosA sin3A $

(a is the side opposite angle A and b....angle B .....)

For question 1 , I could not draw $\displaystyle \displaystyle sin(\dfrac{\pi}{4} - \theta) = \dfrac {5}{13}$ in terms of a triangle in the quadrant and hence I tried to manipulate it and got to $\displaystyle cos\theta - sin\theta = \dfrac{5(\sqrt2)}{13}$ (cheated a little by changing radians to degrees here) and got stuck. I am totally lost on this from here on and my approach is wrong too i guess,since we need to find the value of $\displaystyle cos2\theta $ .

For question 2, i tried to use the law of cosines (to find length of side) to see if i could derive anything but i am stuck with $\displaystyle 2b^2 = c^2 + a^2 cos2A $

Any help would be greatly appreciated.