Find angle using the sine and cosine formulas Given that $\cos A = \frac{8}{17}$ find

a) $\sin 2A$

b) $\cos 2A$

Using $\cos A$ I can tell that $\sin A = \frac{15}{17}$

a) $\sin 2A = 2\sin A\cos A = \frac{240}{289}$

Is this correct?

b) $\cos 2A$

For this part, there are 3 formulas for expanding $\cos 2A$, is there any one I should favour over the others?

Say I chose $1 - 2\sin^2 A$, how could I find $A$?
For a) 

No, it is not correct. If $\cos A=8/17$, then from $\cos^2A+\sin^2A=1$
$$\sin A=\color{red}{\pm}\sqrt{1-\left(\frac{8}{17}\right)^2}=\color{red}{\pm}\frac{15}{17}$$
and so
$$\sin 2A=2\sin A\cos A=\color{red}{\pm}\frac{240}{289}$$

For b)

You can use whichever you like :
$$\cos 2A=2\cos^2A-1=1-2\sin^2A=\cos^2A-\sin^2A$$
By the way, note that $\cos^2A=(\cos A)^2$ and that $\sin^2A=(\sin A)^2$ : you don't need to find $A$.
Draw a right angled triangle. This example uses Pythagorean triplets so integral sides occur.Only positive sign of $cos$ considered here. For negative sign change sign.

$$ c = 8/ 17 ;\ s  = 15/17 $$

$$ s_2 = 2 \cdot s \cdot c = =\frac {240}{289} = \pm 240/289 $$,

$$ c_2 =  c^2 -s^2 =  \frac {8^2-15^2}{289} = \pm \frac {-161}{289} .  $$