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 $A$?
 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 $A$.
A: Since $\cos A = \frac{8}{17}$, you can use the identity $$\cos 2A = 2 cos^2 A -1 = 2\big( \frac{8}{17} \big)^2 -1 = -\frac{161}{289}$$.
It makes no difference which identity you choose to use for $\cos 2A$.  Try all three and prove it to yourself.
A: You have that $\cos A = \frac{8}{17}$ and have deduced from this that $\sin A = \frac{15}{17}$. For part (a), as you have correctly stated, $$\sin 2A = 2\sin A \cos A = 2\left(\frac{15}{17}\times \frac{8}{17}\right) = \frac{240}{289}.$$
For part (b) you could use any one of the identites, I personally would use $$\cos2A = \cos^2A - \sin^2A,$$
but this is up to you. All that you must remember is that $$\cos^2A = \left( \cos A \right)^2,$$
and, of course, the same for the sine. Now you're all set.
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.
$$ c = 8/ 17 ;\ s  = 15/17 $$
$$ s_2 = 2 \cdot s \cdot c = =\frac {240}{289} = \pm 240/289 $$,
$$ c_2 =  c^2 -s^2 =  \frac {8^2-15^2}{289} = \pm \frac {-161}{289} .  $$
