<s>[INST] Double Angle Trigonometry Question So there is this question which consists of 2 parts.
$$
a) \text{ Simplify } \frac{\sin2x}{1+\cos2x} \\
b) \text{ Hence, find the exact value of tan 15.}
$$
So far I've discovered that $ \text{a)} \tan x $
But I have no idea how to begin on part $b$, although I'm guessing the answer's correlated with a specific part of the working for part $a$. Can someone help me? Thanks in advance!
 [/INST] We have: $\dfrac{\sin(2x)}{1+\cos(2x)}$
$=\hspace{12 mm}\dfrac{2\sin(x)\cos(x)}{1+2\cos^{2}(x)-1}$
$=\hspace{12 mm}\dfrac{2\sin(x)\cos(x)}{2\cos^{2}(x)}$
$=\hspace{12 mm}\dfrac{\sin(x)}{\cos(x)}$
$=\hspace{12 mm}\tan(x)$
Then, we want to evaluate $\tan(15)$.
We can do this using the original expression $\dfrac{\sin(2x)}{1+\cos(2x)}$:
$\Rightarrow \tan(15)=\dfrac{\sin(2\times15)}{1+\cos(2\times15)}$
$\hspace{19.5 mm}=\dfrac{\sin(30)}{1+\cos(30)}$
$\hspace{19.5 mm}=\dfrac{\dfrac{1}{2}}{1+\dfrac{\sqrt{3}}{2}}$
$\hspace{19.5 mm}=\dfrac{\dfrac{1}{2}}{\dfrac{2+\sqrt{3}}{2}}$
$\hspace{19.5 mm}=\dfrac{1}{2+\sqrt{3}}$
$\hspace{19.5 mm}=\dfrac{1}{2+\sqrt{3}}\times\dfrac{2-\sqrt{3}}{2-\sqrt{3}}$
$\hspace{19.5 mm}=\dfrac{2-\sqrt{3}}{2^{2}-(\sqrt{3})^{2}}$
$\hspace{19.5 mm}=\dfrac{2-\sqrt{3}}{4-3}$
$\hspace{19.5 mm}=\dfrac{2-\sqrt{3}}{1}$
$\hspace{19.5 mm}=2-\sqrt{3}$
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