Double Angle Trigonometry Question

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!

Answer 1

Hint: $\sin(2\cdot 15°)=\sin(30°)=1/2$ and $\cos(2\cdot 15°)\cos(30°)=\sqrt{3}/2$, this can be easily computed using an equilateral triangle with edge of length $1$.

Answer 2

To simply this you need to know your identities.

As for tan15, the solution above seems to have covered that very well so I don't think I need to expand.

Answer 3

For a), you are asked to simplify $$ \frac{\sin2x}{1+\cos2x} $$ Just use the double angle formulas $$\sin(2x)=2\sin(x)\cos(x)$$ $$\cos(2x)=2\cos^2(x)-1$$ So....

Answer 4

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 $\dfrac{\sin(2x)}{1+\cos(2x)}$:

$\Rightarrow \tan(15)=\dfrac{\sin(2\times15)}{1+\cos(2\times15)}$

$\hspace{19.5 mm}=\dfrac{\sin(30)}{1+\cos(30)}$

$\hspace{19.5 mm}=\dfrac{\dfrac{1}{2}}{1+\dfrac{\sqrt{3}}{2}}$

$\hspace{19.5 mm}=\dfrac{\dfrac{1}{2}}{\dfrac{2+\sqrt{3}}{2}}$

$\hspace{19.5 mm}=\dfrac{1}{2+\sqrt{3}}$

$\hspace{19.5 mm}=\dfrac{1}{2+\sqrt{3}}\times\dfrac{2-\sqrt{3}}{2-\sqrt{3}}$

$\hspace{19.5 mm}=\dfrac{2-\sqrt{3}}{2^{2}-(\sqrt{3})^{2}}$

$\hspace{19.5 mm}=\dfrac{2-\sqrt{3}}{4-3}$

$\hspace{19.5 mm}=\dfrac{2-\sqrt{3}}{1}$

$\hspace{19.5 mm}=2-\sqrt{3}$