Find the exact value of tan(15°) and prove sin(3x) = 3sin(x) − 4sin³(x).

$$= \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ)\tan(30^\circ)}$$ $$= \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}}$$ $$= \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$$ $$= \frac{\sqrt{3}-1}{\sqrt{3}+1}$$

$$= \frac{\sqrt{3}-1}{\sqrt{3}+1} \cdot \frac{\sqrt{3}-1}{\sqrt{3}-1}$$ $$= \frac{(\sqrt{3})^2 - 2\sqrt{3} + 1^2}{(\sqrt{3})^2 - 1^2}$$ $$= \frac{3 - 2\sqrt{3} + 1}{3 - 1}$$ $$= \frac{4 - 2\sqrt{3}}{2}$$ $$= 2 - \sqrt{3}$$

The final answer is $\boxed{2 - \sqrt{3}}$.

11. Prove that $\sin(3x) = 3\sin(x) - 4\sin^3(x)$.

Start with the left-hand side (LHS):

$$\sin(3x) = \sin(2x+x)$$

Apply the sum formula for sine: $\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$.

$$= \sin(2x)\cos(x) + \cos(2x)\sin(x)$$

Apply double angle formulas: $\sin(2x) = 2\sin(x)\cos(x)$ and $\cos(2x) = 1-2\sin^2(x)$.

$$= (2\sin(x)\cos(x))\cos(x) + (1-2\sin^2(x))\sin(x)$$ $$= 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^3(x)$$

Substitute $\cos^2(x) = 1-\sin^2(x)$.

$$= 2\sin(x)(1-\sin^2(x)) + \sin(x) - 2\sin^3(x)$$ $$= 2\sin(x) - 2\sin^3(x) + \sin(x) - 2\sin^3(x)$$

Combine like terms.

$$= 3\sin(x) - 4\sin^3(x)$$

This matches the right-hand side (RHS). The identity is proven.