Unsolved Challenge: Trigonometric Identity

[anemone]

Prove $\tan 3x=\tan \left(\dfrac{\pi}{3}-x\right) \tan x \tan \left(\dfrac{\pi}{3}+x\right)$ geometrically.

 

[jvicens]

Hello! Great question. To prove this statement geometrically, we can use the definition of tangent as the ratio of the opposite side to the adjacent side in a right triangle.

Let's start by drawing a right triangle with angle $3x$ and label the sides as shown in the diagram below:

[insert diagram of right triangle with labeled sides and angle 3x]

From the definition of tangent, we know that $\tan 3x = \dfrac{opposite}{adjacent}$. So, the length of the opposite side is $x\tan 3x$ and the length of the adjacent side is $x$.

Next, let's draw a line from the vertex of angle $3x$ to the opposite side, creating two smaller triangles. Label the angles and sides as shown in the diagram below:

[insert diagram of two smaller triangles created by drawing a line from vertex of angle 3x to opposite side]

From the definition of tangent, we know that $\tan \left(\dfrac{\pi}{3}-x\right) = \dfrac{opposite}{adjacent}$. So, the length of the opposite side in the smaller triangle on the left is $x\tan \left(\dfrac{\pi}{3}-x\right)$ and the length of the adjacent side is $x$. Similarly, in the smaller triangle on the right, the length of the opposite side is $x\tan \left(\dfrac{\pi}{3}+x\right)$ and the length of the adjacent side is $x$.

Now, let's look at the big triangle again. Using the Pythagorean theorem, we can find the length of the hypotenuse, which is $\sqrt{x^2 + (x\tan 3x)^2}$.

[insert diagram highlighting the hypotenuse and labeling the length]

Using the definition of tangent again, we know that $\tan 3x = \dfrac{opposite}{adjacent}$, so the length of the opposite side is $x\tan 3x$.

Now, let's look at the smaller triangles again. Using the Pythagorean theorem, we can find the length of the hypotenuse in each smaller triangle, which is $\sqrt{x^2 + (x\tan \left(\dfrac{\pi}{3}-x\right))^2}$ and $\sqrt{x^2 + (x\t