Question

What is the formula of $\tan 2x$?

Answer 1

Determine the formula of $\tan 2x$.

It is known that tangent function is the ratio of the sine function and cosine function,

$\tan 2x=\frac{\sin 2x}{\cos 2x}$

Apply the trigonometric formulas, $\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{sin}\mathbf{}\mathbf{xcos}\mathbf{}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{=}{\mathbf{cos}}^{\mathbf{2}}\mathbf{x}\mathbf{-}{\mathbf{sin}}^{\mathbf{2}}\mathbf{x}$:

$\tan 2x=\frac{2\sin x\cos x}{{\cos }^{2}x-{\sin }^{2}x}$

Divide the numerator and denominator of the right side by ${\cos }^{2}x$:

$\begin{array}{rcl}\tan 2x& =& \frac{{\displaystyle \frac{2\sin x\cos x}{{\cos }^{2}x}}}{{\displaystyle \frac{{\cos }^{2}x-{\sin }^{2}x}{{\cos }^{2}x}}}\\ & \Rightarrow & \tan 2x=\frac{{\displaystyle \frac{2\sin x}{\cos x}}}{1-{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}}\\ & \Rightarrow & \tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}\end{array}$

Therefore, the formula is $\begin{array}{rcl}\mathbf{tan}\mathbf{}\mathbf{2}\mathbf{x}& \mathbf{=}& \frac{\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{x}}{\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}}\mathbf{=}\frac{\mathbf{2}\mathbf{tan}\mathbf{}\mathbf{x}}{\mathbf{1}\mathbf{-}{\mathbf{tan}}^{\mathbf{2}}\mathbf{x}}\end{array}$.