## Trigonometry (11th Edition) Clone

$\dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}=\sin{2x}\\\\$
Use a graphing utility to graph the given expression. (Refer to the graph below,) Notice that the graph is identical with the graph of $\sin{2x}$. This means that $\dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}=\sin{(2x)}$. Work on the left side of the equation above by writing each function using $\sin{x}$ and $\cos{x}$ only: \begin{align*} \require{cancel} \dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}&=\dfrac{4\left(\frac{\sin{x}}{\cos{x}}(\cos^2{x})\right)-2\left(\frac{\sin{x}}{\cos{x}}\right)}{1-\left(\frac{\sin{x}}{\cos{x}}\right)^2}\\\\ &=\dfrac{4\sin{x}\cos{x}-\left(\frac{2\sin{x}}{\cos{x}}\right)}{1-\frac{\sin^2{x}}{\cos^2{x}}}\\\\ &=\dfrac{\frac{(4\sin{x}\cos{x})\cdot \cos{x}}{\cos{x}}-\frac{2\sin{x}}{\cos{x}}}{\frac{\cos^2{x}}{\cos^2{x}}-\frac{\sin^2{x}}{\cos^2{x}}}\\\\ &=\dfrac{\frac{4\sin{x}\cos^2{x}}{\cos{x}}-\frac{2\sin{x}}{\cos{x}}}{\frac{\cos^2{x}-\sin^2{x}}{\cos^2{x}}}\\\\ &=\dfrac{\frac{4\sin{x}\cos^2{x}-2\sin{x}}{\cos{x}}}{\frac{\cos^2{x}-\sin^2{x}}{\cos^2{x}}}\\\\ &=\dfrac{4\sin{x}\cos^2{x}-2\sin{x}}{\cos{x}} \cdot \frac{\cos^2{x}}{\cos^2{x}-\sin^2{x}}\\\\ &=\dfrac{4\sin{x}\cos^2{x}-2\sin{x}}{\cancel{\cos{x}}} \cdot \frac{\cos^\cancel{2}{x}}{\cos^2{x}-\sin^2{x}}\\\\ &=\left(4\sin{x}\cos^2{x}-2\sin{x}\right) \cdot \frac{\cos{x}}{\cos^2{x}-\sin^2{x}}\\\\ \end{align*} Factor out $(2\sin{x})$ to obtain: \begin{align*} \dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}&=2\sin{x}\left(2\cos^2{x}-1\right) \cdot \frac{\cos{x}}{\cos^2{x}-\sin^2{x}}\\\\ \end{align*} Recall: (1) $2\cos^2{A}-1=\cos{(2A)}\\$ (2) $\cos^2{A}-\sin^2{A}=\cos{(2A)}$ Use the rules above to obtain: \begin{align*} \require{cancel} \dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}&=2\sin{x}\left(\cos{2x}\right) \cdot \frac{\cos{x}}{\cos{2x}}\\\\ \dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}&=2\sin{x}\left(\cancel{\cos{2x}}\right) \cdot \frac{\cos{x}}{\cancel{\cos{2x}}}\\\\ \dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}&=2\sin{x}\cos{x}\\\\ \end{align*} Recall: $\sin{2x}=2\sin{x}\cos{x}$ Thus, using the rule above gives: $$\dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}=\sin{2x}\\\\$$