# Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 237: 56

$\dfrac{\cot^2{x}-1}{2\cot{x}}=\cot{2x}$

#### Work Step by Step

Use a graphing utility to graph the given expression. (Refer to the graph below,) Notice that the graph is identical with the graph of $\cot{2x}$. This means that $\dfrac{\cot^2{x}-1}{2\cot{x}}=\cot{(2x)}$. 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{\cot^2{x}-1}{2\cot{x}}&=\dfrac{\frac{\cos^2{x}}{\sin^2{x}}-1}{2\left(\frac{\cos{x}}{\sin{x}}\right)}\\\\ &=\dfrac{\frac{\cos^2{x}}{\sin^2{x}}-\frac{\sin^2{x}}{\sin^2{x}}}{\frac{2\cos{x}}{\sin{x}}}\\\\ &=\dfrac{\frac{\cos^2{x}-\sin^2{x}}{\sin^2{x}}}{\frac{2\cos{x}}{\sin{x}}}\\\\ &=\frac{\cos^2{x}-\sin^2{x}}{\sin^2{x}}\cdot \frac{\sin{x}}{2\cos{x}}\\\\ &=\frac{\cos^2{x}-\sin^2{x}}{\sin^\cancel{2}{x}}\cdot \frac{\cancel{\sin{x}}}{2\cos{x}}\\\\ &=\frac{\cos^2{x}-\sin^2{x}}{\sin{x}}\cdot \frac{1}{2\cos{x}}\\\\ &=\frac{\cos^2{x}-\sin^2{x}}{2\sin{x}\cos{x}}\\\\ \end{align*} Recall: (1) $\cos^2{A}-\sin^2{A}=\cos{(2A)}\\$ (2) $2\sin{A}\cos{A}=\sin{(2A)}$ Use the rules above to obtain: \begin{align*} \require{cancel} \dfrac{\cot^2{x}-1}{2\cot{x}}&=\frac{\cos^2{x}-\sin^2{x}}{2\sin{x}\cos{x}}\\\\ &=\frac{\cos{2x}}{\sin{2x}}\\\\ &=\cot{2x} \end{align*}

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