Precalculus (6th Edition) Blitzer

One of the sum-to-product formulas is $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$. Therefore, $\sin 2x+\sin 4x$ can be written as given below: $\sin 2x+\sin 4x=2\sin \frac{2x+4x}{2}\cos \frac{2x-4x}{2}$ One of the sum-to-product formulas is $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$. Therefore, $\cos 2x+\cos 4x$ can be written as provided below: $\cos 2x+\cos 4x=2\cos \frac{2x+4x}{2}\cos \frac{2x-4x}{2}$ Now, consider the left side of the provided expression: $\frac{\sin 2x+\sin 4x}{\cos 2x+\cos 4x}$ The expression can be simplified as provided below: \begin{align} & \frac{\sin 2x+\sin 4x}{\cos 2x+\cos 4x}=\frac{2\sin \left( \frac{2x+4x}{2} \right)\cos \left( \frac{2x-4x}{2} \right)}{2\cos \left( \frac{2x+4x}{2} \right)\cos \left( \frac{2x-4x}{2} \right)} \\ & =\frac{2\sin \left( \frac{6x}{2} \right)\cos \left( \frac{-2x}{2} \right)}{2\cos \left( \frac{6x}{2} \right)\cos \left( \frac{-2x}{2} \right)} \\ & =\frac{2\sin 3x\cos \left( -x \right)}{2\cos 3x\cos \left( -x \right)} \\ & =\frac{\sin 3x}{\cos 3x} \end{align} Now, by using one of the quotient identities of trigonometry, which is $\tan x=\frac{\sin x}{\cos x}$, the expression can be further written as: $\frac{\sin 3x}{\cos 3x}=\tan 3x$ Hence, the left side of the expression is equal to the right side, which is $\frac{\sin 2x+\sin 4x}{\cos 2x+\cos 4x}=\tan 3x$