## Precalculus (6th Edition) Blitzer

Some of the identities that are verified using sum-to product formulas are: Verified identities are: \begin{align} & \tan x=\frac{\cos 3x-\cos 5x}{\sin 3x+\sin 5x} \\ & -\tan x=\frac{\cos 3x-\cos x}{\sin 3x+\sin x} \\ & -\cot 2x=\frac{\sin 3x-\sin x}{\cos 3x-\cos x} \\ & \tan 3x=\frac{\sin 2x+\sin 4x}{\cos 2x+\cos 4x} \end{align} Other identities are: \begin{align} & \frac{\sin x-\sin y}{\sin x+\sin y}=\tan \frac{x-y}{2}\cot \frac{x+y}{2} \\ & \frac{\sin x+\sin y}{\sin x-\sin y}=\tan \frac{x+y}{2}\cot \frac{x-y}{2} \\ & \frac{\sin x-\sin y}{\cos x-\cos y}=-\cot \frac{x+y}{2} \end{align} And, \begin{align} & \frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\tan 2x \\ & \frac{\cos 4x-\cos 2x}{\sin 2x-\sin 4x}=\tan 3x \end{align} Sum-to product formulas are given below: \begin{align} & \sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2} \\ & \sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2} \\ & \cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2} \\ & \cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2} \end{align}
Let us consider the identity: $\frac{\cos 3x-\cos 5x}{\sin 3x+\sin 5x}=\tan x$ By using the sum-to product formula $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$ to numerator and sum-to product formula $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side. Then, \begin{align} & \frac{\cos 3x-\cos 5x}{\sin 3x+\sin 5x}=\frac{-2\sin \frac{3x+5x}{2}\sin \frac{3x-5x}{2}}{2\sin \frac{3x+5x}{2}\cos \frac{3x-5x}{2}} \\ & =\frac{-2\sin \frac{8x}{2}\sin \frac{-2x}{2}}{2\sin \frac{8x}{2}\cos \frac{-2x}{2}} \\ & =\frac{-2\sin 4x\sin \left( -x \right)}{2\sin 4x\cos \left( -x \right)} \\ & =-\tan \left( -x \right) \end{align} Thus, the result will be $\tan x$. Hence, verified. Also, consider the identity: $\frac{\cos 3x-\cos x}{\sin 3x+\sin x}=-\tan x$ And use the sum-to product formula $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$ to numerator and sum-to product formula $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side. Then, \begin{align} & \frac{\cos 3x-\cos x}{\sin 3x+\sin x}=\frac{-2\sin \frac{3x+x}{2}\sin \frac{3x-x}{2}}{2\sin \frac{3x+x}{2}\cos \frac{3x-x}{2}} \\ & =\frac{-2\sin \frac{4x}{2}\sin \frac{2x}{2}}{2\sin \frac{4x}{2}\cos \frac{2x}{2}} \\ & =\frac{-2\sin 2x\sin x}{2\sin 2x\cos x} \\ & =-\tan x \end{align} Hence verified. Also, consider the identity: $\frac{\sin 3x-\sin x}{\cos 3x-\cos x}=-\cot 2x$ By using the sum-to product formula $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$ to numerator and $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$ to denominator of left hand side. Then, \begin{align} & \frac{\sin 3x-\sin x}{\cos 3x-\cos x}=\frac{2\sin \frac{3x-x}{2}\cos \frac{3x+x}{2}}{-2\sin \frac{3x+x}{2}\sin \frac{3x-x}{2}} \\ & =\frac{2\sin \frac{2x}{2}\cos \frac{4x}{2}}{-2\sin \frac{4x}{2}\sin \frac{2x}{2}} \\ & =-\frac{\cos 2x}{\sin 2x} \\ & =-\cot 2x \end{align} Hence verified. And also consider the identity: $\frac{\sin x-\sin y}{\sin x+\sin y}=\tan \frac{x-y}{2}\cot \frac{x+y}{2}$ Also, use the sum-to product formula $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$ to numerator and $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side. Then, \begin{align} & \frac{\sin x-\sin y}{\sin x+\sin y}=\frac{2\sin \frac{x-y}{2}\cos \frac{x+y}{2}}{2\sin \frac{x+y}{2}\cos \frac{x-y}{2}} \\ & =\tan \frac{x-y}{2}\cot \frac{x+y}{2} \end{align} Hence verified. Also, consider the identity: $\frac{\sin x+\sin y}{\sin x-\sin y}=\tan \frac{x+y}{2}\cot \frac{x-y}{2}$ And use the sum-to product formula $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$ to denominator and $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to numerator of left hand side. Then, \begin{align} & \frac{\sin x+\sin y}{\sin x-\sin y}=\frac{2\sin \frac{x+y}{2}\cos \frac{x-y}{2}}{2\sin \frac{x-y}{2}\cos \frac{x+y}{2}} \\ & =\tan \frac{x+y}{2}\cot \frac{x-y}{2} \end{align} Hence verified. And consider the identity: $\frac{\sin x+\sin y}{\cos x+\cos y}=\tan \frac{x+y}{2}$ Then use the sum-to product formula $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to numerator and $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side. Then, \begin{align} & \frac{\sin x+\sin y}{\cos x+\cos y}=\frac{2\sin \frac{x+y}{2}\cos \frac{x-y}{2}}{2\cos \frac{x+y}{2}\cos \frac{x-y}{2}} \\ & =\tan \frac{x+y}{2} \end{align} Hence verified. Also, consider the identity: $\frac{\sin x-\sin y}{\cos x-\cos y}=-\cot \frac{x+y}{2}$ And use the sum-to product formula $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$ to numerator and $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$ to denominator of left hand side. Then, \begin{align} & \frac{\sin x-\sin y}{\cos x-\cos y}=\frac{2\sin \frac{x-y}{2}\cos \frac{x+y}{2}}{-2\sin \frac{x+y}{2}\sin \frac{x-y}{2}} \\ & =-\cot \frac{x+y}{2} \end{align} Hence verified. Also, consider the identity; $\frac{\sin 2x+\sin 4x}{\cos 2x+\cos 4x}=\tan 3x$ And use the sum-to product formula $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to numerator and $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side. Then, \begin{align} & \frac{\sin 2x+\sin 4x}{\cos 2x+\cos 4x}=\frac{2\sin \frac{2x+4x}{2}\cos \frac{2x-4x}{2}}{2\cos \frac{2x+4x}{2}\cos \frac{2x-4x}{2}} \\ & =\frac{2\sin \frac{6x}{2}\cos \frac{-2x}{2}}{2\cos \frac{6x}{2}\cos \frac{-2x}{2}} \\ & =\frac{\sin 3x}{\cos 3x} \\ & =\tan 3x \end{align} Hence verified. Also, consider the identity: $\frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\tan 2x$ And use the sum-to product formula $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to numerator and $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side. Then, \begin{align} & \frac{\sin x+\sin 3x}{\cos 3x+\cos 3x}=\frac{2\sin \frac{x+3x}{2}\cos \frac{x-3x}{2}}{2\cos \frac{x+3x}{2}\cos \frac{x-3x}{2}} \\ & =\frac{2\sin \frac{4x}{2}\cos \frac{-2x}{2}}{2\cos \frac{4x}{2}\cos \frac{-2x}{2}} \\ & =\frac{\sin 2x}{\cos 2x} \\ & =\tan 2x \end{align} Hence verified. Also, consider the identity: $\frac{\cos 4x-\cos 2x}{\sin 2x-\sin 4x}=\tan 3x$ And use the sum-to product formula $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$ to denominator and $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$ to numerator of left hand side. Then, \begin{align} & \frac{\cos 4x-\cos 2x}{\sin 2x-\sin 4x}=\frac{-2\sin \frac{4x+2x}{2}\sin \frac{4x-2x}{2}}{2\sin \frac{2x-4x}{2}\cos \frac{4x+2x}{2}} \\ & =\frac{-2\sin \frac{6x}{2}\sin \frac{2x}{2}}{2\sin \frac{-2x}{2}\cos \frac{6x}{2}} \\ & =\frac{\sin 3x}{\cos 3x} \\ & =\tan 3x \end{align} Hence, verified.