What is `(cos7x-cos3x)/(sin7x-2sin5x+sin3x)` equal to?

Options

A. tan x
B. cot x
C. tan 2x
D. cot 2x

Answer

To solve the expression \((\cos 7x - \cos 3x) / (\sin 7x - 2\sin 5x + \sin 3x)\), we will use trigonometric identities to simplify both the numerator and the denominator.

### Step 1: Simplifying the Numerator
We will use the cosine difference identity:
\[ \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]
For our case, let \(A = 7x\) and \(B = 3x\):
\[ \cos 7x - \cos 3x = -2 \sin\left(\frac{7x + 3x}{2}\right) \sin\left(\frac{7x - 3x}{2}\right) \]
Calculating the averages:
\[ = -2 \sin(5x) \sin(2x) \]

### Step 2: Simplifying the Denominator
Now, we simplify the denominator \(\sin 7x - 2\sin 5x + \sin 3x\). We can use the sine addition formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
We will group \(\sin 7x\) and \(\sin 3x\):
\[ \sin 7x + \sin 3x = 2 \sin\left(\frac{7x + 3x}{2}\right) \cos\left(\frac{7x - 3x}{2}\right) = 2 \sin(5x) \cos(2x) \]
Thus, the denominator becomes:
\[ \sin 7x - 2\sin 5x + \sin 3x = 2 \sin(5x) \cos(2x) - 2\sin(5x) \]
Factoring out \(2\sin(5x)\):
\[ = 2\sin(5x)(\cos(2x) - 1) \]

### Step 3: Putting it All Together
Now we can substitute our simplified numerator and denominator back into the expression:
\[ \frac{-2 \sin(5x) \sin(2x)}{2\sin(5x)(\cos(2x) - 1)} \]
We can cancel \(2\sin(5x)\) from the numerator and denominator (assuming \(\sin(5x) \neq 0\)):
\[ = \frac{-\sin(2x)}{\cos(2x) - 1} \]

### Step 4: Further Simplification
Using the identity \(\cos(2x) - 1 = -2\sin^2(x)\):
\[ = \frac{-\sin(2x)}{-2\sin^2(x)} = \frac{\sin(2x)}{2\sin^2(x)} \]
Using the identity \(\sin(2x) = 2\sin(x)\cos(x)\):
\[ = \frac{2\sin(x)\cos(x)}{2\sin^2(x)} = \frac{\cos(x)}{\sin(x)} = \cot(x) \]

### Final Result
Thus, the expression \(\frac{\cos 7x - \cos 3x}{\sin 7x - 2\sin 5x + \sin 3x}\) simplifies to:
\[ \cot(x) \]