The period of the function `f(x)=(sin x+sin 2x+sin 4x+sin 5x)/(cos x+cos 2x+cos 4x+cos 5x)` is :

Options

A. `pi/3`
B. `pi/4`
C. `pi`
D. None of these

Answer

To find the period of the function

\[ f(x) = \frac{\sin x + \sin 2x + \sin 4x + \sin 5x}{\cos x + \cos 2x + \cos 4x + \cos 5x}, \]

we can follow these steps:

### Step 1: Simplify the Function
We will use the sum-to-product identities to simplify the numerator and denominator. The identities are:
- \(\sin a + \sin b = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right)\)
- \(\cos a + \cos b = 2 \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right)\)

### Step 2: Apply the Identities
Let's apply the identities to the numerator and denominator:

1. **Numerator:**
- Combine \(\sin x + \sin 5x\):
\[ \sin x + \sin 5x = 2 \sin\left(\frac{6x}{2}\right) \cos\left(\frac{4x}{2}\right) = 2 \sin(3x) \cos(2x) \]
- Combine \(\sin 2x + \sin 4x\):
\[ \sin 2x + \sin 4x = 2 \sin\left(\frac{6x}{2}\right) \cos\left(\frac{2x}{2}\right) = 2 \sin(3x) \cos(x) \]

Thus, the numerator becomes:
\[ 2 \sin(3x) \cos(2x) + 2 \sin(3x) \cos(x) = 2 \sin(3x) (\cos(2x) + \cos(x)) \]

2. **Denominator:**
- Combine \(\cos x + \cos 5x\):
\[ \cos x + \cos 5x = 2 \cos\left(\frac{6x}{2}\right) \cos\left(\frac{4x}{2}\right) = 2 \cos(3x) \cos(2x) \]
- Combine \(\cos 2x + \cos 4x\):
\[ \cos 2x + \cos 4x = 2 \cos(3x) \cos(x) \]

Thus, the denominator becomes:
\[ 2 \cos(3x) (\cos(2x) + \cos(x)) \]

### Step 3: Simplify the Entire Function
Now, substituting back, we have:
\[ f(x) = \frac{2 \sin(3x) (\cos(2x) + \cos(x))}{2 \cos(3x) (\cos(2x) + \cos(x))} = \frac{\sin(3x)}{\cos(3x)} = \tan(3x) \]

### Step 4: Determine the Period
The period of \(\tan(kx)\) is given by:
\[ \text{Period} = \frac{\pi}{k} \]
In our case, \(k = 3\), so:
\[ \text{Period} = \frac{\pi}{3} \]

### Final Answer
Thus, the period of the function \(f(x)\) is \(\frac{\pi}{3}\).

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