Find the period of following function
` sin ((pi x)/(3)) + sin ((pi x)/(4))`

Answer

To find the period of the function \( f(x) = \sin\left(\frac{\pi x}{3}\right) + \sin\left(\frac{\pi x}{4}\right) \), follow these steps:

### Step 1: Identify the periods of the individual sine functions
The period of the sine function \( \sin(kx) \) is given by the formula:
\[ \text{Period} = \frac{2\pi}{k} \]

#### For the first term \( \sin\left(\frac{\pi x}{3}\right) \):
Here, \( k = \frac{\pi}{3} \).
\[ \text{Period} = \frac{2\pi}{\frac{\pi}{3}} = 2\pi \cdot \frac{3}{\pi} = 6 \]

#### For the second term \( \sin\left(\frac{\pi x}{4}\right) \):
Here, \( k = \frac{\pi}{4} \).
\[ \text{Period} = \frac{2\pi}{\frac{\pi}{4}} = 2\pi \cdot \frac{4}{\pi} = 8 \]

### Step 2: Find the least common multiple (LCM) of the periods
Now, we need to find the LCM of the two periods we calculated:
- Period of \( \sin\left(\frac{\pi x}{3}\right) \) is 6
- Period of \( \sin\left(\frac{\pi x}{4}\right) \) is 8

To find the LCM of 6 and 8:
- The prime factorization of 6 is \( 2 \times 3 \)
- The prime factorization of 8 is \( 2^3 \)

The LCM is found by taking the highest power of each prime factor:
- For \( 2 \): maximum power is \( 2^3 \)
- For \( 3 \): maximum power is \( 3^1 \)

Thus, the LCM is:
\[ \text{LCM}(6, 8) = 2^3 \times 3^1 = 8 \times 3 = 24 \]

### Step 3: Conclusion
The period of the function \( f(x) = \sin\left(\frac{\pi x}{3}\right) + \sin\left(\frac{\pi x}{4}\right) \) is:
\[ \text{Period} = 24 \]

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