Fundamental period of the function `f (x) = sin^4 x+ cos^4 x=`

Options

A. `pi`
B. ` 2pi`
C. ` pi//2`
D. None

Answer

To find the fundamental period of the function \( f(x) = \sin^4 x + \cos^4 x \), we can follow these steps:

### Step 1: Rewrite the Function
We start with the function:
\[ f(x) = \sin^4 x + \cos^4 x \]
We can use the identity \( a^2 + b^2 = (a + b)^2 - 2ab \) to rewrite this. Let \( a = \sin^2 x \) and \( b = \cos^2 x \):
\[ f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x \]
Since \( \sin^2 x + \cos^2 x = 1 \):
\[ f(x) = 1 - 2\sin^2 x \cos^2 x \]

### Step 2: Use the Double Angle Identity
We can express \( \sin^2 x \cos^2 x \) in terms of \( \sin 2x \):
\[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x \]
Thus,
\[ f(x) = 1 - \frac{1}{2} \sin^2 2x \]

### Step 3: Rewrite \( \sin^2 2x \)
Using the identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \):
\[ \sin^2 2x = \frac{1 - \cos 4x}{2} \]
Substituting this back into our function:
\[ f(x) = 1 - \frac{1}{2} \cdot \frac{1 - \cos 4x}{2} \]
This simplifies to:
\[ f(x) = 1 - \frac{1}{4} + \frac{1}{4} \cos 4x \]
\[ f(x) = \frac{3}{4} + \frac{1}{4} \cos 4x \]

### Step 4: Determine the Period
The function \( f(x) = \frac{3}{4} + \frac{1}{4} \cos 4x \) consists of a constant term and a cosine term. The period of the cosine function \( \cos kx \) is given by:
\[ \text{Period} = \frac{2\pi}{k} \]
For \( \cos 4x \), \( k = 4 \):
\[ \text{Period of } \cos 4x = \frac{2\pi}{4} = \frac{\pi}{2} \]

### Conclusion
Thus, the fundamental period of the function \( f(x) = \sin^4 x + \cos^4 x \) is:
\[ \frac{\pi}{2} \]