`(2)/(sqrt(2) + sqrt(2) + sqrt(2+2 cos 4x))` equals

Options

A. `sec"" (x)/(2)`
B. `sec x`
C. `cosec x`
D. 1

Answer

To solve the expression \( \frac{2}{\sqrt{2} + \sqrt{2} + \sqrt{2 + 2 \cos 4x}} \), we will simplify the denominator step by step.

### Step 1: Simplify the denominator
The denominator is \( \sqrt{2} + \sqrt{2} + \sqrt{2 + 2 \cos 4x} \).
We can combine the first two terms:
\[ \sqrt{2} + \sqrt{2} = 2\sqrt{2} \]
So, we rewrite the denominator as:
\[ 2\sqrt{2} + \sqrt{2 + 2 \cos 4x} \]

### Step 2: Simplify \( \sqrt{2 + 2 \cos 4x} \)
We can factor out a 2 from the expression inside the square root:
\[ \sqrt{2 + 2 \cos 4x} = \sqrt{2(1 + \cos 4x)} = \sqrt{2} \sqrt{1 + \cos 4x} \]
Using the trigonometric identity \( 1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right) \), we have:
\[ 1 + \cos 4x = 2 \cos^2(2x) \]
Thus,
\[ \sqrt{1 + \cos 4x} = \sqrt{2 \cos^2(2x)} = \sqrt{2} \cos(2x) \]
So,
\[ \sqrt{2 + 2 \cos 4x} = \sqrt{2} \cdot \sqrt{2} \cos(2x) = 2 \cos(2x) \]

### Step 3: Substitute back into the denominator
Now substituting back into the denominator:
\[ 2\sqrt{2} + 2\cos(2x) \]
We can factor out a 2:
\[ 2(\sqrt{2} + \cos(2x)) \]

### Step 4: Rewrite the entire expression
Now we can rewrite the original expression:
\[ \frac{2}{2(\sqrt{2} + \cos(2x))} = \frac{1}{\sqrt{2} + \cos(2x)} \]

### Final Result
Thus, the simplified expression is:
\[ \frac{1}{\sqrt{2} + \cos(2x)} \]

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