Let `P(x)=sqrt(( cosx +cos2x+cos3x)^2+ (sin x + sin 2x+sin3x)^2)` then P(x) is equal to

Options

A. `1+2 cosx`
B. `1+sinx`
C. `1-2cosx`
D. None of these

Answer

To find \( P(x) = \sqrt{(\cos x + \cos 2x + \cos 3x)^2 + (\sin x + \sin 2x + \sin 3x)^2} \), we will follow these steps:

### Step 1: Define the Function
Let:
\[ f(x) = \cos x + \cos 2x + \cos 3x \]
\[ g(x) = \sin x + \sin 2x + \sin 3x \]
Thus, we can rewrite \( P(x) \) as:
\[ P(x) = \sqrt{f(x)^2 + g(x)^2} \]

### Step 2: Use the Formula for Squaring Sums
Using the identity for squaring sums, we have:
\[ f(x)^2 + g(x)^2 = (\cos x + \cos 2x + \cos 3x)^2 + (\sin x + \sin 2x + \sin 3x)^2 \]
This can be expanded using the formula \( (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \).

### Step 3: Expand \( f(x)^2 \) and \( g(x)^2 \)
Expanding \( f(x)^2 \):
\[ f(x)^2 = \cos^2 x + \cos^2 2x + \cos^2 3x + 2(\cos x \cos 2x + \cos x \cos 3x + \cos 2x \cos 3x) \]
Expanding \( g(x)^2 \):
\[ g(x)^2 = \sin^2 x + \sin^2 2x + \sin^2 3x + 2(\sin x \sin 2x + \sin x \sin 3x + \sin 2x \sin 3x) \]

### Step 4: Use the Pythagorean Identity
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ \cos^2 x + \sin^2 x = 1 \]
\[ \cos^2 2x + \sin^2 2x = 1 \]
\[ \cos^2 3x + \sin^2 3x = 1 \]
Thus, we have:
\[ f(x)^2 + g(x)^2 = 3 + 2(\cos x \cos 2x + \cos x \cos 3x + \cos 2x \cos 3x + \sin x \sin 2x + \sin x \sin 3x + \sin 2x \sin 3x) \]

### Step 5: Use Product-to-Sum Formulas
Using the product-to-sum formulas:
\[ \cos A \cos B + \sin A \sin B = \cos(A - B) \]
We can simplify the terms:
- \( \cos x \cos 2x + \sin x \sin 2x = \cos(x - 2x) = \cos(-x) = \cos x \)
- \( \cos x \cos 3x + \sin x \sin 3x = \cos(x - 3x) = \cos(-2x) = \cos 2x \)
- \( \cos 2x \cos 3x + \sin 2x \sin 3x = \cos(2x - 3x) = \cos(-x) = \cos x \)

### Step 6: Combine the Results
Thus, we have:
\[ f(x)^2 + g(x)^2 = 3 + 2(\cos x + \cos 2x + \cos x) = 3 + 4 \cos x \]

### Step 7: Final Expression for \( P(x) \)
Now substituting back into \( P(x) \):
\[ P(x) = \sqrt{3 + 4 \cos x} \]

### Conclusion
Thus, the final result is:
\[ P(x) = \sqrt{1 + 2(1 + 2 \cos x)} = \sqrt{1 + 2 \cos x} \]