If ` cos x + cos^(2) x =1 ` then the value of ` sin^(12) x + 3 sin^(10) x + 3 sin^(8) x + sin^(6) x -1` is

Options

A. 2
B. 1
C. `-1`
D. 0

Explanation

To solve the equation \( \cos x + \cos^2 x = 1 \) and find the value of \( \sin^{12} x + 3 \sin^{10} x + 3 \sin^8 x + \sin^6 x - 1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos x + \cos^2 x = 1 \]
We can rearrange this to isolate \( \cos^2 x \):
\[ \cos^2 x = 1 - \cos x \]

### Step 2: Use the Pythagorean identity
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can express \( \cos^2 x \) in terms of \( \sin^2 x \):
\[ \cos^2 x = 1 - \sin^2 x \]
Substituting this into our rearranged equation gives:
\[ 1 - \sin^2 x = 1 - \cos x \]
This simplifies to:
\[ \sin^2 x = \cos x \]

### Step 3: Substitute \( \sin^2 x \) into the expression
Now we need to evaluate:
\[ \sin^{12} x + 3 \sin^{10} x + 3 \sin^8 x + \sin^6 x - 1 \]
Since \( \sin^2 x = \cos x \), we can substitute \( \sin^2 x \) into the expression:
Let \( y = \sin^2 x \), then:
- \( \sin^6 x = y^3 \)
- \( \sin^8 x = y^4 \)
- \( \sin^{10} x = y^5 \)
- \( \sin^{12} x = y^6 \)

Thus, we rewrite the expression as:
\[ y^6 + 3y^5 + 3y^4 + y^3 - 1 \]

### Step 4: Factor the expression
The expression \( y^6 + 3y^5 + 3y^4 + y^3 \) can be factored. Notice that:
\[ y^6 + 3y^5 + 3y^4 + y^3 = (y^3)(y^3 + 3y^2 + 3y + 1) = (y^3)(y + 1)^3 \]
So we can rewrite the expression as:
\[ (y^3)(y + 1)^3 - 1 \]

### Step 5: Evaluate at \( y = \cos x \)
Since \( y = \sin^2 x = \cos x \), we substitute \( y = \cos x \):
\[ (\cos^3 x)(\cos x + 1)^3 - 1 \]
From our earlier equation \( \cos x + \cos^2 x = 1 \), we know that \( \cos x + 1 = 1 - \cos^2 x + 1 = 2 - \cos^2 x \).

### Step 6: Find the final value
Given that \( \cos^2 x = 1 - \sin^2 x \) and substituting back, we find:
\[ \cos^3 x + 3 \cos^2 x + 3 \cos x + 1 - 1 = 0 \]
Thus, the final value is:
\[ 0 \]

### Final Answer
The value of \( \sin^{12} x + 3 \sin^{10} x + 3 \sin^8 x + \sin^6 x - 1 \) is \( \boxed{0} \).