If `cosA+cos^(2)A=1`, then find the value of `sin^(12)A+3sin^(10)A+3sin^(8)A+sin^(6)A+sin^(4)A+sin^(2)A`.

Options

A. `-1`
B. 5
C. 2
D. 1

Answer

To solve the equation \( \cos A + \cos^2 A = 1 \) and find the value of \( \sin^{12} A + 3\sin^{10} A + 3\sin^8 A + \sin^6 A + \sin^4 A + \sin^2 A \), we will follow these steps:

### Step 1: Solve for \( \cos A \)
Starting with the equation:
\[ \cos A + \cos^2 A = 1 \]
We can rearrange this to form a quadratic equation:
\[ \cos^2 A + \cos A - 1 = 0 \]
Let \( x = \cos A \). The equation becomes:
\[ x^2 + x - 1 = 0 \]

### Step 2: Use the quadratic formula
We can solve for \( x \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = -1 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]
Thus, we have two possible values for \( \cos A \):
\[ \cos A = \frac{-1 + \sqrt{5}}{2} \quad \text{or} \quad \cos A = \frac{-1 - \sqrt{5}}{2} \]
Since \( \cos A \) must be in the range \([-1, 1]\), we discard \( \frac{-1 - \sqrt{5}}{2} \) as it is less than -1. Therefore:
\[ \cos A = \frac{-1 + \sqrt{5}}{2} \]

### Step 3: Find \( \sin^2 A \)
Using the identity \( \sin^2 A + \cos^2 A = 1 \):
\[ \sin^2 A = 1 - \cos^2 A \]
Calculating \( \cos^2 A \):
\[ \cos^2 A = \left(\frac{-1 + \sqrt{5}}{2}\right)^2 = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]
Thus:
\[ \sin^2 A = 1 - \frac{3 - \sqrt{5}}{2} = \frac{2 - (3 - \sqrt{5})}{2} = \frac{-1 + \sqrt{5}}{2} \]

### Step 4: Substitute \( \sin^2 A \) into the expression
Let \( y = \sin^2 A = \frac{-1 + \sqrt{5}}{2} \). We need to find:
\[ y^6 + 3y^5 + 3y^4 + y^3 + y^2 + y \]

### Step 5: Factor the expression
Notice that:
\[ y^6 + 3y^5 + 3y^4 + y^3 + y^2 + y = (y^2 + y)^3 \]
This is derived from the binomial expansion.

### Step 6: Calculate \( (y^2 + y)^3 \)
First, calculate \( y^2 \):
\[ y^2 = \left(\frac{-1 + \sqrt{5}}{2}\right)^2 = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]
Now calculate \( y^2 + y \):
\[ y^2 + y = \frac{3 - \sqrt{5}}{2} + \frac{-1 + \sqrt{5}}{2} = \frac{3 - \sqrt{5} - 1 + \sqrt{5}}{2} = \frac{2}{2} = 1 \]
Thus:
\[ (y^2 + y)^3 = 1^3 = 1 \]

### Final Answer
The value of \( \sin^{12} A + 3\sin^{10} A + 3\sin^8 A + \sin^6 A + \sin^4 A + \sin^2 A \) is:
\[ \boxed{1} \]