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

Options

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

Answer

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

### Step 1: Solve for \( \sin \theta \)
From the equation \( \sin \theta + \sin^2 \theta = 1 \), we can rearrange it:
\[ \sin^2 \theta + \sin \theta - 1 = 0 \]
This is a quadratic equation in terms of \( \sin \theta \).

### Step 2: Use the quadratic formula
Using the quadratic formula \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 1, c = -1 \):
\[ \sin \theta = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]
Since \( \sin \theta \) must be between -1 and 1, we take the positive root:
\[ \sin \theta = \frac{-1 + \sqrt{5}}{2} \]

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

### Step 4: Substitute \( x = \cos^2 \theta \)
Let \( x = \cos^2 \theta \). We need to evaluate:
\[ x^6 + 3x^5 + 3x^4 + x^3 - 1 \]
This can be factored as:
\[ x^6 + 3x^5 + 3x^4 + x^3 = (x^2 + 1)^3 - 1 \]

### Step 5: Substitute \( x \)
Now substituting \( x = \frac{-1 + \sqrt{5}}{2} \):
\[ x^2 + 1 = \frac{-1 + \sqrt{5}}{2} + 1 = \frac{-1 + \sqrt{5} + 2}{2} = \frac{1 + \sqrt{5}}{2} \]
Now we calculate:
\[ \left(\frac{1 + \sqrt{5}}{2}\right)^3 - 1 \]

### Step 6: Calculate the cube
Calculating \( \left(\frac{1 + \sqrt{5}}{2}\right)^3 \):
\[ \left(\frac{1 + \sqrt{5}}{2}\right)^3 = \frac{(1 + \sqrt{5})^3}{8} = \frac{1 + 3\sqrt{5} + 3 \cdot 5 + \sqrt{5}^3}{8} = \frac{1 + 3\sqrt{5} + 15 + 5\sqrt{5}}{8} = \frac{16 + 8\sqrt{5}}{8} = 2 + \sqrt{5} \]
Thus,
\[ \left(\frac{1 + \sqrt{5}}{2}\right)^3 - 1 = (2 + \sqrt{5}) - 1 = 1 + \sqrt{5} \]

### Final Step: Conclusion
The final value is:
\[ \cos^{12} \theta + 3 \cos^{10} \theta + 3 \cos^{8} \theta + \cos^{6} \theta - 1 = 1 + \sqrt{5} - 1 = \sqrt{5} \]