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

Options

A. `-1`
B. 5
C. 0
D. 1

Answer

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

### Step 1: Rearranging the equation
Start with the given equation:
\[ \cos A + \cos^2 A = 1 \]
Rearranging gives:
\[ \cos^2 A + \cos A - 1 = 0 \]

### Step 2: Solving the quadratic equation
This is a quadratic equation in terms of \( \cos A \). We can apply the quadratic formula:
\[ \cos A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = -1 \). Plugging in these values:
\[ \cos A = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 3: Finding \( \sin^2 A \)
Using the identity \( \sin^2 A + \cos^2 A = 1 \), we can express \( \sin^2 A \) in terms of \( \cos A \):
\[ \sin^2 A = 1 - \cos^2 A \]
Now, we need to find \( \cos^2 A \):
\[ \cos^2 A = \left(\frac{-1 + \sqrt{5}}{2}\right)^2 = \frac{(-1 + \sqrt{5})^2}{4} = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]
Thus,
\[ \sin^2 A = 1 - \frac{3 - \sqrt{5}}{2} = \frac{2 - (3 - \sqrt{5})}{2} = \frac{-1 + \sqrt{5}}{2} \]

### Step 4: Calculating \( \sin^4 A \)
Now we calculate \( \sin^4 A \):
\[ \sin^4 A = \left(\sin^2 A\right)^2 = \left(\frac{-1 + \sqrt{5}}{2}\right)^2 = \frac{(-1 + \sqrt{5})^2}{4} = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]

### Step 5: Calculating \( \sin^6 A \) and \( \sin^8 A \)
Next, we find \( \sin^6 A \):
\[ \sin^6 A = \sin^4 A \cdot \sin^2 A = \left(\frac{3 - \sqrt{5}}{2}\right) \cdot \left(\frac{-1 + \sqrt{5}}{2}\right) = \frac{(3 - \sqrt{5})(-1 + \sqrt{5})}{4} \]
Calculating this gives:
\[ = \frac{-3 + 3\sqrt{5} + \sqrt{5} - 5}{4} = \frac{-8 + 4\sqrt{5}}{4} = -2 + \sqrt{5} \]

Now, for \( \sin^8 A \):
\[ \sin^8 A = \sin^4 A \cdot \sin^4 A = \left(\frac{3 - \sqrt{5}}{2}\right)^2 = \frac{(3 - \sqrt{5})^2}{4} = \frac{9 - 6\sqrt{5} + 5}{4} = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2} \]

### Step 6: Putting it all together
Now we can compute \( \sin^8 A + 2 \sin^6 A + \sin^4 A \):
\[ \sin^8 A + 2 \sin^6 A + \sin^4 A = \left(\frac{7 - 3\sqrt{5}}{2}\right) + 2(-2 + \sqrt{5}) + \left(\frac{3 - \sqrt{5}}{2}\right) \]
Calculating this gives:
\[ = \frac{7 - 3\sqrt{5}}{2} - 4 + 2\sqrt{5} + \frac{3 - \sqrt{5}}{2} \]
Combining terms:
\[ = \frac{(7 + 3) - 4 \cdot 2 + (-3\sqrt{5} + 2\sqrt{5} - \sqrt{5})}{2} = \frac{10 - 4}{2} = \frac{6}{2} = 3 \]

### Final Answer
Thus, the value of \( \sin^8 A + 2 \sin^6 A + \sin^4 A \) is:
\[ \boxed{3} \]