If `costheta+cos^(2)theta=1`, then find the value of `sin^(4)theta+sin^(2)theta`.

Options

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

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos \theta + \cos^2 \theta = 1 \]

### Step 2: Let \( x = \cos \theta \)
Substituting \( x \) for \( \cos \theta \), we can rewrite the equation as:
\[ x + x^2 = 1 \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ x^2 + x - 1 = 0 \]

### Step 4: Solve the quadratic equation
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 1, c = -1 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ x = \frac{-1 \pm \sqrt{1 + 4}}{2} \]
\[ x = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 5: Find \( \cos \theta \)
Thus, we have two possible values for \( \cos \theta \):
\[ \cos \theta = \frac{-1 + \sqrt{5}}{2} \quad \text{or} \quad \cos \theta = \frac{-1 - \sqrt{5}}{2} \]
Since \( \cos \theta \) must be in the range \([-1, 1]\), we discard \( \frac{-1 - \sqrt{5}}{2} \) as it is less than -1. Therefore:
\[ \cos \theta = \frac{-1 + \sqrt{5}}{2} \]

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

### Step 7: Find \( \sin^4 \theta \)
Now, we need to find \( \sin^4 \theta \):
\[ \sin^4 \theta = (\sin^2 \theta)^2 = \left(\frac{-1 + \sqrt{5}}{2}\right)^2 = \frac{(-1 + \sqrt{5})^2}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]

### Step 8: Calculate \( \sin^4 \theta + \sin^2 \theta \)
Now we can find:
\[ \sin^4 \theta + \sin^2 \theta = \frac{3 - \sqrt{5}}{2} + \frac{-1 + \sqrt{5}}{2} \]
Combining the fractions:
\[ = \frac{(3 - \sqrt{5}) + (-1 + \sqrt{5})}{2} = \frac{2}{2} = 1 \]

### Final Answer
Thus, the value of \( \sin^4 \theta + \sin^2 \theta \) is:
\[ \boxed{1} \]