If `theta=18^(@)` , then what is the value of `4sin^(2)theta+2sintheta`?

Options

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

Explanation

To solve the problem, we need to find the value of \( 4\sin^2 \theta + 2\sin \theta \) when \( \theta = 18^\circ \).

### Step-by-Step Solution:

1. **Identify the expression**: We start with the expression \( 4\sin^2 \theta + 2\sin \theta \).

2. **Factor the expression**: Notice that \( 2\sin \theta \) is common in both terms. We can factor it out:
\[ 4\sin^2 \theta + 2\sin \theta = 2\sin \theta (2\sin \theta + 1) \]

3. **Substitute \( \theta \)**: Now, we substitute \( \theta = 18^\circ \):
\[ 2\sin(18^\circ)(2\sin(18^\circ) + 1) \]

4. **Find \( \sin(18^\circ) \)**: We know from trigonometric identities that:
\[ \sin(18^\circ) = \frac{\sqrt{5} - 1}{4} \]

5. **Substitute \( \sin(18^\circ) \)**: Plug this value into the expression:
\[ 2\left(\frac{\sqrt{5} - 1}{4}\right)\left(2\left(\frac{\sqrt{5} - 1}{4}\right) + 1\right) \]

6. **Simplify the expression**: First, simplify \( 2\left(\frac{\sqrt{5} - 1}{4}\right) + 1 \):
\[ 2\left(\frac{\sqrt{5} - 1}{4}\right) = \frac{\sqrt{5} - 1}{2} \]
Therefore,
\[ 2\left(\frac{\sqrt{5} - 1}{4}\right) + 1 = \frac{\sqrt{5} - 1}{2} + 1 = \frac{\sqrt{5} - 1 + 2}{2} = \frac{\sqrt{5} + 1}{2} \]

7. **Combine the terms**: Now substitute back into the expression:
\[ 2\left(\frac{\sqrt{5} - 1}{4}\right)\left(\frac{\sqrt{5} + 1}{2}\right) \]

8. **Multiply the terms**: Multiply the fractions:
\[ = \frac{2(\sqrt{5} - 1)(\sqrt{5} + 1)}{8} = \frac{(\sqrt{5})^2 - (1)^2}{4} = \frac{5 - 1}{4} = \frac{4}{4} = 1 \]

### Final Answer:
Thus, the value of \( 4\sin^2(18^\circ) + 2\sin(18^\circ) \) is \( 1 \).

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