If `theta + varphi = (2)/(3) pi and cos theta = ( sqrt3)/(2),` then what is the value of `sin varphi` ?

Options

A. 0
B. `1//2`
C. `1//sqrt2`
D. 1

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
- \( \theta + \varphi = \frac{2}{3} \pi \)
- \( \cos \theta = \frac{\sqrt{3}}{2} \)

2. **Finding \( \theta \)**:
- We know that \( \cos \theta = \frac{\sqrt{3}}{2} \) corresponds to angles where \( \theta = \frac{\pi}{6} \) or \( \theta = \frac{11\pi}{6} \).
- However, since \( \theta + \varphi = \frac{2}{3} \pi \) and \( \frac{2}{3} \pi \) is in the range of \( [0, \pi] \), we can take \( \theta = \frac{\pi}{6} \).

3. **Substituting \( \theta \) into the first equation**:
- Substitute \( \theta \) into the equation:
\[ \frac{\pi}{6} + \varphi = \frac{2}{3} \pi \]

4. **Solving for \( \varphi \)**:
- Rearranging gives:
\[ \varphi = \frac{2}{3} \pi - \frac{\pi}{6} \]
- To subtract these fractions, we need a common denominator. The least common multiple of 3 and 6 is 6.
- Convert \( \frac{2}{3} \pi \) to sixths:
\[ \frac{2}{3} \pi = \frac{4}{6} \pi \]
- Now we can subtract:
\[ \varphi = \frac{4}{6} \pi - \frac{1}{6} \pi = \frac{3}{6} \pi = \frac{\pi}{2} \]

5. **Finding \( \sin \varphi \)**:
- Now we need to find \( \sin \varphi \):
\[ \sin \varphi = \sin\left(\frac{\pi}{2}\right) = 1 \]

Thus, the value of \( \sin \varphi \) is **1**.