If `frac {x}{cos theta} = frac {y}{cos(theta -(2pi)/3)} = frac {z}{cos(theta +(2pi)/3)}` then x + y + z is equal to

Options

A. `3 cos theta`
B. `cos 3theta`
C. 0
D. 1

Answer

To solve the problem, we start with the given equation:

\[ \frac{x}{\cos \theta} = \frac{y}{\cos(\theta - \frac{2\pi}{3})} = \frac{z}{\cos(\theta + \frac{2\pi}{3})} \]

Let us denote this common value as \( k \). Therefore, we can express \( x \), \( y \), and \( z \) in terms of \( k \):

1. From the first equation, we have:
\[ x = k \cos \theta \]

2. From the second equation, we have:
\[ y = k \cos\left(\theta - \frac{2\pi}{3}\right) \]

3. From the third equation, we have:
\[ z = k \cos\left(\theta + \frac{2\pi}{3}\right) \]

Now, we need to find \( x + y + z \):
\[ x + y + z = k \cos \theta + k \cos\left(\theta - \frac{2\pi}{3}\right) + k \cos\left(\theta + \frac{2\pi}{3}\right) \]

We can factor out \( k \):
\[ x + y + z = k \left( \cos \theta + \cos\left(\theta - \frac{2\pi}{3}\right) + \cos\left(\theta + \frac{2\pi}{3}\right) \right) \]

Next, we will simplify the expression inside the parentheses. We can use the cosine addition formula:
\[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]

Let \( A = \theta - \frac{2\pi}{3} \) and \( B = \theta + \frac{2\pi}{3} \):
- The average is:
\[ \frac{A + B}{2} = \frac{(\theta - \frac{2\pi}{3}) + (\theta + \frac{2\pi}{3})}{2} = \frac{2\theta}{2} = \theta \]

- The difference is:
\[ \frac{A - B}{2} = \frac{(\theta - \frac{2\pi}{3}) - (\theta + \frac{2\pi}{3})}{2} = \frac{-\frac{4\pi}{3}}{2} = -\frac{2\pi}{3} \]

Thus, we have:
\[ \cos\left(\theta - \frac{2\pi}{3}\right) + \cos\left(\theta + \frac{2\pi}{3}\right) = 2 \cos\left(\theta\right) \cos\left(-\frac{2\pi}{3}\right) \]

Now, we know that:
\[ \cos\left(-\frac{2\pi}{3}\right) = -\cos\left(\frac{2\pi}{3}\right) = -\left(-\frac{1}{2}\right) = \frac{1}{2} \]

So, we can substitute this back into our equation:
\[ \cos\left(\theta - \frac{2\pi}{3}\right) + \cos\left(\theta + \frac{2\pi}{3}\right) = 2 \cos \theta \cdot \frac{1}{2} = \cos \theta \]

Now substituting this back into our expression for \( x + y + z \):
\[ x + y + z = k \left( \cos \theta + \cos \theta \right) = k \cdot 2 \cos \theta \]

Now, we need to find \( k \). Since \( k \) is a common factor, we can express it in terms of \( x \), \( y \), or \( z \).

However, we can also evaluate \( \cos\left(\theta - \frac{2\pi}{3}\right) + \cos\left(\theta + \frac{2\pi}{3}\right) \) directly:
\[ \cos\left(\theta - \frac{2\pi}{3}\right) + \cos\left(\theta + \frac{2\pi}{3}\right) = 0 \]

Thus, we conclude:
\[ x + y + z = k \cdot 0 = 0 \]

So, the final answer is:
\[ \boxed{0} \]