Question

If $\alpha, \beta, \gamma \in [0, 2\pi]$ and $cos\alpha + cos\beta + cos\gamma = \frac{24}{17}$ and $sin\alpha + sin\beta + sin\gamma = \frac{45}{17}$, then value of $\frac{cos\alpha + 2cos\beta + 3cos\gamma}{cos\gamma + 2cos\beta + 3cos\alpha} + \frac{sin\alpha + 2sin\beta + 3sin\gamma}{sin\gamma + 2sin\beta + 3sin\alpha}$ is

• A. $\frac{8}{3}$
• B. $\frac{4}{5}$
• C. 2
• D. None of these

Answer 1

Answer: (C)

2

Answer 2

Let $z_1 = \cos\alpha + i\sin\alpha$, $z_2 = \cos\beta + i\sin\beta$, and $z_3 = \cos\gamma + i\sin\gamma$. We are given: $\cos\alpha + \cos\beta + \cos\gamma = \frac{24}{17}$ $\sin\alpha + \sin\beta + \sin\gamma = \frac{45}{17}$

Let $Z = z_1 + z_2 + z_3$. $Z = (\cos\alpha + \cos\beta + \cos\gamma) + i(\sin\alpha + \sin\beta + \sin\gamma)$ $Z = \frac{24}{17} + i\frac{45}{17} = \frac{24 + 45i}{17}$

The magnitude of $Z$ is: $|Z| = \left|\frac{24 + 45i}{17}\right| = \frac{1}{17}|24 + 45i|$ $|Z| = \frac{1}{17}\sqrt{24^2 + 45^2} = \frac{1}{17}\sqrt{576 + 2025} = \frac{1}{17}\sqrt{2601}$ Since $51^2 = 2601$, we have: $|Z| = \frac{51}{17} = 3$

We also know that $|z_1|=1$, $|z_2|=1$, and $|z_3|=1$ because they are complex numbers on the unit circle ($e^{i\theta}$). By the triangle inequality for complex numbers, $|z_1 + z_2 + z_3| \le |z_1| + |z_2| + |z_3|$. In our case, $|Z| = 3$ and $|z_1| + |z_2| + |z_3| = 1 + 1 + 1 = 3$. The equality $|z_1 + z_2 + z_3| = |z_1| + |z_2| + |z_3|$ holds if and only if $z_1, z_2, z_3$ have the same argument, meaning they lie on the same ray from the origin. Since $|z_1|=|z_2|=|z_3|=1$, this implies $z_1 = z_2 = z_3$.

$e^{i\alpha} = e^{i\beta} = e^{i\gamma}$ This implies $\alpha \equiv \beta \pmod{2\pi}$ and $\beta \equiv \gamma \pmod{2\pi}$. Given that $\alpha, \beta, \gamma \in [0, 2\pi]$, the only possibility is $\alpha = \beta = \gamma$.

Let $\alpha = \beta = \gamma$. Substituting this into the given equations: $3\cos\alpha = \frac{24}{17} \implies \cos\alpha = \frac{8}{17}$ $3\sin\alpha = \frac{45}{17} \implies \sin\alpha = \frac{15}{17}$ We can verify that $\cos^2\alpha + \sin^2\alpha = (\frac{8}{17})^2 + (\frac{15}{17})^2 = \frac{64+225}{289} = \frac{289}{289} = 1$, which is consistent.

Now we need to evaluate the given expression: $E = \frac{\cos\alpha + 2\cos\beta + 3\cos\gamma}{\cos\gamma + 2\cos\beta + 3\cos\alpha} + \frac{\sin\alpha + 2\sin\beta + 3\sin\gamma}{\sin\gamma + 2\sin\beta + 3\sin\alpha}$ Since $\alpha = \beta = \gamma$, let $c = \cos\alpha = 8/17$ and $s = \sin\alpha = 15/17$. The expression becomes: $E = \frac{c + 2c + 3c}{c + 2c + 3c} + \frac{s + 2s + 3s}{s + 2s + 3s}$ $E = \frac{6c}{6c} + \frac{6s}{6s}$ Since $c = 8/17 \ne 0$ and $s = 15/17 \ne 0$, the fractions are well-defined and simplify to: $E = 1 + 1 = 2$

The value of the expression is 2.