Question

$\frac{cos6\theta + 6cos4\theta + 15cos2\theta + 10}{cos5\theta + 5cos3\theta + 10cos\theta} = \lambda cos\theta$

$\lambda = ?$

Answer 1

2

Answer 2

To solve the given problem, we need to simplify the expression on the left-hand side and compare it with $\lambda \cos\theta$. The key to simplifying this expression lies in recognizing the numerator and denominator as expansions of powers of cosine in terms of multiple angles. These expansions are derived from De Moivre's theorem.

Let's recall the general formulas for expressing powers of $\cos\theta$ in terms of multiple angles: Using $2\cos\theta = e^{i\theta} + e^{-i\theta}$, we have $(2\cos\theta)^n = (e^{i\theta} + e^{-i\theta})^n$. By binomial expansion and pairing terms, we get:

1. If $n$ is an odd integer: $2^{n-1}\cos^n\theta = \binom{n}{0}\cos n\theta + \binom{n}{1}\cos(n-2)\theta + \dots + \binom{n}{(n-1)/2}\cos\theta$ This can be written as: $2^{n-1}\cos^n\theta = \sum_{k=0}^{(n-1)/2} \binom{n}{k} \cos((n-2k)\theta)$

2. If $n$ is an even integer: $2^{n-1}\cos^n\theta = \binom{n}{0}\cos n\theta + \binom{n}{1}\cos(n-2)\theta + \dots + \binom{n}{n/2 - 1}\cos2\theta + \frac{1}{2}\binom{n}{n/2}$ This can be written as: $2^{n-1}\cos^n\theta = \sum_{k=0}^{n/2-1} \binom{n}{k} \cos((n-2k)\theta) + \frac{1}{2}\binom{n}{n/2}$

Let's analyze the denominator: $D = \cos5\theta + 5\cos3\theta + 10\cos\theta$. Comparing this with the formula for odd $n$, we can see that for $n=5$: $2^{5-1}\cos^5\theta = \binom{5}{0}\cos(5\theta) + \binom{5}{1}\cos(5-2)\theta + \binom{5}{2}\cos(5-4)\theta$ $16\cos^5\theta = 1\cos5\theta + 5\cos3\theta + 10\cos\theta$ Thus, the denominator $D = 16\cos^5\theta$.

Now, let's analyze the numerator: $N = \cos6\theta + 6\cos4\theta + 15\cos2\theta + 10$. Comparing this with the formula for even $n$, we can see that for $n=6$: $2^{6-1}\cos^6\theta = \binom{6}{0}\cos(6\theta) + \binom{6}{1}\cos(6-2)\theta + \binom{6}{2}\cos(6-4)\theta + \frac{1}{2}\binom{6}{6/2}$ $32\cos^6\theta = 1\cos6\theta + 6\cos4\theta + 15\cos2\theta + \frac{1}{2}\binom{6}{3}$ We know $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$. So, $32\cos^6\theta = \cos6\theta + 6\cos4\theta + 15\cos2\theta + \frac{1}{2}(20)$ $32\cos^6\theta = \cos6\theta + 6\cos4\theta + 15\cos2\theta + 10$ Thus, the numerator $N = 32\cos^6\theta$.

Now, substitute these simplified expressions for the numerator and denominator back into the original equation: $\frac{N}{D} = \frac{32\cos^6\theta}{16\cos^5\theta}$ $\frac{32\cos^6\theta}{16\cos^5\theta} = 2\cos\theta$ We are given that $\frac{cos6\theta + 6cos4\theta + 15cos2\theta + 10}{cos5\theta + 5cos3\theta + 10cos\theta} = \lambda cos\theta$. Comparing our result with the given equation: $2\cos\theta = \lambda \cos\theta$ Assuming $\cos\theta \neq 0$ (as the expression would be undefined otherwise), we can cancel $\cos\theta$ from both sides: $\lambda = 2$