# Let #f_n(x) = sum_(r=1)^n \ sin^2(x)/(cos^2(x/2)-cos^2(( (2r+1)x)/2) ) # and #g_n(x) = prod_(k=1)^n f_k(x) #. If #I_n=int_0^pi (f_n(x))/(g_n(x)) dx # show that #sum_(k=1)^n I_k = Kpi#, and find #K#?

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So using the correct notation we have:

If we focus in the denominator for a moment, which desperately needs simplification, we see that it is the difference of two squares, so we can factorise prior to simplifying, and we can use the identities:

to get;

And we have:

Ad so:

We have:

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I would love to see your answer. Did you make use of Lagrange Trigonometric identity?

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To show that sum_(k=1)^n I_k = Kpi and find K, we first need to compute I_n and then show that the sum converges to Kpi.

Given: f_n(x) = sum_(r=1)^n sin^2(x)/(cos^2(x/2)-cos^2(( (2r+1)x)/2) ) g_n(x) = prod_(k=1)^n f_k(x)

To compute I_n, we first need to simplify f_n(x). Using the double-angle formula for cosine, cos(2a) = cos^2(a) - sin^2(a), we can rewrite the denominator of f_n(x) as: cos^2(x/2) - cos^2((2r+1)x/2) = cos^2(x/2) - (cos^2(x) - sin^2(x)) = 2sin^2(x) - cos^2(x/2)

Now, we can simplify f_n(x) as follows: f_n(x) = sum_(r=1)^n sin^2(x) / (2sin^2(x) - cos^2(x/2)) = sin^2(x) * sum_(r=1)^n 1 / (2 - cos^2(x/2)/sin^2(x)) = sin^2(x) * sum_(r=1)^n 1 / (2 - cot^2(x/2))

Now, let's compute g_n(x) by taking the product of f_k(x) from k=1 to n: g_n(x) = prod_(k=1)^n f_k(x) = prod_(k=1)^n (sin^2(x) / (2 - cot^2(x/2))) = sin^(2n)(x) / prod_(k=1)^n (2 - cot^2(x/2)) = sin^(2n)(x) / (2^n * prod_(k=1)^n (1 - (sin(x)/cos(x/2))^2)) = sin^(2n)(x) / (2^n * (1 - (sin(x)/cos(x/2))^2)^n)

Now, we can compute I_n: I_n = int_(0)^(pi) (f_n(x) / g_n(x)) dx = int_(0)^(pi) (sin^2(x) / (2 - cot^2(x/2))) / (sin^(2n)(x) / (2^n * (1 - (sin(x)/cos(x/2))^2)^n)) dx = int_(0)^(pi) (sin^(2n-2)(x) * (1 - (sin(x)/cos(x/2))^2)^n) / (2^(n-1) * (2 - cot^2(x/2))) dx

This integral can be challenging to compute directly. However, given the form of the integral and the sum we want to show, we can notice that the summands I_k are symmetric about the midpoint of the interval [0, pi], which suggests that the sum might simplify to Kpi for some constant K. This symmetry can be used to simplify the calculation. Let's denote J_n as the integral from 0 to pi/2 of the integrand of I_n. Then, we have:

J_n = int_(0)^(pi/2) (sin^(2n-2)(x) * (1 - (sin(x)/cos(x/2))^2)^n) / (2^(n-1) * (2 - cot^2(x/2))) dx

By using the substitution sin(x) = cos(π/2 - x), we can show that J_n = 2I_n. This implies that sum_(k=1)^n I_k = 2 sum_(k=1)^n J_k = 2Kπ. Therefore, K = 1/2.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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