(a) The claim is made for all positive integers n, so we start at n = 1. We have
[tex]\displaystyle \sin(x) \sum_{k=1}^1 \sin(2kx) = \sin(x) \sin(2x) = \sin(1x) \sin((1+1)x)[/tex]
so the claim is true for the base case.
Suppose the claim is true for n = m, so that
[tex]\displaystyle \sin(x) \sum_{k=1}^m \sin(2kx) = \sin(mx) \sin((m+1)x)[/tex]
We want to use this to establish the identity for n = m + 1. That is, we want to prove that
[tex]\displaystyle \sin(x) \sum_{k=1}^{m+1} \sin(2kx) = \sin((m+1)x) \sin((m+2)x)[/tex]
Working with the left side, we remove the last term from the sum and we have
[tex]\displaystyle \sin(x) \sum_{k=1}^{m+1} \sin(2kx) = \sin(x) \sum_{k=1}^m \sin(2kx) + \sin(x) \sin(2(m+1)x) \\\\\\ = \sin(mx) \sin((m+1)x) + \sin(x) \sin(2(m+1))x[/tex]
Recall the angle sum identities:
[tex]\sin(x + y) = \sin(x) \cos(y) + \cos(x) \sin(y)[/tex]
[tex]\sin(x - y) = \sin(x) \cos(y) - \cos(x) \sin(y)[/tex]
[tex]\cos(x + y) = \cos(x) \cos(y) - \sin(x) \sin(y)[/tex]
[tex]\cos(x - y) = \cos(x) \cos(y) + \sin(x) \sin(y)[/tex]
Then
[tex]\sin(2(m+1)x) = 2 \sin((m+1)x) \cos((m+1)x)[/tex]
so we can remove a factor of sin((m + 1) x) :
[tex]\displaystyle \sin(x) \sum_{k=1}^{m+1} \sin(2kx) = \sin((m+1)x) \bigg(\sin(mx) + 2 \sin(x) \cos((m+1)x)\bigg)[/tex]
and we also have
[tex]2 \sin(x) \cos((m+1)x) = \sin(x+(m+1)x) + \sin(x - (m+1)x) \\\\ = \sin((m+2)x) + \sin(-mx) \\\\ = \sin((m+2)x) - \sin(mx)[/tex]
Then the sin(mx) terms cancel, and we're left with what we wanted:
[tex]\displaystyle \sin(x) \sum_{k=1}^{m+1} \sin(2kx) = \sin((m+1)x) \sin((m+2)x)[/tex]
and the induction proof is complete.
(b) From the identities mentioned earlier, one has
[tex]\sin\left(x - \dfrac\pi4\right) = \sin(x) \cos\left(\dfrac\pi4\right) - \cos(x) \sin\left(\dfrac\pi4\right) \\\\ = \dfrac{\sin(x) - \cos(x)}{\sqrt2}[/tex]
Then
[tex]\sin^2\left(x - \dfrac\pi4\right) = \dfrac{(\sin(x) - \cos(x))^2}2 \\\\ = \dfrac{\sin^2(x) - 2 \sin(x) \cos(x) + \cos^2(x)}2 \\\\ = \dfrac{1 - \sin(2x)}2[/tex]
We can then rewrite the sum as
[tex]\displaystyle \sum_{k=1}^{369} \sin^2\left(\dfrac{k\pi}5 - \dfrac\pi4\right) = \frac12 \sum_{k=1}^{369} \left(1 - \sin\left(\frac{2k\pi}5\right)\right) \\\\ = \frac12 \sum_{k=1}^{369} 1 - \frac12 \sum_{k=1}^{369} \sin\left(\frac{2k\pi}5\right)[/tex]
Recall that
[tex]\displaystyle \sum_{k=1}^n 1 = 1 + 1 + \cdots + 1 = n[/tex]
For the sum involving sine, let x = π/5. Then using the result from part (a),
[tex]\displaystyle \sin\left(\frac\pi5\right) \sum_{k=1}^{369} \sin\left(\dfrac{2k\pi}5\right) = \sin\left(\dfrac{369\pi}5\right) \sin\left(\dfrac{370\pi}5\right)[/tex]
and this sum vanishes, since sin(370π/5) = sin(74π) = 0.
It follows that
[tex]\displaystyle \sum_{k=1}^{369} \sin^2\left(\dfrac{k\pi}5 - \dfrac\pi4\right) = \frac12 \sum_{k=1}^{369} 1 = \boxed{\frac{369}2}[/tex]
