Using the formulae for $\sin(A \pm B)$, prove that
$\sin(2r + 1)\theta - \sin(2r - 1)\theta = 2 \cos 2r\theta \sin \theta .$ Hence find a formula for $\sum_{r=1}^{n} \cos 2r\theta$, where $0 < \theta < \pi$, in terms of $\sin(2n + 1)\theta$ and $\sin \theta$. Using the formula found in part (b), show that the sum of the series
$\sin^2 10\theta + \sin^2 11\theta + \sin^2 12\theta + \dots + \sin^2 20\theta, \text{ for } 0 < \theta < \pi$
is $k - \frac{\sin(41\theta) - \sin(19\theta)}{4 \sin \theta}$, where $k$ is a constant to be determined.

1. Apply the compound angle formulae for sine to both terms on the left hand side.

Let $A = 2r\theta$ and $B = \theta$.
$\sin(2r\theta + \theta) = \sin 2r\theta \cos \theta + \cos 2r\theta \sin \theta$
$\sin(2r\theta - \theta) = \sin 2r\theta \cos \theta - \cos 2r\theta \sin \theta$

2. Subtract the two expanded expressions and simplify to show the required result.

$\sin(2r + 1)\theta - \sin(2r - 1)\theta$
$= (\sin 2r\theta \cos \theta + \cos 2r\theta \sin \theta) - (\sin 2r\theta \cos \theta - \cos 2r\theta \sin \theta)$
$= 2 \cos 2r\theta \sin \theta \text{ (shown)}$

1. Rearrange the identity given in part (a) to make $\cos 2r\theta$ the subject.

From part (a), we have:
$\sin(2r+1)\theta - \sin(2r-1)\theta = 2 \cos 2r\theta \sin \theta$

Rearranging to isolate $\cos 2r\theta$:
$\cos 2r\theta = \frac{\sin(2r+1)\theta - \sin(2r-1)\theta}{2 \sin \theta}$

2. Apply the summation operator to both sides and factor out constants independent of $r$.

$\sum_{r=1}^{n} \cos 2r\theta = \sum_{r=1}^{n} \frac{\sin(2r+1)\theta - \sin(2r-1)\theta}{2 \sin \theta}$

Since $2 \sin \theta$ does not depend on $r$, we can factor it out:
$\sum_{r=1}^{n} \cos 2r\theta = \frac{1}{2 \sin \theta} \sum_{r=1}^{n} \left[ \sin(2r+1)\theta - \sin(2r-1)\theta \right]$

3. Expand the sum and use the Method of Differences to cancel terms.

Listing out the terms from $r = 1$ to $r = n$:
$\sum_{r=1}^{n} \left[ \sin(2r+1)\theta - \sin(2r-1)\theta \right]$
$= (\sin 3\theta - \sin \theta)$
$+ (\sin 5\theta - \sin 3\theta)$
$+ (\sin 7\theta - \sin 5\theta)$
$+ \dots$
$+ (\sin(2n-1)\theta - \sin(2n-3)\theta)$
$+ (\sin(2n+1)\theta - \sin(2n-1)\theta)$

Notice that the terms cancel diagonally, leaving only the first negative term and the last positive term:
$= \sin(2n+1)\theta - \sin \theta$

4. Combine the results to state the final formula.

Substitute the evaluated sum back into the expression from Step 2:
$\sum_{r=1}^{n} \cos 2r\theta = \frac{\sin(2n+1)\theta - \sin \theta}{2 \sin \theta}$

1. Establish the result of part (b) by using the method of differences on the identity from part (a).

From part (a), we have $2 \cos 2r\theta \sin \theta = \sin(2r + 1)\theta - \sin(2r - 1)\theta$.
Summing from $r=1$ to $n$:
$\sum_{r=1}^{n} 2 \cos 2r\theta \sin \theta = \sum_{r=1}^{n} (\sin(2r + 1)\theta - \sin(2r - 1)\theta)$
$2 \sin \theta \sum_{r=1}^{n} \cos 2r\theta = (\sin 3\theta - \sin \theta) + (\sin 5\theta - \sin 3\theta) + \dots + (\sin(2n+1)\theta - \sin(2n-1)\theta)$
The sum telescopes to leave:
$2 \sin \theta \sum_{r=1}^{n} \cos 2r\theta = \sin(2n+1)\theta - \sin \theta$
$\therefore \sum_{r=1}^{n} \cos 2r\theta = \frac{\sin(2n+1)\theta - \sin \theta}{2 \sin \theta}$

2. Rewrite the target series using the double angle formula for cosine.

Using the trigonometric identity $\sin^2 x = \frac{1 - \cos 2x}{2}$, we can rewrite the series:
$\sum_{r=10}^{20} \sin^2 r\theta = \sum_{r=10}^{20} \frac{1 - \cos 2r\theta}{2}$
$= \frac{1}{2}\sum_{r=10}^{20} 1 - \frac{1}{2}\sum_{r=10}^{20} \cos 2r\theta$

3. Express the sum of the cosine terms as the difference of two series starting from $r=1$, and apply the formula from part (b).

We express $\sum_{r=10}^{20} \cos 2r\theta$ as $\sum_{r=1}^{20} \cos 2r\theta - \sum_{r=1}^{9} \cos 2r\theta$.
Using the formula $\sum_{r=1}^{n} \cos 2r\theta = \frac{\sin(2n+1)\theta - \sin \theta}{2 \sin \theta}$ for $n=20$ and $n=9$:
$\sum_{r=10}^{20} \cos 2r\theta = \frac{\sin(41\theta) - \sin \theta}{2 \sin \theta} - \frac{\sin(19\theta) - \sin \theta}{2 \sin \theta}$
$= \frac{\sin(41\theta) - \sin \theta - \sin(19\theta) + \sin \theta}{2 \sin \theta}$
$= \frac{\sin(41\theta) - \sin(19\theta)}{2 \sin \theta}$

4. Evaluate the constant part of the sum and determine the value of $k$.

The number of terms from $r=10$ to $r=20$ is $20 - 10 + 1 = 11$, so $\sum_{r=10}^{20} 1 = 11$.
Substituting back into the full expression:
$\sum_{r=10}^{20} \sin^2 r\theta = \frac{1}{2}(11) - \frac{1}{2} \left[ \frac{\sin(41\theta) - \sin(19\theta)}{2 \sin \theta} \right]$
$= \frac{11}{2} - \frac{\sin(41\theta) - \sin(19\theta)}{4 \sin \theta}$
Comparing this to the given expression $k - \frac{\sin(41\theta) - \sin(19\theta)}{4 \sin \theta}$, we identify the constant:
$k = \frac{11}{2}$