- Let $P(n)$ be the given statement.
For $n=1$,
$\begin{array}{cl}
& \text{LS} \\
= & \sin \theta \dsum_{k=1}^1 \sin2k\theta \\
= & \sin \theta \sin 2\theta
\end{array}$Also,
$\begin{array}{cl}
& \text{RS} \\
= & \sin (1\theta)\sin (1+1)\theta \\
= & \sin \theta \sin 2\theta
\end{array}$$\therefore \text{LS}=\text{RS}$.
Hence, $P(1)$ is true.
Assume that $P(m)$ is true for some positive integers $m$.
i.e. $\sin \theta \dsum_{k=1}^m\sin2k\theta = \sin m\theta\sin(m+1)\theta$.
For $n=m+1$,
$\begin{array}{cl}
& \text{LS} \\
= & \sin \theta \dsum_{k=1}^{m+1}\sin 2k\theta \\
= & \sin \theta \dsum_{k=1}^m \sin 2k\theta +\sin\theta\sin2(m+1)\theta \\
= & \sin m\theta \sin(m+1)\theta+\sin\theta\sin2(m+1)\theta \\
= & \dfrac{1}{2}\{\cos [(m+1)\theta-m\theta]-\cos[(m+1)\theta+m\theta]\} \\
& \ \ \ \ +\dfrac{1}{2}\{\cos [2(m+1)\theta-\theta]-\cos[2(m+1)\theta+\theta]\} \\
= & \dfrac{1}{2}(\cos \theta -\cos (2m+1)\theta+\cos(2m+1)\theta-\cos(2m+3)\theta) \\
= & \dfrac{1}{2} (\cos \theta -\cos(2m+3)\theta) \\
= & \dfrac{-1}{2}\left(2\sin \dfrac{\theta+(2m+3)\theta}{2} \sin\dfrac{\theta-(2m+3)\theta}{2}\right) \\
= & -\sin(m+2)\theta\sin(-m-1)\theta \\
= & \sin(m+1)\theta\sin(m+2)\theta \\
= & \text{RS}
\end{array}$$\therefore P(m+1)$ is also true.
Therefore by Mathematics Induction, $P(n)$ is true for all positive integers $n$.
-
$\begin{array}{cl}
& \dsum_{k=1}^{111} \sin\dfrac{k\pi}{11}\cos\dfrac{k\pi}{11} \\
= & \dsum_{k=1}^{111}\dfrac{1}{2}\sin 2\left(\dfrac{k\pi}{11}\right) \\
= & \dfrac{1}{2}\dfrac{\sin\dfrac{111\pi}{11}\sin\dfrac{(111+1)\pi}{11}}{\sin\dfrac{\pi}{11}} \text{ , by (a), for } \theta=\dfrac{\pi}{11} \\
= & \dfrac{\sin\left(10\pi+\dfrac{\pi}{11}\right)\sin\left(10\pi+\dfrac{2\pi}{11}\right)}{2\sin\dfrac{\pi}{11}} \\
= & \dfrac{\sin\dfrac{\pi}{11}\sin\dfrac{2\pi}{11}}{2\sin\dfrac{\pi}{11}} \\
= & \dfrac{1}{2}\sin\dfrac{2}{11}\pi
\end{array}$Therefore, $a=\dfrac{1}{2}$ and $b=\dfrac{2}{11}$.
2023-M2-08
Ans: (b) $a=\dfrac{1}{2}$, $b=\dfrac{2}{11}$
