Prove that $\displaystyle P(n):|\sum_{k=1}^{n} \sin(k) \sin(k^2)| \leq 1, \forall n\geq 1$ is true
What I've tried: For $n=1 \implies \left|\sin(1)\sin(1^2)\right| \leq1$ is true.
Suppose that $P(n)$ is true.
Then, we need to show that $P(n+1): |\displaystyle\sum_{k=1}^{n+1} \sin(k) \sin(k^2)| \leq 1$ is true $\forall n\geq 1$.
$$\left|\displaystyle\sum_{k=1}^{n+1} \sin(k) \sin(k^2)\right| \\= \left|\sum_{k=1}^{n} \sin(k) \sin(k^2) + \sin(n+1)\sin((n+1)^2) \right| \\ \leq \left|\sum_{k=1}^{n} \sin(k) \sin(k^2)\right| + \left|\sin(n+1)\sin((n+1)^2 \right| \\ \leq 1 + \left|\sin(n+1)\right|\left| \sin((n+1)^2)\right| \\\leq 1 + 1*1 = 1+1 = 2$$
But how can I obtain that it's $\leq 1$ ?
sin(k)
in equations because it looks ugly: $sin(k)$ looks like multiplication between $s,i,n$ and $k$. Instead, use\sin(k)
to have it appear normally as $\sin(k)$. Also, replace<=
with\leq
. $\endgroup$