$\displaystyle P(n):|\sum_{k=1}^{n} \sin(k)\sin(k^2)| \leq 1, \forall n\geq 1$

Question

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 $\leq 1$ ?

Answer 1

$$S=\sum_{k=1}^n\sin k\sin (k^2)=\frac12\left(\sum_{k=1}^n\cos\left({k^2-k}\right)-\cos\left(k^2+k\right)\right)$$$$= \frac12\Bigg((\cos0-\cos 2)+(\cos 2-\cos 6)+(\cos 6-\cos 12)+…+(\cos(n^2-n)-\cos(n^2+n))\Bigg)$$$$=\frac12(\cos 0-\cos(n^2+n))$$$$=\frac12(1-\cos(n^2+n))$$

Now, $1\geq\cos(n^2+n)\geq-1$ so $0\leq 1-\cos(n^2+n)\leq 2$ so $0\leq S\leq 1$ and hence $|S|=S\leq 1$.