trigonometry - Prove that series with $(-1)^kcos^3(k)$ has sequence of partial sums bounded.

I found that $(-1)^k\cos^3(k)$ equals $$(1-sin^2(k))cos((\pi+1)k)$$ or $$cos((\pi+3)k) + sin(k)sin((\pi+2)k) + sin(k)cos(k)sin((\pi+1)k)$$.

I wanted to flatten it into such a sum of terms that each has bounded partial sums, so the final sequence of partial sums is bounded too.

But I got the $sin(k)$ and $sin(k)cos(k)$.

I should use that $(-1)^k$ equals $cos(\pi k)$ and the trigonometric identities.

Answer

$$
\begin{align}
&\left|\,\sum_{k=0}^{n-1}(-1)^k\cos^3(k)\,\right|\\
&=\left|\,\sum_{k=0}^{n-1}\frac{(-1)^k}8\left(e^{ik}+e^{-ik}\right)^3\,\right|\\
&=\left|\,\sum_{k=0}^{n-1}\frac{(-1)^k}8\left(e^{3ik}+3e^{ik}+3e^{-ik}+e^{-3ik}\right)\,\right|\\
&\le\frac18\left|\,\frac{1-\left(-e^{3i}\right)^n}{1+e^{3i}}\,\right|+\frac38\left|\,\frac{1-\left(-e^{i}\right)^n}{1+e^{i}}\,\right|+\frac38\left|\,\frac{1-\left(-e^{-i}\right)^n}{1+e^{-i}}\,\right|+\frac18\left|\,\frac{1-\left(-e^{-3i}\right)^n}{1+e^{-3i}}\,\right|\\
&\le\frac1{4\cos\left(\frac32\right)}+\frac3{4\cos\left(\frac12\right)}

\end{align}
$$