# Give upper and lower bounds for $\left|\frac{\sin^3{na}-2}{\left(1+\frac{\cos b}{2}\right)^n}\right|$

Give upper and lower bounds for $$\left|\frac{\sin^3{na}-2}{\left(1+\frac{\cos b}{2}\right)^n}\right|$$

The only way that book explained up to this point in other examples uses this kind of technique:

$$|\sin^3{na}|\le1$$; $$-3\le\sin^3{na}-2\le-1$$

$$-1\le\cos{b}\le1$$; $$-1/2\le(\cos{b})/2\le1/2$$;$$1/2\le1+(\cos{b})/2\le3/2$$; $$2/3\le1/(1+(\cos{b})/2)\le2$$; $$(2/3)^n\le(1/(1+(\cos{b})/2))^n\le(2)^n$$

So, $$-1(2^n)\le\frac{\sin^3{na}-2}{\left(1+\frac{\cos b}{2}\right)^n}\le-3\left(\frac23\right)^n$$

However, this doesn't make sense.

There is n error in the last inequality.Let $$A=\sin^3(n\,a)-2,\quad B=\frac{1}{\Bigl(1+\dfrac{\cos b}{2}\Bigr)^n}.$$ You have shown that $$-3\le A\le-1\quad\text{and}\quad \Bigl(\frac{2}{3}\Bigr)^n\le B\le2^n.$$ Since $$B>0$$, we have $$-3\,B\le A\,B\le -B$$ and $$-3\times2^n\le A\,B\le-\Bigl(\frac{2}{3}\Bigr)^n.$$ This makes perfectly good sense. Taking absolute values we get $$\Bigl(\frac{2}{3}\Bigr)^n\le| A\,B|\le3\times2^n.$$