# Problem about absolute value of percentage error

$$(a)$$ Bound the error in the approximation $$\sin(x)\approx x$$ for $$-\frac{\pi}{4}\le x \le \frac{\pi}4$$.

$$(b)$$ Since this is a good approximation for small values of $$x$$, also consider the "percentage error"

$$\frac{\sin(x)-x}{\sin(x)}\approx\frac{\sin(x)-x}{x}$$ Bound the absolute value of the latter quantity for $$-\delta\le x\le\delta$$. Pick $$\delta$$ to make the absolute value of the percentage error less than $$1$$%.

I've successfully solved part $$a$$. for part $$b$$ I think I supposed to solve the problem with Taylor Remainder theorem (like first part):

$$|\frac{\sin(x)-x}{x}|\le0.01$$ For $$\sin(x)$$ we have: $$R_{2n}=\frac{x^{2n+1}}{(2n+1)!}\times\cos(c),\quad 0\le c\le x$$ Substitute it in the inequality:

$$0.99\le\frac{x^{2n}}{(2n+1)!}\times\cos(c)\le1.01,\quad0\le c \le x$$ I don't know how to find $$\delta$$ in original question from the above inequality.

If you have the Taylor series $$\sin(x)= x -\frac1{3!}x^3+\frac1{5!}x^5-\cdots$$
then $$\left|\frac{\sin(x)-x}{x}\right|\le \frac1{3!}x^2$$
so $$|x|<\sqrt{0.06}\approx 0.24$$ will give $$\left|\frac{\sin(x)-x}{x}\right|\le 0.01$$