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$(a)$ 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.

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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 $\left|\frac{\sin(x)-x}{x}\right|\le 0.01$

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