What is the maximal value of $\theta$ for which $\sin\theta\approx \theta$ holds to within $10\%$ error?

Question

question:

maximum value of $\theta$ untill which the approximation $\sin\theta\approx \theta$ holds to within $10\%$ error is

$(a)10^{\circ}$

$(b)18^{\circ}$

$(c)50^{\circ}$

$(d)90^{\circ}$

my attempt:

i calculated percentage error for each of 4 options and got $\theta = 50 $degree

but is there any quick method to arrive at answer without verifying all options

one by one . because it is MCQ there will be very less time availaible per question to solve it.

thank you

Answer 1

For small theta $$\frac{\sin\theta}\theta\approx1-\frac{\theta^2}6.$$ So we get a $10\%$ error about where $\theta^2/6\approx 0.1$, that is $\theta\approx\sqrt{0.6}\approx0.8$. A radian is about $57$ degrees, so that's about $50$ degrees or so.

Answer 2

The next best approximation is $$\sin x \approx x-\frac 16x^3 =x\cdot \left(1-\frac16x^2\right)$$ (and this is an underestimation), so we look for $\frac16x^2\approx 0.1$, $x^2\approx 0.6\approx 0.64=0.8^2$. Finally, $0.8\,\text{rad}=0.8\cdot\frac{180}\pi\,^\circ$, wich from the available choices seems to be closest to $50^\circ$.

Answer 3

This is similar to the other two answers but another way of thinking about it. By definition, the true $\theta$ is at $\left|\frac{\sin(\theta)-\theta}{\sin(\theta)}\right|=\frac{10\%}{100\%}$. This would give us $\theta=0.749=42.9^\circ$. We can use Taylor series to approximate the LHS as $\left|\frac{\theta-\frac{\theta^3}{6}-\theta}{\theta-\frac{\theta^3}{6}}\right|=\left|\frac{\theta^2}{6-\theta^2}\right|=0.1$. This solves for $\theta\approx42^\circ$, which is around $50^\circ$.