So i found this problem in my textbook
Approximate the following
$$a) \sin(29°)$$
I don't understand how this realtes to the topic it was under - differentiation. Could you show me how this kinds of problems with specifically this kind of functions are solved using derivatives. Thank you very much.
hint
$$f(a-h)\approx f(a)-hf'(a)$$
$$\sin(a-h)\approx \sin(a)-h\cos(a)$$
$$\sin(29°)=\sin(30°-1°)=$$ $$\sin(\frac{\pi}{6}-\frac{\pi}{180})$$
$$\arctan(1.05)=\arctan(1+0.05)$$ $$\approx \arctan(1)+0.05\frac{1}{1+(1)^2}$$
Remember that the first derivative of a function is the gradient of the tangents to that function, and if the function is smooth enough then the tangent will be a nice linear approximation to the function if you're close enough to the tangent point - or in other words,
$$f(x + \delta x) \approx f(x) + \delta x f'(x)$$
when $\delta x \approx 0$ (there are results in more advanced calculus to tell you how good that approximation might be).
So for both of the given questions, we can find a nearby point where we know the value of the function, and use what we know about the derivative to approximate the required value. For example, if we were asked to find an approximation for $\sin 0.26 \pi$ we could use the fact that $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ and do:
$$\begin{eqnarray}\sin(x + \delta x) & \approx & \sin(x) + \delta x \cos x \\ \sin\left(\frac{\pi}{4} + \delta x\right)& \approx & \sin\frac{\pi}{4} + \delta x \cos \frac{\pi}{4} \\ & = & \frac{\sqrt{2}}{2}\left(1 + \delta x \right) \\ \sin(0.26 \pi) & \approx & \frac{\sqrt{2}}{2}(1 + 0.01 \pi) \approx 0.729\end{eqnarray}$$
And this turns out to be right to 2 decimal places.
For your examples, you'll need to be careful about what happens when you switch between radians and degrees, but otherwise the principle is roughly the same.
