I found an exam question that I managed to solve via calculator but not by using only pen and paper. Is there a solution to this?
Prove that there is an $x$ satisfying $10x-9 = 9\sin x-10\cos x$ and $0.654<x<0.6545$.
I thought I need Taylor polynomial but how can I estimate that the error is small enough.
You can get pretty close given that you have a hint about where to start. If $x=u+\frac{\pi}4$ you get to $$10u-\frac{19}{\sqrt2}+\frac1{\sqrt2}(\cos u-1)=9-\frac{5\pi}2-\frac1{\sqrt2}$$ That is roughly $$-3.435u-0.3536u^2=0.4359$$ Or if $x=v+\frac{\pi}5$, then $$10v-(10\sin\frac{\pi}5+9\cos\frac{\pi}5)\sin v+(10\cos\frac{\pi}5-9\sin\frac{\pi}5)(\cos v-1)=9-2\pi-(10\cos\frac{\pi}5-9\sin\frac{\pi}5)$$ $$-3.159v-1.400v^2=-0.0833$$ Of course, you had to calculate $\cos\frac{\pi}5=\frac{\sqrt5+1}4$, $\sin\frac{\pi}5=\frac{\sqrt{10-2\sqrt5}}4$. If you were really fast and accurate at pencil and paper calculation, this might be survivable. However, although I consider such a skill to be a personal strength I have to admit that I would be in a lot of trouble.
EDIT: I think I know where this problem comes from. There are pedagogues that want you to remember various angles for different reasons. One obvious example is $\cos^{-1}\left(-\frac13\right)=109.47°$ (although they usually teach it $109.5°$) but another one that gets taught is $\sin^{-1}0.8=53.13°$. The reason for this is that a lot of problems use the pythagorean triple $3^2+4^2=5^2$ to simplify the geometry. Especially mechanical engineers carry that angle around in their heads.. $90°-53.13°=37.87°=6.345\text{ rad}$ if you use a $5$-digit value for $\pi$ to get $115.83$ and then divide by $9$ and $20$ successively to get $12.87$ and $0.6435\text{ rad}$, and this is real close to the values of $0.654$ and $0.6545$ which are to be tested. Now if you let $x=0.6435+u$, then $$10(0.6435+u)-9=9(0.8\sin u+0.6\cos u)-10(-0.6\sin u+0.8\cos u)=13.2\sin u-2.6\cos u=13.2u-2.6+1.3u^2$$ To the accuracy of the problem because the next term is about $2\times10^{-6}$, and for the cost of only $1$ $4\times5$-digit multiplication and carrying around that funny angle in our heads. Then in terms of $u$, $$f(u)=-3.2u-1.3u^2=-0.0350$$ When you plug in $x=0.654$, $u=0.0105$, $f(u)=-0.0337$, whereas if $x=0.6545$, $u=0.011$, $f(u)=-0.0354$, so we have found numbers bracketing the roots. One problem with this approach is that if all you know is $\sin^{-1}0.8=53.13°$, for all you know it could be as little as $53.126°$, in which case the same procedure gets to $0.64356\text{ rad}$ and the bigger $u=0.01094$ and this is not big enough to make $f(u)<-0.0356$, which is our new target value. So unless you knew that $53.13°$ was one of those mysterious numbers that has more accuracy than memorized, like $\frac{355}{113}$ or $0.30103$ or even $0.511$, this proof would be invalid. In fact $\sin^{-1}0.8=53.1301°$ so we actually do have significant breathing room.
If $f(x)$ is the left side minus the right side, then $f(0.654) \approx 0.001273888$ while $f(0.6545) \approx -0.000339703$. The Intermediate Value Theorem then shows that there is a zero between $0.654$ and $0.6545$. Those values of $f$ were, of course, not computed by hand. I would be extremely surprised if there were a way to compute sufficiently accurate approximations by hand in a reasonable time.
