Showing that $\cos (\sqrt x)=e^x-2$ has solution in $(0,1)$

Question

I rephrased this problem as finding a solution of $f(x)=e^x-2-\cos(\sqrt x)$ in $(0,1)$. I apply the intermediate value theorem: $f(0)=e^0-2-\cos0=-2$. The goal is to show $f(1)>0$ (which is true) or to find some other number $c\in (0,1)$ for which it is obvious that $f(c)>0$.

However, $f(1)=e-2-\cos 1$. Without using power series expansions of $e^x,\cos x$ (e.g., explaining this problem to a student in differential calculus) how can I show $f(1)>0$? I tried reasoning using the unit circle to estimate $\cos 1,$ however it is not obvious that $(e-2)>\cos 1$.

Can someone suggest a nicer way to go about this problem (e.g., find a value to plug in that works better than 1 or explain why $(e-2)>\cos 1$ is clear?

Answer 1

Hint: Cosine is decreasing on $[0,\pi/2]$, and $\pi/4<1$, so $$ \frac{\sqrt{2}}{2}=\cos\pi/4>\cos 1. $$ Can you take it from here?

Edit

Note that $$ \frac{\sqrt{2}}{2}\approx 0.707 $$ while $$ e\approx 2.718. $$ I assumed these (rather classical) approximate values were known.

Answer 2

For lack of a better idea, use $\dfrac12=\cos\dfrac\pi3\simeq\cos\bigg[\dfrac13\bigg(3+\dfrac17\bigg)\bigg]=\cos\bigg(1+\dfrac1{21}\bigg)$, along with

$\cos(a+b)=\cos a\cos b-\sin a\sin b$, and $\lim\limits_{t\to0}\dfrac{\sin t}t=1$, implying that $\sin t\simeq t$ for very small

values of t $\bigg($such as $t=\dfrac1{21}$, for instance$\bigg)$. Together with $\sin^2x+\cos^2x=1$, we have obtain

the stunningly beautiful and mesmerizingly elegant expression $\cos1\simeq\dfrac{441+\sqrt{1327}}{884}$ . Now,

just like $\pi\simeq\dfrac{22}7$, we also have $e\simeq\dfrac{19}7$, which is very easy to remember, considering that it's my

birthday in the old Roman calendar, which started in March. If, however, the last statement might seem either too egocentric, or too European-centric, just remember that the Jewish calendar also has a Metonic cycle of $19$ years, $7$ of which are leap years. If even this might sound a tad eccentric, or perhaps just a little too exotic, recall that Leonhard Euler, who discovered the constant, passed away on September $18$, or $18/7$, if once again we are to take into consideration the etymological interpretation of the names of the months of the Roman calendar. So you are basically left with proving that $\bigg(\dfrac57\cdot884-441\bigg)^2>1327$, which is equivalent to showing that $(5\cdot884-7\cdot441)^2$ $>49\cdot1327$, which is trivial.