MHB Can the Sum of Two Trigonometric Functions Be Less than pi/2?

[anemone]

Prove that $\cos(\sin x))+\cos(\cos x))<\dfrac{\pi}{2}$.

 

[topsquark]

[sp]There's a minimum of calculator work here. I'm assuming that's okay since I don't know an exact expression for cos(1).

Let [math]f(x) = cos( sin(x) ) + cos( cos(x) )[/math] (Let [math]0 \leq x < 2 \pi[/math] ).

Then [math]f'(x) = sin(sin(x)) \cdot ( 1 + cos( cos(x) ) )[/math] will give relative maxs and mins.

The maxima are found to be at [math]x = 0, ~ \pi[/math]. We can easily show that [math]f(0) = f( \pi ) = 1 + cos(1) \approx 1.5032 < \dfrac{ \pi }{2}[/math]
[/sp]

-Dan

 

[I like Serena]

Spoiler: My attempt

Let $y=\sin x$ and $z=\cos x$.
Then we want to know if the maximum value of $\cos(\cos x)+\cos(\sin x)=\cos y+\cos z$ is less than $\frac\pi 2$ with the boundary condition $y^2+z^2=1$.

We can find the extrema with Lagrange multipliers.
Define $\Lambda(y,z,\lambda)=\cos y+\cos z-\lambda(y^2+z^2-1). \tag 1$

The extrema occur for $(y,z)$ with:
$$\begin{cases}\Lambda_y=-\sin y -2\lambda y = 0\\
\Lambda_z=-\sin z -2\lambda z = 0 \\
\Lambda_\lambda = -(y^2+z^2-1) = 0 \end{cases} \implies
\begin{cases}-2\lambda = \frac{\sin y}{y} \lor y=0 \\
-2\lambda = \frac{\sin z}{z} \lor z = 0 \\
y^2+z^2=1\end{cases} \implies
\begin{cases}\frac{\sin y}{y} = \frac{\sin z}{z} \lor y=0 \lor z=0 \\
y^2+z^2=1\end{cases} \tag 2
$$
So $(y,z)$ is on the unit circle and $-1\le y,z \le +1$.
Note that $\frac{\sin t}{t}$ is an even function that is bijective on the interval $(0,1]$. Therefore $\frac{\sin y}{y} = \frac{\sin z}{z} \implies y=\pm z$.
It follows that the extrema are when $(y,z)$ is $\left(\pm \frac 1{\sqrt 2},\pm \frac 1{\sqrt 2}\right)$ and/or $(0,\,\pm 1)$ and/or $(\pm 1,\,0)$.

Thus: $\cos(\sin x)+\cos(\cos x) \le \max\left[ \cos\left(\pm \frac 1{\sqrt 2}\right) + \cos\left(\pm \frac 1{\sqrt 2}\right),\,\cos(0) + \cos(\pm 1)\right]
=\max\left[2\cos\left(\frac 1{\sqrt 2}\right),\, 1+\cos 1\right] \tag 3$

From the Taylor expansion we get that $\cos x \le 1-\frac {x^2}2+\frac{x^4}{4!}$.
So $2\cos\left(\frac 1{\sqrt 2}\right) \le 2\left(1-\frac {1/2}2 + \frac {(1/2)^2}{24}\right)=1.5+\frac 1{48} \le 1.5 + \frac 1{20} = 1.55$ and $1+\cos 1 \le 1+ \left(1-\frac 12 + \frac 1{24}\right) \le 1.5 + \frac 1{20}=1.55$.

Finally we have that $\frac\pi 2 > \frac{3.14}{2} = 1.57$, which completes the proof since $1.55 < 1.57$.

 

[Opalg]

Spoiler: My solution

Using $\cos x \leqslant 1 - \frac{x^2}2 + \frac{x^4}{24}$ (as in Klaas van Aarsen's solution), $$\begin{aligned}\cos(\sin x) + \cos(\cos x) &\leqslant 1 - \frac{\sin^2x}2 + \frac{\sin^4x}{24} + 1 - \frac{\cos^2x}2 + \frac{\cos^4x}{24} \\ &= 2 - \frac{\sin^2x + \cos^2x}2 + \frac{\sin^4x + \cos^4x}{24} \\ &< \frac32 + \frac{(\sin^2x + \cos^2x)^2}{24} = \frac{37}{24} < 1.55 < 1.57 <\frac{\pi}2.\end{aligned}$$