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

In summary, a trigonometric inequality is an inequality involving trigonometric functions that compares the values of two trigonometric expressions. To solve a trigonometric inequality, you must isolate the trigonometric function and use algebraic techniques to solve for the variable. The key concepts to understand are the properties of trigonometric functions, the unit circle, and the relationships between them. Trigonometric inequalities are used in real life to solve problems involving angles and measurements. When solving these inequalities, it is important to consider the period of the trigonometric function and check for extraneous solutions.
  • #1
anemone
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Prove that $\cos(\sin x))+\cos(\cos x))<\dfrac{\pi}{2}$.
 
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  • #2
[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 \(\displaystyle f(x) = cos( sin(x) ) + cos( cos(x) )\) (Let \(\displaystyle 0 \leq x < 2 \pi\) ).

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

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

-Dan
 
  • #3
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 $1.55 < 1.57$.
 
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  • #4
Using $\cos x \leqslant 1 - \frac{x^2}2 + \frac{x^4}{24}$ (as in Klaas van Aarsen's solution), $$\begin{aligned}\cos(\sin x) + \cos(\cos x) &\leqslant 1 - \frac{\sin^2x}2 + \frac{\sin^4x}{24} + 1 - \frac{\cos^2x}2 + \frac{\cos^4x}{24} \\ &= 2 - \frac{\sin^2x + \cos^2x}2 + \frac{\sin^4x + \cos^4x}{24} \\ &< \frac32 + \frac{(\sin^2x + \cos^2x)^2}{24} = \frac{37}{24} < 1.55 < 1.57 <\frac{\pi}2.\end{aligned}$$
 

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

Can the sum of two trigonometric functions be less than pi/2?

Yes, it is possible for the sum of two trigonometric functions to be less than pi/2. This depends on the specific values of the functions and the angle being used.

What conditions must be met for the sum of two trigonometric functions to be less than pi/2?

The sum of two trigonometric functions can be less than pi/2 if the values of the functions are both less than pi/4 and the angle being used is less than pi/2.

Is there a limit to how small the sum of two trigonometric functions can be?

Yes, there is a limit to how small the sum of two trigonometric functions can be. The smallest possible value is 0, which occurs when both functions have a value of 0 and the angle is 0.

Can the sum of two trigonometric functions ever be greater than pi/2?

No, the sum of two trigonometric functions can never be greater than pi/2. This is because the maximum value of any trigonometric function is 1, and the sum of two 1's is 2 which is greater than pi/2.

How can knowing the sum of two trigonometric functions be useful in scientific research?

Knowing the sum of two trigonometric functions can be useful in various fields of scientific research, such as physics, engineering, and astronomy. It can help in calculating the amplitude or phase of a wave, determining the position of an object, or analyzing the behavior of a system.

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