Which is Greater: $\cos(\sin\ x)$ or $\sin (\cos \ x)$

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  • Thread starter kaliprasad
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In summary, the main difference between $\cos(\sin\ x)$ and $\sin (\cos \ x)$ is their composition, with inputs and outputs being interchanged. It is not possible to determine which function is greater for all values of x, but it is possible to find their maximum and minimum values. By using the properties of trigonometric functions and their graphs, we can determine that the maximum value of $\cos(\sin\ x)$ is 1 and the minimum value is -1, while the maximum value of $\sin (\cos \ x)$ is 1 and the minimum value is -1. These functions can be equal for certain values of x, but not for all values. Calculus can be used to
  • #1
kaliprasad
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which is greater of the 2

$\cos(\sin\ x) $ or $\sin (\cos \ x) $
 
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  • #2
kaliprasad said:
which is greater of the 2

$\cos(\sin\ x) $ or $\sin (\cos \ x) $

Suppose they intersect somewhere.
Then:
\begin{array}{lcll}
\cos(\sin x)&=&\sin(\cos x) \\
\cos(\sin x)&=&\cos(\frac \pi 2 - \cos x) \\
\sin x &=& \pm(\frac \pi 2 - \cos x) + 2\pi k \\
\sin x \pm \cos x &=& \pm \frac \pi 2 + 2\pi k \\
\sqrt 2 \sin(x+\phi) &=&\pm \frac \pi 2 + 2\pi k & \text{ where }\phi =\text{atan2}(\pm 1, 1) \\
\end{array}
Since $\sqrt 2 < \frac\pi 2$ it follows that there are no intersections.

Filling in $x = 0$ tells us that:
$$\cos(\sin 0)=\cos 0 = 1 > \sin 1=\sin(\cos 0)$$
Therefore:
$$\cos(\sin x)>\sin(\cos x)$$
 
  • #3
I like Serena said:
Suppose they intersect somewhere.
Then:
\begin{array}{lcll}
\cos(\sin x)&=&\sin(\cos x) \\
\cos(\sin x)&=&\cos(\frac \pi 2 - \cos x) \\
\sin x &=& \pm(\frac \pi 2 - \cos x) + 2\pi k \\
\sin x \pm \cos x &=& \pm \frac \pi 2 + 2\pi k \\
\sqrt 2 \sin(x+\phi) &=&\pm \frac \pi 2 + 2\pi k & \text{ where }\phi =\text{atan2}(\pm 1, 1) \\
\end{array}
Since $\sqrt 2 < \frac\pi 2$ it follows that there are no intersections.

Filling in $x = 0$ tells us that:
$$\cos(\sin 0)=\cos 0 = 1 > \sin 1=\sin(\cos 0)$$
Therefore:
$$\cos(\sin x)>\sin(\cos x)$$

good solution. Here is another

we have

$\cos(\sin\, x)-\sin(\cos\, x)$

= $\cos(\sin\, x)-\cos(\dfrac{\pi}{2}- \cos\, x)$

= $\cos(\sin\, x)+\cos(\dfrac{\pi}{2}+ \cos\, x)$

= $2 \cos \dfrac{\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2} \cos \dfrac{-\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$

= $2 \cos\ t \ \cos \ s$

where $t= \dfrac{\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$

and $s= \dfrac{-\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$is product of 2 terms and we can show that both t and s are between 0 and $\frac{\pi}{2}$

we have $\sin\, x\pm \cos\, x= \sqrt{2}\sin ( x\pm \dfrac{\pi}{4})$ so between $- \sqrt{2}$ and $\sqrt{2}$hence s and t both are positive and less than $\frac{\pi}{2}$ so the difference $\cos(\sin\, x)-\sin(\cos\, x)\gt \ 0 $ and hence

$\cos(\sin\, x) \gt \sin(\cos\, x)$
 

FAQ: Which is Greater: $\cos(\sin\ x)$ or $\sin (\cos \ x)$

What is the difference between $\cos(\sin\ x)$ and $\sin (\cos \ x)$?

The main difference between these two trigonometric functions is their composition. $\cos(\sin\ x)$ is a composition of the sine function with the cosine function, while $\sin (\cos \ x)$ is a composition of the cosine function with the sine function. This means that the inputs and outputs of these functions are being interchanged.

Which is greater, $\cos(\sin\ x)$ or $\sin (\cos \ x)$, for all values of x?

For all values of x, it is impossible to determine which function is greater as they both have varying values depending on the input. However, it is possible to determine the maximum and minimum values for each function.

How do you find the maximum and minimum values of $\cos(\sin\ x)$ and $\sin (\cos \ x)$?

To find the maximum and minimum values of these functions, we can use the properties of trigonometric functions and their graphs. The maximum value of $\cos(\sin\ x)$ is 1 and the minimum value is -1, while the maximum value of $\sin (\cos \ x)$ is 1 and the minimum value is -1.

Can $\cos(\sin\ x)$ and $\sin (\cos \ x)$ ever be equal?

Yes, it is possible for these functions to be equal for certain values of x. For example, when x = $\frac{\pi}{2}$, both functions will be equal to 1. However, there are infinitely many other values of x for which these functions will not be equal.

How can we use calculus to compare $\cos(\sin\ x)$ and $\sin (\cos \ x)$?

By taking the derivative of each function and analyzing the resulting expressions, we can compare the rates of change for $\cos(\sin\ x)$ and $\sin (\cos \ x)$. This can also help us identify the maximum and minimum points for each function, which can be used to determine which function is greater for certain values of x.

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