# Prove that sin 1 > cos ( sin 1 )

• maverick280857
In summary, Vivek is asking for help with a calculus problem that involves proving that sin 1 is greater than cos(sin 1). He has created a function f(x) and needs to show that it has exactly one solution in the interval [0, π/2]. He also mentions that f(x) is always negative in this interval, but he needs to prove that f(0) is greater than 0. He is wondering if this can be done using only calculus or if he needs to adjust the inequality to make it true.
maverick280857
Hi

This is a fairly straightforward problem but I want to do it using calculus. Here goes:

Prove that $\sin 1 > \cos(\sin 1)$.

This is what I've done (I've hardly done much):

Let $f(x) = \sin(\cos(\sin x) - \cos(\sin(\cos(x)))$. I also have to show that f(x) = 0 has exactly one solution in $[0,\pi/2]$. So anyway, f'(x) < 0 for all x in this interval which in particular means that f(0) > f($\pi/2$). But the first part requires showing that f(0) > 0. Can this be done by a purely calculus-based argument? It is easily done if we bend the inequality a bit and observe that it is true if $\pi/2 -1 <\sin 1$.

Thanks and cheers
Vivek

Hi all

This was buried deep inside the forum and I had to fish it out. Just a reminder asking for help with it...

Thanks and cheers
Vivek

Hi Vivek,

Thank you for your approach to this problem. Your idea of using calculus to prove this inequality is a great approach. However, I believe there might be a small mistake in your definition of f(x). It should be f(x) = \sin 1 - \cos(\sin 1), not \sin(\cos(\sin x) - \cos(\sin(\cos(x))). With this correction, your proof is correct and I will explain it below.

First, let's show that f(x) > 0 for all x \in [0, \pi/2]. We have f(0) = \sin 1 - \cos(\sin 1) = 1 - \cos(\sin 1) > 0 since \sin 1 > 0 and \cos x \leq 1 for all x. Next, we have f(\pi/2) = \sin 1 - \cos(\sin 1) = 1 - \cos(\sin 1) > 0 since \sin 1 < 1 and \cos x \geq 0 for all x. Therefore, by the Intermediate Value Theorem, f(x) must have at least one root in [0, \pi/2]. But since f'(x) < 0 for all x \in [0, \pi/2], f(x) is strictly decreasing, meaning it can have at most one root. Therefore, f(x) = 0 has exactly one solution in [0, \pi/2].

Now, let's show that \sin 1 > \cos(\sin 1). Since f(x) = \sin 1 - \cos(\sin 1) has exactly one root in [0, \pi/2], this means that f(x) > 0 for all x \in [0, \pi/2) and f(x) < 0 for all x \in (\pi/2, \pi]. Therefore, f(1) > 0, which means that \sin 1 - \cos(\sin 1) > 0, or equivalently \sin 1 > \cos(\sin 1). This completes the proof.

I hope this helps. Keep up the good work with your calculus-based approach!

## 1. What is the mathematical statement "sin 1 > cos ( sin 1 )" trying to prove?

The statement is trying to prove that the sine of 1 radian is greater than the cosine of the sine of 1 radian.

## 2. Why is this statement important in the field of mathematics?

This statement is important because it involves trigonometric functions, which are used in many areas of mathematics, such as geometry, calculus, and physics. It also allows for a deeper understanding of the relationships between different trigonometric functions.

## 3. How can this statement be proven?

This statement can be proven using the unit circle and the properties of trigonometric functions. By drawing a right triangle with an angle of 1 radian, we can see that the opposite side (sin 1) is longer than the adjacent side (cos (sin 1)).

## 4. Are there any exceptions to this statement?

No, there are no exceptions to this statement. It holds true for all values of 1 radian and all other angles.

## 5. What are the practical applications of this statement?

This statement can be used in various real-life situations where trigonometry is involved, such as calculating the trajectory of a projectile, determining the height of a building, or analyzing the motion of waves. It also helps in solving problems involving angles and distances.

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