This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

The graph of $\tan (\sin \theta ) - \sin ( \tan \theta )$ is given below:

alt text

This provides a lot of information, for example, that $\tan ( \sin \theta) - \sin (\tan \theta)$ is not an increasing function. Furthermore, your explanation must be able to provide a reason why the value approach back to 0 as $\theta$ approaches $\pi$.

You can't write anything involving $\tan \dfrac{\pi}{2},$ since it's undefined! Even though $\sin x \le 1,$ you can't take the sine of an undefined number.

As we know from looking at the graph of the $\sin\theta$ and $\tan\theta$ functions, we know that the $\tan\theta$ graph starts increasing at an increasing rate after $\theta=0$ and the $\sin\theta$ graph starts increasing at a decreasing rate. At $\theta=0$, they are both equal. This only applies for $0<\theta<\frac{\pi}{2}$, because from $\frac{\pi}{2}<\theta<\pi$, $\tan\theta$ becomes negative. now that we have established that, we should know that from about $0<\theta\leq \frac{\pi}{4}$, we can take $\tan\theta≈\sin\theta$. Therefore $\sin(\tan\theta$) will be lesser than $\tan(\sin\theta)$, as $\tan\theta$ is greater than $\sin\theta$ for $0<\theta<\frac{\pi}{2}$.

from $\frac{\pi}{4}<\theta<\frac{\pi}{2}$, $\tan\theta$ begins to become much greater than $\sin\theta$. Despite that, $\sin(\tan\theta)$ can only have values for $-1\leq \sin(\tan\theta)\leq 1$. From $\frac{\pi}{4}<\theta\leq\frac{\pi}{2}$, the minimum value of $\sin\theta$ is $\frac{\sqrt{2}}{2}$ and the maximum is $1$. hence the minimum for $\tan(\sin\theta$ is 0.854 and 1.56. since we know that $\tan1≈\frac{\pi}{2}$, we know that $\frac{\pi}{4}≈1$, we can deduce that $\sin(\tan\theta)$ will decrease from its maximum of 1, so $\tan(\sin\theta)$ can increase from its minimum of 0.854, and it can increase beyond 1, and therefore pass the reach of $\sin(\tan\theta$.

Since trigonometric graphs are periodical, $\tan(\sin\theta)$ is also periodical, and knowing it hits its maximum at $\theta = \frac{\pi}{2}$, it must go back to 0 at $\theta=\pi$. since $\tan(\sin\theta)$ started its period at the same time as $sin(\tan\theta)$, then both of them will hit 0 $\theta=\pi$. However using differentiation, we can deduce that $\tan(\sin\theta)$ is decreasing at $\theta=\pi$, whereas $\sin(\tan\theta)$ is increasing. Hence, $\sin(\tan\theta)<\tan(\sin\theta)$ from $0<\theta<\pi$

This is a good start, with several relevant ideas. However, you need to be really careful with your arguments. They are currently too 'handwavy' and do not provide any proper arguments.

In the first paragraph, if all we know is that $\tan \theta \approx \sin \theta$, how does that give us the result? You cannot say that $\sin (\tan \theta) \approx \sin (\sin \theta) < \tan (\sin \theta)$, because in the first 'approximation', we actually have $\sin (\tan \theta) > \sin (\sin \theta)$.

In your second paragraph, you seem to be arguing that for $\frac{\pi}{4} < \theta < \frac{\pi}{2}$, we have $\tan ( \sin \theta) \geq 0.854$ (approx). However, this is not enough to conclude that $\tan ( \sin \theta) \geq 1 \geq \sin ( \tan \theta)$, which seems to be your argument. Note that $\sin ( \tan \theta)$ in the range $\frac{\pi}{4} < \theta < \frac{\pi}{2}$ is a highly oscillating function, which explains the middle portion of the graph that I attached above. (This occurs because $\sin$ is periodic, and $\tan \theta$ goes towards infinity. It is not a constantly decreasing function.

The difficulty caused by oscillating $\sin$ also extends to the range $\frac{\pi}{2} < \theta < \frac{3\pi}{4}$, as $\tan \theta$ increases from $- \infty$. The argument that you provided doesn't hold over the rest of the range, but could hold over a reduced range. In particular, you should carry out the differentiation and state the ranges in which your conditions hold, and ensure that it overlaps with the reduced range.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThe graph of $\tan (\sin \theta ) - \sin ( \tan \theta )$ is given below:

alt text

This provides a lot of information, for example, that $\tan ( \sin \theta) - \sin (\tan \theta)$ is not an increasing function. Furthermore, your explanation must be able to provide a reason why the value approach back to 0 as $\theta$ approaches $\pi$.

Log in to reply

But $\tan x$ is undefined at $x= \frac{\pi}{2}$

Log in to reply

But still taking x=$\frac {\pi}{2}$ we get

$\sin$ ($\tan$ $\frac {\pi}{2}$) and $\tan$ ($\sin$ $\frac {\pi}{2}$)

Let $\tan$ $\frac {\pi}{2}$ be $\alpha$

As per range of $\sin \theta$ it cannot exceed 1

so $\sin \alpha$ < 1 or =1

But $\tan$ ($\sin$ $\frac {\pi}{2}$) = $\tan$(1) > 1

And it is applicable for $\frac {\pi}{2}$

Log in to reply

You can't write anything involving $\tan \dfrac{\pi}{2},$ since it's undefined! Even though $\sin x \le 1,$ you can't take the sine of an undefined number.

Log in to reply

Log in to reply

As we know from looking at the graph of the $\sin\theta$ and $\tan\theta$ functions, we know that the $\tan\theta$ graph starts increasing at an increasing rate after $\theta=0$ and the $\sin\theta$ graph starts increasing at a decreasing rate. At $\theta=0$, they are both equal. This only applies for $0<\theta<\frac{\pi}{2}$, because from $\frac{\pi}{2}<\theta<\pi$, $\tan\theta$ becomes negative. now that we have established that, we should know that from about $0<\theta\leq \frac{\pi}{4}$, we can take $\tan\theta≈\sin\theta$. Therefore $\sin(\tan\theta$) will be lesser than $\tan(\sin\theta)$, as $\tan\theta$ is greater than $\sin\theta$ for $0<\theta<\frac{\pi}{2}$.

from $\frac{\pi}{4}<\theta<\frac{\pi}{2}$, $\tan\theta$ begins to become much greater than $\sin\theta$. Despite that, $\sin(\tan\theta)$ can only have values for $-1\leq \sin(\tan\theta)\leq 1$. From $\frac{\pi}{4}<\theta\leq\frac{\pi}{2}$, the minimum value of $\sin\theta$ is $\frac{\sqrt{2}}{2}$ and the maximum is $1$. hence the minimum for $\tan(\sin\theta$ is 0.854 and 1.56. since we know that $\tan1≈\frac{\pi}{2}$, we know that $\frac{\pi}{4}≈1$, we can deduce that $\sin(\tan\theta)$ will decrease from its maximum of 1, so $\tan(\sin\theta)$ can increase from its minimum of 0.854, and it can increase beyond 1, and therefore pass the reach of $\sin(\tan\theta$.

Since trigonometric graphs are periodical, $\tan(\sin\theta)$ is also periodical, and knowing it hits its maximum at $\theta = \frac{\pi}{2}$, it must go back to 0 at $\theta=\pi$. since $\tan(\sin\theta)$ started its period at the same time as $sin(\tan\theta)$, then both of them will hit 0 $\theta=\pi$. However using differentiation, we can deduce that $\tan(\sin\theta)$ is decreasing at $\theta=\pi$, whereas $\sin(\tan\theta)$ is increasing. Hence, $\sin(\tan\theta)<\tan(\sin\theta)$ from $0<\theta<\pi$

Log in to reply

This is a good start, with several relevant ideas. However, you need to be really careful with your arguments. They are currently too 'handwavy' and do not provide any proper arguments.

In the first paragraph, if all we know is that $\tan \theta \approx \sin \theta$, how does that give us the result? You cannot say that $\sin (\tan \theta) \approx \sin (\sin \theta) < \tan (\sin \theta)$, because in the first 'approximation', we actually have $\sin (\tan \theta) > \sin (\sin \theta)$.

In your second paragraph, you seem to be arguing that for $\frac{\pi}{4} < \theta < \frac{\pi}{2}$, we have $\tan ( \sin \theta) \geq 0.854$ (approx). However, this is not enough to conclude that $\tan ( \sin \theta) \geq 1 \geq \sin ( \tan \theta)$, which seems to be your argument. Note that $\sin ( \tan \theta)$ in the range $\frac{\pi}{4} < \theta < \frac{\pi}{2}$ is a highly oscillating function, which explains the middle portion of the graph that I attached above. (This occurs because $\sin$ is periodic, and $\tan \theta$ goes towards infinity. It is not a constantly decreasing function.

The difficulty caused by oscillating $\sin$ also extends to the range $\frac{\pi}{2} < \theta < \frac{3\pi}{4}$, as $\tan \theta$ increases from $- \infty$. The argument that you provided doesn't hold over the rest of the range, but could hold over a reduced range. In particular, you should carry out the differentiation and state the ranges in which your conditions hold, and ensure that it overlaps with the reduced range.

Log in to reply

Thanks for your feedback!

Log in to reply

Thats an nice attempt but can we not prove it by any other way I mean not by graphs .

Log in to reply