Prove the inequality of sin and tan

sin(tanθ)\sin (\tan \theta) < tan(sinθ)\tan (\sin \theta)

For all θ\theta in (0, π\pi ) except π2\frac {\pi}{2}

Note by Tarun Garg
5 years, 8 months ago

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

alt text alt text

This provides a lot of information, for example, that tan(sinθ)sin(tanθ) \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 θ \theta approaches π\pi.

Calvin Lin Staff - 5 years, 8 months ago

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But tanx\tan x is undefined at x=π2x= \frac{\pi}{2}

Jaydee Lucero - 5 years, 8 months ago

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But still taking x=π2\frac {\pi}{2} we get

sin\sin (tan\tan π2\frac {\pi}{2} ) and tan\tan (sin\sin π2\frac {\pi}{2} )

Let tan\tan π2\frac {\pi}{2} be α\alpha

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

so sinα\sin \alpha < 1 or =1

But tan\tan (sin\sin π2\frac {\pi}{2} ) = tan\tan (1) > 1

And it is applicable for π2\frac {\pi}{2}

Tarun Garg - 5 years, 8 months ago

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You can't write anything involving tanπ2,\tan \dfrac{\pi}{2}, since it's undefined! Even though sinx1,\sin x \le 1, you can't take the sine of an undefined number.

Michael Tang - 5 years, 8 months ago

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@Michael Tang Sorry for that I thought this could be done.

Tarun Garg - 5 years, 8 months ago

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As we know from looking at the graph of the sinθ\sin\theta and tanθ\tan\theta functions, we know that the tanθ\tan\theta graph starts increasing at an increasing rate after θ=0\theta=0 and the sinθ\sin\theta graph starts increasing at a decreasing rate. At θ=0\theta=0, they are both equal. This only applies for 0<θ<π20<\theta<\frac{\pi}{2}, because from π2<θ<π\frac{\pi}{2}<\theta<\pi, tanθ\tan\theta becomes negative. now that we have established that, we should know that from about 0<θπ40<\theta\leq \frac{\pi}{4}, we can take tanθsinθ\tan\theta≈\sin\theta. Therefore sin(tanθ\sin(\tan\theta) will be lesser than tan(sinθ)\tan(\sin\theta), as tanθ\tan\theta is greater than sinθ\sin\theta for 0<θ<π20<\theta<\frac{\pi}{2}.

from π4<θ<π2\frac{\pi}{4}<\theta<\frac{\pi}{2}, tanθ\tan\theta begins to become much greater than sinθ\sin\theta. Despite that, sin(tanθ)\sin(\tan\theta) can only have values for 1sin(tanθ)1-1\leq \sin(\tan\theta)\leq 1. From π4<θπ2\frac{\pi}{4}<\theta\leq\frac{\pi}{2}, the minimum value of sinθ\sin\theta is 22\frac{\sqrt{2}}{2} and the maximum is 11. hence the minimum for tan(sinθ\tan(\sin\theta is 0.854 and 1.56. since we know that tan1π2\tan1≈\frac{\pi}{2}, we know that π41\frac{\pi}{4}≈1, we can deduce that sin(tanθ)\sin(\tan\theta) will decrease from its maximum of 1, so tan(sinθ)\tan(\sin\theta) can increase from its minimum of 0.854, and it can increase beyond 1, and therefore pass the reach of sin(tanθ\sin(\tan\theta.

Since trigonometric graphs are periodical, tan(sinθ)\tan(\sin\theta) is also periodical, and knowing it hits its maximum at θ=π2\theta = \frac{\pi}{2}, it must go back to 0 at θ=π\theta=\pi. since tan(sinθ)\tan(\sin\theta) started its period at the same time as sin(tanθ)sin(\tan\theta), then both of them will hit 0 θ=π\theta=\pi. However using differentiation, we can deduce that tan(sinθ)\tan(\sin\theta) is decreasing at θ=π\theta=\pi, whereas sin(tanθ)\sin(\tan\theta) is increasing. Hence, sin(tanθ)<tan(sinθ)\sin(\tan\theta)<\tan(\sin\theta) from 0<θ<π0<\theta<\pi

Saad Haider - 5 years, 8 months ago

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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 tanθsinθ \tan \theta \approx \sin \theta , how does that give us the result? You cannot say that sin(tanθ)sin(sinθ)<tan(sinθ) \sin (\tan \theta) \approx \sin (\sin \theta) < \tan (\sin \theta) , because in the first 'approximation', we actually have sin(tanθ)>sin(sinθ) \sin (\tan \theta) > \sin (\sin \theta) .

In your second paragraph, you seem to be arguing that for π4<θ<π2 \frac{\pi}{4} < \theta < \frac{\pi}{2} , we have tan(sinθ)0.854 \tan ( \sin \theta) \geq 0.854 (approx). However, this is not enough to conclude that tan(sinθ)1sin(tanθ) \tan ( \sin \theta) \geq 1 \geq \sin ( \tan \theta) , which seems to be your argument. Note that sin(tanθ) \sin ( \tan \theta) in the range π4<θ<π2 \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 \sin is periodic, and tanθ \tan \theta goes towards infinity. It is not a constantly decreasing function.

The difficulty caused by oscillating sin\sin also extends to the range π2<θ<3π4 \frac{\pi}{2} < \theta < \frac{3\pi}{4} , as tanθ \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.

Calvin Lin Staff - 5 years, 8 months ago

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Thanks for your feedback!

Saad Haider - 5 years, 8 months ago

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

Tarun Garg - 5 years, 8 months ago

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