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# Prove the inequality of sin and tan

$$\sin (\tan \theta)$$ < $$\tan (\sin \theta)$$

For all $$\theta$$ in (0, $$\pi$$) except $$\frac {\pi}{2}$$

Note by Tarun Garg
3 years, 10 months ago

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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 $$\pi$$. Staff · 3 years, 10 months ago

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 $$0<\theta<\pi$$ · 3 years, 10 months ago

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. Staff · 3 years, 10 months ago

Thanks for your feedback! · 3 years, 10 months ago

Thats an nice attempt but can we not prove it by any other way I mean not by graphs . · 3 years, 10 months ago

But $$\tan x$$ is undefined at $$x= \frac{\pi}{2}$$ · 3 years, 10 months ago

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 $$\frac {\pi}{2}$$ · 3 years, 10 months ago

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. · 3 years, 10 months ago