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I am trying to solve a homework problem that has to do with deciding which of two trigonometric functions is greater. This would be simple to do with a calculator, but the instructions explicitly say not to, so I've hit a wall.

For example, I have to decide whether $\sin 30^{\circ}$ or $\tan 30^{\circ}$ is greater, but I do not understand how to do this. I know that $\sin$ is $y/r$ and $\tan$ is $y/x$, but the only thing I can think of is that $\sin$ would have to be smaller because r is always larger than x.

However, when the situation becomes $\cos 26^{\circ}$ or $\cos 27^{\circ}$, I can only guess that $\cos 27^{\circ}$ would be larger because, as $\theta$ comes closer to terminating at 90 degrees, y becomes larger and x becomes smaller, but that really doesn't help too much.

Am I doing any of this correctly in terms of logic? I am not sure I understand the concept very well - my class's lessons are usually finished in ten to fifteen minutes...

Thanks for any and all help.

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    $\begingroup$ You are doing fine. But $\cos 27^\circ$ is smaller than $\cos 26^\circ$. How one explains it depends on how you visualize the trig functions. I think of $r$ as constant at $1$. Then as the angle increases, the adjacent side $x$ decreases. Or else increase the angle from $26^\circ$ to $27^\circ$, keeping the adjacent side unchanged. Then "$r$" increases, so $\cos$ decreases. $\endgroup$ Commented Jan 13, 2012 at 22:47
  • $\begingroup$ @AndréNicolas I think that is worthy of an answer! $\endgroup$
    – nmagerko
    Commented Jan 13, 2012 at 22:50
  • $\begingroup$ Might it make sense that those who answer a question should also up-vote it? (I did and two others haven't yet.) If it's worth answering, then it's good, isn't it? $\endgroup$ Commented Jan 14, 2012 at 0:46
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    $\begingroup$ @MichaelHardy you make a great point. $\endgroup$
    – nmagerko
    Commented Jan 14, 2012 at 3:55

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On the question about $\sin 30^\circ$ and $\tan 30^\circ$, I don't know why the post says that you do not understand how to do this. You have in fact given a full justification of the fact that $\sin 30^\circ <\tan 30^\circ$.

You wrote, correctly, that $\sin 30^\circ=\frac{y}{r}$ and $\tan 30^\circ=\frac{y}{x}$, and that $r>x$. So when you divide $y$ by $r$, you must get something smaller than when you divide $y$ by $x$. The desired result follows.

That is probably the best way to view things. But you may also know that $\tan \theta=\frac{\sin\theta}{\cos \theta}$. So to get $\tan\theta$, you divide $\sin\theta$ by $\cos\theta$. Let's assume that $\cos\theta$ is positive. (This is true as long as $\theta<90^\circ$.) Note that $\cos\theta <1$, because $x$ is always less than the hypotenuse $r$. When you divide a number $a$ by a positive number $b<1$, the answer is $>a$.

For the second problem, the idea was fine, but there was a problem with the details. We will show that $\cos 27^\circ$ is less than $\cos 26^\circ$. How one explains it depends on how you visualize the trigonometric functions. If you increase the angle from $26^\circ$ to $27^\circ$, keeping $x$ unchanged, then (draw a picture!) "$r$" will increase, so the cosine $\frac{x}{r}$ will decrease.

Or else keep $r$ constant at $1$. Then as your angle increases from $26^\circ$ to $27^\circ$, the number "$x$" will decrease, so cosine will decrease. As $\theta$ increases from $0^\circ$ to $90^\circ$, $\cos\theta$ steadily decreases. It starts at $1$ and ends up at $0$.

A roughly similar argument shows that as $\theta$ increases from $0^\circ$ to $90^\circ$, $\sin\theta$ steadily increases. It starts at $0$ and ends up at $1$.

The behaviour of $\tan\theta$ is much wilder. As $\theta$ increases from $0^\circ$ to $90^\circ$, $\tan\theta$ steadily increases, after a while very rapidly. It becomes enormously large as $\theta$ approaches $90^\circ$.

For the long run, it will be very useful to have a good mental image of the shape of the curves $y=\sin x$, $y=\cos x$, and $y=\tan x$.

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This may shed some light on whether the tangent is bigger than the sine:

enter image description here

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Consider the graphs of sin and cos. On the interval $[0,\frac{\pi}{2}]$ you know that cos is decreasing monotonically ($x > y \Rightarrow \cos(x) < \cos(y)$), whereas sin is increasing monotonically ($x > y \Rightarrow \sin(x) > \sin(y)$).

Thus, for the case $\cos(26^{\circ}$) vs. $\cos(27^{\circ}$), we know that $\cos(26^{\circ}$) > $\cos(27^{\circ}$) since $26^{\circ}$ < $27^{\circ}$.

For the case sin($30^{\circ}$) vs. tan($30^{\circ}$), we know that $\tan(x)=\frac{\sin(x)}{\cos(x)}$, so when comparing sin($30^{\circ}$) and

tan($30^{\circ}$)=$\frac{\sin(30^{\circ})}{\cos(30^{\circ})}$, we only need to note that $\cos(30^{\circ})=\frac{\sqrt{3}}{2} < 1$, and we can conclude that tan($30^{\circ}$) > $\sin(30^{\circ})$.

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