Finding the Value of a Trigonometric Function 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.
 A: 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 $y=\tan x$.
A: This may shed some light on whether the tangent is bigger than the sine:

A: 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($30^{\circ}$) > $\sin(30^{\circ})$.
