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I ended up in the wrong math class (trigonometry) for my level but am trying to survive by catching up on some more basic principles. I'm wondering if the same principle (and if so, what is it) is applying in the second and third part of each equation. In the first equation, how does it get reduced to $_y^x$ if, in the middle part, x was on the bottom and y was on the top and what happened to r

$$ \frac{\sin\theta}{\cos\theta} = \frac{\frac yr}{\frac xr} = \frac yx$$

In the bottom equation, I understand that the equation in the middle (fraction divided by a fraction) is rewritten as multiplication by making the denominator the numerator etc. Is that's what's happening in the top equation?

$$ {\tan\theta} = \frac{\frac{\sqrt{15}}{4}}{\frac 14} = \frac{\sqrt{15}}{4} \times \frac 41 $$

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    $\begingroup$ You might want to try \frac{\sin \theta}{\cos \theta} for fractions, and similarly elsewhere. Also, \sqrt{15} will look better than \sqrt 15. $\endgroup$ – Muphrid Feb 6 '16 at 22:53
  • $\begingroup$ Thank you, how to do the fraction over a fraction? $\endgroup$ – Michael Feb 6 '16 at 23:00
  • $\begingroup$ In general $(a/b)/(c/d) = (a/b) \times (d/c)$, (i.e., $\frac{\frac ab}{\frac cd}=\frac ab \times \frac dc$), which is how the bottom equation simplifies, and is how the top equation should simplify. But there appears to be a typo in the top equation. I would have expected something more like $\sin\theta / \cos\theta = (y/r) / (x/r) = y/x$. $\endgroup$ – grand_chat Feb 6 '16 at 23:00
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    $\begingroup$ @Michael You just nest them. e.g. \frac{ \frac{a}{b} }{c} gives $\frac{\frac{a}{b} }{c}$. $\endgroup$ – Muphrid Feb 6 '16 at 23:01
  • $\begingroup$ Any point in the unit circle gives you the tangent equal to $\frac yx$ $\endgroup$ – Piquito Feb 6 '16 at 23:49
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When dividing fractions you can make use of the fact that:$$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}$$


In your case, your first equation is most likely more like:$$\require{cancel}\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{y}{r}}{\frac{x}{r}}=\frac{y}{r}\div\frac{x}{r}=\frac{y}{\cancel{r}}\times\frac{\cancel{r}}{x}=\frac{y}{x}$$

and your second equation is:$$\require{cancel}\tan{\theta}=\frac{\frac{\sqrt{15}}{4}}{\frac{1}{4}}=\frac{\sqrt{15}}{4}\div\frac{1}{4}=\frac{\sqrt{15}}{\cancel{4}}\times\frac{\cancel{4}}{1}=\sqrt{15}$$

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  • $\begingroup$ so the top equation's just a typo then as @grand_chat said $\endgroup$ – Michael Feb 6 '16 at 23:04
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1) $\require {cancel}\frac {\frac y r}{\frac x r} = \frac {r*\frac y r}{r*\frac x r} =\frac {\cancel r*\frac y {\cancel r}}{\cancel r*\frac x {\cancel r}} \frac y x$

2) $\frac {\frac y r}{\frac x r} = \frac {\frac 1 r y}{\frac 1 r x} = \frac{ \cancel {\frac 1 r} y}{ \cancel {\frac 1 r} x}= \frac y x$

3) $\frac {\frac y r}{\frac x r} = \frac y r \cdot \frac r x = \frac y {\cancel r} \cdot \frac{\cancel r} x = \frac y x$

Take your pick. Ultimately these should become very quick and automatic

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