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What is the reciprocal of sin(x), or what is 1/sin(x) equal to in terms of trigonometric ratios? Please answer simply, as I am a high school student, not a mathematician.

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    $\begingroup$ Say you have an angle $x$ (which is acute) inside a right triangle. Then $sin(x)=\frac{\mbox{opp}}{\mbox{hyp}}$, then $\frac{1}{\sin x}$ (which is called $\csc x$, the cosecant), would be $\frac{\mbox{hyp}}{\mbox{opp}}$, this would probably work for a high student... $\endgroup$ Commented Mar 21, 2012 at 23:42
  • $\begingroup$ You may also find it useful to notice that due to $\tan(x)=\frac{\sin(x)}{\cos(x)}$, $\sin(x)=\tan(x)\cos(x)$ (And, as a result. . .) $\frac{1}{\sin(x)}=\frac{1}{\tan(x)\cos(x)}=\cot(x)\sec(x)$. You can find more than enough information at en.wikipedia.org/wiki/Trigonometric_functions Much of trigonometry is founded in repetition of the same thing in different ways. $\endgroup$
    – 000
    Commented Mar 22, 2012 at 1:14
  • $\begingroup$ $$r = \frac{a}{b} \iff \frac{1}{r} = \frac{b}{a}$$ $\endgroup$
    – user2468
    Commented Mar 22, 2012 at 1:33

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Once you said it's $\dfrac{1}{\sin x}$, you've already got a "trigonometric ratio". It's also equal to $\dfrac{\sec x}{\tan x}$. Or you can say it's $\dfrac{\text{hypotenuse}}{\text{opposite}}$. All of those are ratios.

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