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I'm trying to understand the solution of a trigonometry problem. One of the steps of the solution says that:

$\frac{\sqrt2}{2} = \sin x$

And then directly deduces that:

$\sqrt2 = \frac{1}{\sin x}$

I wonder how this equivalence works. It looks like they multiply both sides of the equation by 2. When I check with a calculator, $\frac{1}{\sin x}$ is indeed equal to $2 \sin x$ for the value of $x$ used in the exercise, which happens to be $\frac{\pi}{4}$, but it doesn't seem to be the case of other values of $x$. What am I missing here?

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    $\frac{a^b}{a^c}=a^{b-c}$. Now for $a=2$, $b=1$ and $c=\frac{1}{2}$ this gives $\frac{2}{\sqrt{2}}=\sqrt{2}$.2013-07-25

4 Answers 4

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Recall that $2=\sqrt 2\cdot\sqrt 2$ and therefore:

$\sin x=\frac{\sqrt 2}{2}=\frac{\sqrt 2}{\sqrt 2\cdot\sqrt 2} = \frac{1}{\sqrt 2}$

Now multiply by $\frac{\sqrt 2}{\sin x}$ both sides and you have as needed.

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    @Alex: Taking reciprocal is informally to say "multiply by $\frac{\sqrt 2}{\sin x}$".2011-10-12
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$\frac1{\frac{\sqrt2}2}=\sqrt2. $

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The equation in the title can be rewritten as $\sin(x)^2 = \frac{1}{2}$, which has the solutions $x = (2k+1) \frac{\pi}{4}$. That's why it isn't true for other values of $x$. Did's answer explains the algebraic manipulation in the step; the way I think of it is: you can always put the root on top.

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    The $k$ is just a way of expressing _all_ of the solutions to the equation. Set $k$ equal to any integer and you will have a solution. Trigonometric functions are periodic so there are an infinite number of inputs yielding any particular output.2011-10-25
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$\frac{1}{\sin x}=2 \sin x$ has the solutions $x=n\pi \pm \frac{\pi}{4}$ for integer $n$, since you have $\sin^2 x = \frac{1}{2}$ and so $\sin x = \pm \frac{1}{\sqrt 2} = \pm \frac{\sqrt 2}{2}$ and if you insist $\text{cosec}\; x =\frac{1}{\sin x} = \pm {\sqrt 2}.$