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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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$\frac{\sqrt2}{2} = \frac 1 {\sqrt2}$ – Mark Bennet Oct 11 '11 at 7:20
They didn't multiply both sides of the equation by $2$ but took the reciprocal on both sides. – Christian Blatter Oct 11 '11 at 8:15
$\frac{a^b}{a^c}=a^{b-c}$. Now for $a=2$, $b=1$ and $c=\frac{1}{2}$ this gives $\frac{2}{\sqrt{2}}=\sqrt{2}$. – NikolajK Jul 25 '13 at 13:57
up vote 10 down vote accepted

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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Thanks for the downvote!!! Can the downvoter point my mistake? – Asaf Karagila Oct 11 '11 at 13:21
Between your answer, Didier's answer and Christian's comment, I now understand how the equivalence works. Taking the reciprocal sounds simpler to me than multiplying by $\frac{\sqrt{2}}{\sin{x}}$. – Alex Marandon Oct 12 '11 at 3:50
@Alex: Taking reciprocal is informally to say "multiply by $\frac{\sqrt 2}{\sin x}$". – Asaf Karagila Oct 12 '11 at 7:06

$$\frac1{\frac{\sqrt2}2}=\sqrt2. $$

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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 $$\text{cosec}\; x =\frac{1}{\sin x} = \pm {\sqrt 2}.$$

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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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I don't understand where the $k$ is coming from. Could you please tell me what part of algebra or trigonometry I'm missing? (maybe a or Khan Academy link?) – Alex Marandon Oct 12 '11 at 3:54
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. – Dan Brumleve Oct 25 '11 at 5:46

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