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Given that $\sin(2x)=\frac 5 {13}$ and $0^\circ < x < 45^\circ$, I need to find $\cos(x)$ and $\sin(x)$.

If I work it out, I get $$\begin{gather} \frac 5 {13} = 2 \sin(x) \cos(x) \\ \frac 5 {26} = \sin(x) \cos(x) \end{gather}$$

I'm stuck at this point. I understand that $$\cos(x) = \sqrt{1-\sin^2(x)}.$$

The answers should be $\frac {26}{\sqrt{26}}$ and $5\left(\frac {26}{\sqrt{26}} \right)$ but I'm having trouble getting there. Thanks for the help!

dfeuer
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Mike
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    You can edit your question when you have something to add to it -please don't re-post it. – Zev Chonoles Jun 22 '13 at 03:10
  • Tip, you can use the pythagorean theorem to find cos2x With a little modification of the halfangle formula of cos(x/2) and sin(x/2), that both contain cosx, you can find your answers. The modification is to replace x/2 by x in the half angle formulas. Now give it a try. – imranfat Jun 22 '13 at 03:16

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