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I though i got the answer, but i already lost 5 points out of 10, so.

Veteran  Mar 20, 2017
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 #1
avatar+90613 
+3

Hi Veteran

 

I will assume that x is a acute angle, is that ok?

Draw a right angled triangle. Mark one acute angle as x.

The adjacent side is \(2\sqrt6\)

The hypotenuse is  5

 

So the opposite side is  \(\sqrt{(5^2-2\sqrt6)^2}=\sqrt{(25-24)}=1\)

 

\(cos(x)=\frac{2\sqrt6}{5}\\ sin(x)=\frac{1}{5}\\ tan(x)=\frac{sinx}{cosx}=\frac{1}{5}\div \frac{2\sqrt6}{5}=\frac{1}{2\sqrt6}\)

 

\(sin(2x) =2sinxcosx =2*\frac{1}{5}*\frac{2\sqrt6}{5}=\frac{4\sqrt6}{25}\\~\\ cos(2x)=cos^2x-sin^2x=\frac{24}{25}-\frac{1}{25}=\frac{23}{25}\\~\\ tan(2x)\\ =\frac{2tanx}{1-tan^2x}\\ =\frac{2}{2\sqrt6}\div (1-\frac{1}{24)}\\ =\frac{1}{\sqrt6}\div \frac{23}{24}\\ =\frac{24}{23\sqrt6}\\ =\frac{24\sqrt6}{23*6}\\ =\frac{4\sqrt6}{23}\\ \)

Melody  Mar 20, 2017
 #2
avatar+18626 
+3

identities in Quadrant I

\(\begin{array}{|rcll|} \hline \sin(x) &=& \sqrt{1-\cos^2(x)} \quad & | \quad \cos(x) = \frac{2}{5} \sqrt{6} \\ &=& \sqrt{1-\left(\frac{2}{5} \sqrt{6}\right)^2} \\ &=& \sqrt{1- \frac{4}{25} \cdot 6 } \\ &=& \sqrt{1- \frac{24}{25} } \\ &=& \sqrt{\frac{25-24}{25} } \\ &=& \sqrt{\frac{1}{25} } \\ \sin(x) &=& \frac{1}{5} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \sin(2x) &=& 2\cdot \sin(x)\cdot \cos(x) \\ &=& 2\cdot \frac{1}{5}\cdot \frac{2}{5} \sqrt{6} \\ &=& \frac{4}{25}\cdot \sqrt{6} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \cos(2x) &=& 1-2\cdot \sin^2(x) \\ &=& 1-2\cdot \left( \frac{1}{5} \right)^2 \\ &=& 1- \frac{2}{25} \\ &=& \frac{25-2}{25} \\ &=& \frac{23}{25} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \tan(2x) &=& \frac{\sin(2x)}{\cos(2x)} \\ &=& \frac{\frac{4}{25}\cdot \sqrt{6}} {\frac{23}{25}} \\ &=& \frac{4}{23}\cdot \sqrt{6}\\ \hline \end{array} \)

 

laugh

heureka  Mar 20, 2017

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