Respuesta :
[tex]\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
1-2sin^2(\theta)\\
\boxed{2cos^2(\theta)-1}
\end{cases}
\\ \quad \\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\
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[/tex]
[tex]\bf cos(2\theta)=\cfrac{5}{6}\implies 2cos^2(\theta)-1=\cfrac{5}{6}\implies 2cos^2(\theta)=\cfrac{5}{6}+1 \\\\\\ 2cos^2(\theta)=\cfrac{11}{6}\implies cos^2(\theta)=\cfrac{11}{12}\implies cos(\theta)=\pm\sqrt{\cfrac{11}{12}}[/tex]
now, bear in mind, the square root gives us +/- versions, so, which is it? well, we know the angle is in the range of "0 ≤ θ < π/2", that simply means the 1st quadrant, so, we'll use the positive one then
[tex]\bf cos(\theta)=\cfrac{\sqrt{11}}{\sqrt{12}}\implies cos(\theta)=\cfrac{\sqrt{11}}{2\sqrt{3}} \\\\\\ \textit{now, let's rationalize the denominator} \\\\\\ \cfrac{\sqrt{11}}{2\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{11}\cdot \sqrt{3}}{2\sqrt{3^2}}\implies \cfrac{\sqrt{11\cdot 33}}{2\cdot 3}\implies \boxed{\cfrac{\sqrt{33}}{6}}[/tex]
[tex]\bf cos(2\theta)=\cfrac{5}{6}\implies 2cos^2(\theta)-1=\cfrac{5}{6}\implies 2cos^2(\theta)=\cfrac{5}{6}+1 \\\\\\ 2cos^2(\theta)=\cfrac{11}{6}\implies cos^2(\theta)=\cfrac{11}{12}\implies cos(\theta)=\pm\sqrt{\cfrac{11}{12}}[/tex]
now, bear in mind, the square root gives us +/- versions, so, which is it? well, we know the angle is in the range of "0 ≤ θ < π/2", that simply means the 1st quadrant, so, we'll use the positive one then
[tex]\bf cos(\theta)=\cfrac{\sqrt{11}}{\sqrt{12}}\implies cos(\theta)=\cfrac{\sqrt{11}}{2\sqrt{3}} \\\\\\ \textit{now, let's rationalize the denominator} \\\\\\ \cfrac{\sqrt{11}}{2\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{11}\cdot \sqrt{3}}{2\sqrt{3^2}}\implies \cfrac{\sqrt{11\cdot 33}}{2\cdot 3}\implies \boxed{\cfrac{\sqrt{33}}{6}}[/tex]
