When x = 1, then y =

Mathematics

1 answer:

slega [8]3 years ago

5 0

Answer:

y=3..................

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$\bf \textit{Half-Angle Identities}
\\\\
sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}
\qquad 
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-------------------------------\\\\
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\\\\\\
cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}$

$\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}
\\\\\\
therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}$

which simplifies thus far to

$\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}
\\\\\\
\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}$

$\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}
\\\\\\
2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}
\\\\\\
2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}
\\\\\\
2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}
\\\\\\
2\sqrt{2-\sqrt{3}}$

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