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I can't seem to demonstrate how $\frac{\sqrt{\sqrt{3}+2}}{2}$ is equal to $\frac{\sqrt{2}+\sqrt{6}}{4}$... even though my calculator tells me that both are equal $0.965925826$. Hoping someone can show me how this is done.

For context, I was trying to obtain the exact value of $\cos(\frac{\pi}{12})$ using the double angle formula (i.e., $\cos(2 \cdot \frac{\pi}{12})$) which gave me the first result, and then using the subtraction formula (i.e., $\cos(\frac{\pi}{3} - \frac{\pi}{4})$), which gave me the second.

sesandc3123
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    If you square both you get $\frac {2+\sqrt 3}4$. – lulu Mar 23 '24 at 17:23
  • Hint: You know that they are equal. Assume you don't know that $\frac{\sqrt{\sqrt{3}+2}}{2} = \frac{\sqrt{2}+\sqrt{6}}{4}$ but instead you know it's in the form of $\frac{\sqrt{a}+\sqrt{b}}{c}$. Solve for $a$, $b$, and $c$. You should get $2$, $6$, and $4$ respectively. – Wasu Chanyasubkit Mar 23 '24 at 17:24
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    Hint: Multiply the first term by $\sqrt{4}$ in numerator and denominator, and try to perceive it as a perfect square. – Stuti Mar 23 '24 at 17:27

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