Hope Roth

2023-03-22

How to find the exact value of $\mathrm{sec}75$ using the half angle formula?

Barrecajhoe

As $\mathrm{sec}75=\frac{1}{\mathrm{cos}75}\text{, we have to first find the value of}\phantom{\rule{1ex}{0ex}}\mathrm{cos}75$.
The following Half-Angle Formula must be applied:
$\mathrm{cos}\left(\frac{\theta }{2}\right)=±\sqrt{\frac{1+\mathrm{cos}\theta }{2}},$
where, the sign (+ or -) is to be determined in accordance with
$\mathrm{cos}\left(\frac{\theta }{2}\right)$
Taking, $\frac{\theta }{2}=75,i.e.,\theta =150$, and, noting that $75$ lies in the
First Quadrant , where, $\mathrm{cos}\phantom{\rule{1ex}{0ex}}\text{is}\phantom{\rule{1ex}{0ex}}+ve$, we get,
$\mathrm{cos}75=+\sqrt{\frac{1+\mathrm{cos}150}{2}}=\sqrt{\left(1+\frac{\mathrm{cos}\left(180-30\right)}{2}\right)}$
$=\sqrt{\frac{1-\mathrm{cos}30}{2}}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{2-\sqrt{3}}{4}}$
$=\frac{1}{2}\sqrt{2-\sqrt{3}}=\frac{1}{2}\sqrt{2-2\sqrt{\frac{3}{4}}}$
$=\frac{1}{2}\sqrt{\frac{3}{2}+\frac{1}{2}-2\sqrt{\frac{3}{2}\cdot \frac{1}{2}}}$
$=\frac{1}{2}\sqrt{{\sqrt{\frac{3}{2}}}^{2}+{\sqrt{\frac{1}{2}}}^{2}-2\cdot \sqrt{\frac{3}{2}}\cdot \sqrt{\frac{1}{2}}}$
$=\frac{1}{2}\sqrt{{\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right)}^{2}}$
$=\frac{1}{2}\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right)=\frac{\sqrt{3}-1}{2\sqrt{2}}$.
Hence,
$\mathrm{sec}75=\frac{1}{\mathrm{cos}75}=\frac{2\sqrt{2}}{\sqrt{3}-1}$
$=\frac{2\sqrt{2}}{\sqrt{3}-1}×\frac{\sqrt{3}+1}{\sqrt{3}+1}$
$=\sqrt{2}\left(\sqrt{3}-1\right)$
$=\sqrt{6}-\sqrt{2}$.

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