Question

How do you use the angle sum or difference identity to find the exact value of `sec(pi/12)`?

Answer 1

`\sec({\pi}/{12})=\sqrt{6}-\sqrt{2}`.

Explanation:

First render

`\sec({\pi}/{12})=1/{\cos({\pi}/{12})}`

Then, `{\pi}/{12}={\pi}/{4}-{\pi}/{6}` and you have the cosine of a difference:

`\cos({\pi}/{12})=\cos({\pi}/{4})\cos({\pi}/{6})+\sin({\pi}/{4})\sin({\pi}/{6})`

Put in the well known trigonometry values for `\{\pi}/{4}` and `\{\pi}/{6}` to get:

`\cos({\pi}/{12})={\sqrt{6}+\sqrt{2}}/{4}`

And then

`\sec({\pi}/{12})={4}/{\sqrt{6}+\sqrt{2}}`
`={4(\sqrt{6}-\sqrt{2})}/{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}`

From the factoring of a difference of squares, `(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})=6-2=4` and so

`\sec({\pi}/{12})=\sqrt{6}-\sqrt{2}`