Step by step solution to "How to find the exact value of sec⁡(π12)?"

gwestaikd4

Answered question

2023-01-19

How to find the exact value of

\sec (\frac{π}{12})

Giaj5sf

Beginner2023-01-20Added 8 answers

We have

\sec x = \frac{1}{\cos x}

.
Then:

\sec (\frac{π}{12}) = \frac{1}{\cos (\frac{π}{12})}

We know that

\frac{π}{12} = \frac{π}{3} - \frac{π}{4}

. Therefore

\cos (\frac{π}{12}) = \cos (\frac{π}{3} - \frac{π}{4})

The alternative cosine formula is

\cos (A - B) = \cos A \cos B + \sin A \sin B

\cos (\frac{π}{12}) = \cos (\frac{π}{3}) \cos (\frac{π}{4}) + \sin (\frac{π}{3}) \sin (\frac{π}{4})

\cos (\frac{π}{12}) = \frac{\sqrt{3}}{2} (\frac{1}{\sqrt{2}}) + \frac{1}{2} (\frac{1}{\sqrt{2}})

\cos (\frac{π}{12}) = \frac{\sqrt{3} + 1}{2 \sqrt{2}}

You could rewrite this as

\sec (\frac{π}{12}) = \frac{2 \sqrt{2}}{\sqrt{3} + 1}

Then rationalize. The conjugate of

You can't use 'macro parameter character #' in math mode

\sec (\frac{π}{12}) = \frac{2 \sqrt{2}}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1}

\sec (\frac{π}{12}) = \frac{2 \sqrt{6} - 2 \sqrt{2}}{3 - 1}

\sec (\frac{π}{12}) = \frac{2 (\sqrt{6} - \sqrt{2})}{2}

\sec (\frac{π}{12}) = \sqrt{6} - \sqrt{2}

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