Question

How do you evaluate the expression `sec3`?

Answer 1

`sec 3^@ = 16/( sqrt(10+2sqrt(5))(sqrt(6)+sqrt(2)) + (sqrt(5)-1)(sqrt(6)-sqrt(2)))`

Explanation:

I will assume that you mean `sec 3^@`, since that has an interesting answer.

Let's work back from what we want to what we know...

`sec 3^@ = 1/cos 3^@`

`cos 3^@ = cos(18^@ - 15^@)`

`color(white)(cos 3^@) = cos 18^@ cos 15^@ + sin 18^@ sin 15^@`

`cos 15^@ = cos(45^@-30^@)`

`color(white)(cos 15^@) = cos 45^@ cos 30^@ + sin 45^@ sin 30^@`

`color(white)(cos 15^@) = 1/4(sqrt(6)+sqrt(2))`

`sin 15^@ = sin(45^@-30^@)`

`color(white)(sin 15^@) = sin 45^@ cos 30^@ - cos 45^@ sin 30^@`

`color(white)(sin 15^@) = 1/4(sqrt(6)-sqrt(2))`

`cos 18^@ = 1/4 sqrt(10+2sqrt(5))`

`sin 18^@ = 1/4 (sqrt(5)-1)`

(See https://socratic.org/s/aMzLvAhu)

So:

`sec 3^@ = 1/(cos 18^@ cos 15^@ + sin 18^@ sin 15^@)`

`color(white)(sec 3^@) = 1/((1/4 sqrt(10+2sqrt(5)))(1/4(sqrt(6)+sqrt(2))) + (1/4 (sqrt(5)-1))( 1/4(sqrt(6)-sqrt(2))))`

`color(white)(sec 3^@) = 16/( sqrt(10+2sqrt(5))(sqrt(6)+sqrt(2)) + (sqrt(5)-1)(sqrt(6)-sqrt(2)))`

It is possible, but somewhat tedious to rationalise the denominator, and results in a somewhat more complicated expression, so I'll stop here.