Question

How do you evaluate `sec ((5pi)/12)`?

Answer 1

`2/(sqrt(2 - sqrt3))`

Explanation:

sec = 1/cos . Evaluate cos ((5pi)/12)
Trig unit circle, and property of complementary arcs give -->
`cos ((5pi)/12) = cos ((6pi)/12 - (pi)/12) = cos (pi/2 - pi/12) = sin (pi/12)`
Find sin (pi/12) by using trig identity:
`cos 2a = 1 - 2sin^2 a`
`cos (pi/6) = sqrt3/2 = 1 - 2sin^2 (pi/12)`
`2sin^2 (pi/12) = 1 - sqrt3/2 = (2 - sqrt3)/2`
`sin^2 (pi/12) = (2 - sqrt3)/4`
`sin (pi/12) = (sqrt(2 - sqrt3))/2` --> `sin (pi/12)` is positive.
Finally,
`sec ((5pi)/12) = 2/(sqrt(2 - sqrt3))`

You can check the answer by using a calculator.

Answer 2

`sec ((5pi)/12)=sqrt6+sqrt2`

Explanation:

`sec x=1/cosx`

`sec ((5pi)/12)=1/cos((5pi)/12)`

`(5pi)/12 = pi/4 +pi/6`-> Break up into composite Argument

`=1/cos(pi/4 +pi/6)`

->use `cos (A+B)=cosAcosB-sinAsinB`

`=1/(cos(pi/4)cos(pi/6)-sin(pi/4)sin(pi/6))`

`=1/((sqrt2/2)(sqrt3/2)-(sqrt2/2)(-1/2))`

`=1/(sqrt6/4 -sqrt2/4) = 1/ ((sqrt6-sqrt2)/4)=4/(sqrt6-sqrt2)`

`=4/(sqrt6-sqrt2) * (sqrt6+sqrt2)/(sqrt6+sqrt2)`

`=(4(sqrt6+sqrt2))/(6-2) = (4(sqrt6+sqrt2))/4`

`=sqrt6+sqrt2`