How do you find the exact value of #sec(pi/12)#?

1 Answer
Noah G · Alan P.
Feb 13, 2018

See below.

Explanation:

We know that #secx= 1/cosx#.

Therefore:

#sec(pi/12)= 1/cos(pi/12)#

We know that #pi/12 = pi/3 - pi/4#. Thus

#cos(pi/12) = cos(pi/3 - pi/4)#

The difference formula for cosine is #cos(A - B) = cosAcosB + sinAsinB#.

#cos(pi/12) = cos(pi/3)cos(pi/4) + sin(pi/3)sin(pi/4)#

#cos(pi/12) = sqrt(3)/2(1/sqrt(2)) + 1/2(1/sqrt(2))#

#cos(pi/12) = (sqrt(3) + 1)/(2sqrt(2))#

This can be rewritten as

#sec(pi/12) = (2sqrt(2))/(sqrt(3) + 1)#

Now rationalize. The conjugate of #sqrt(3) + 1# is #sqrt(3) - 1#.

#sec(pi/12) = (2sqrt(2))/(sqrt(3) + 1) * (sqrt(3) - 1)/(sqrt(3) - 1)#

#sec(pi/12) = (2sqrt(6) - 2sqrt(2))/(3 - 1)#

#sec(pi/12) = (2(sqrt(6 ) - sqrt(2)))/2#

#sec(pi/12) = sqrt(6) - sqrt(2)#

Hopefully this helps!

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