How do you evaluate #sec 15 #?

1 Answer
Jun 3, 2016

Exact value:

Explanation:

This is one of those rare questions that you can evaluate exactly using the sum and différence formulas.

First, though, let's define #sectheta#. By the reciprocal identities #sectheta = 1/costheta#

#sec15#
#=1/cos15#

Now, #15^@# can be written as #60^@ - 45^@#

By the sum and différence identity #cos(alpha - theta) = cosalphacostheta + sinalphasintheta#

We can therefore state the following:

#1/cos15 = 1/cos(60 - 45)#

Expanding:

#=1/(cos60cos45 + sin60sin45)#

#=1/(1/2 xx 1/sqrt(2) + sqrt(3)/2 xx 1/sqrt(2))#

#= 1/((1/(2sqrt(2)) + sqrt(3)/(2sqrt(2)))#

#= 1/((1 + sqrt(3))/(2sqrt2))#

#= (2sqrt(2))/(1 + sqrt(3))#

Rationalizing the denominator:

#= (2sqrt(2))/(1 + sqrt(3)) xx (1 - sqrt(3))/(1 - sqrt(3))#

#=(2sqrt(2) - 2sqrt(6))/-2#

#=(2(sqrt(2) - sqrt(6)))/-2#

#= sqrt6 - sqrt(2)#

Therefore, #sec15 = sqrt(6) - sqrt(2)#

Hopefully this helps!