What is the value of # Cos(pi/8)#?

2 Answers
Oct 18, 2016

#cos(pi/8)=sqrt(2+sqrt(2))/2 ~~0.92388#

Explanation:

Version 1
Things to remember:
#color(white)("XXX")pi/8 = (pi/4)/2#

#color(white)("XXX")cos(pi/4)= sqrt(2)/2#

#color(white)("XXX")cos(theta/2)=+-sqrt((1+cos(theta))/2#

#rArr cos(pi/8) = sqrt((1+sqrt(2)/2)/2)# (note the negative possibility can be ignored since #pi/8# is in Q I)

#color(white)("XXXXXXX")=sqrt(2+sqrt(2))/2#

using a calculator
#color(white)("XXXXXXX")~~0.92388#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Version 2 (if you're going to use a calculator anyway.
Use calculator to evaluate:
#color(white)("XXX")cos(pi/8)~~0.92388#

#cos(pi/8)=sqrt(2+sqrt2)/2#

Explanation:

Let us assume #pi/8=A# and then #2A=pi/4#

Now as #cos2A=2cos^2A-1#,

we have #cos(pi/4)=2cos^2A-1#

or #2cos^2A=1+1/sqrt2=(sqrt2+1)/sqrt2#

or #2cos^2A=(sqrt2+1)/2sqrt2#

Multiplying numerator and denominator by #sqrt2# on RHS

#cos^2A=(2+sqrt2)/4#

or #cosA=sqrt((2+sqrt2)/4)#

or #cos(pi/8)=sqrt(2+sqrt2)/2#