How do you evaluate #cos((13pi)/8)#?

2 Answers
P dilip_k
May 14, 2016

#=1/2sqrt((2-sqrt2)#

Explanation:

using formula #costheta=sqrt(1/2(1+cos(2theta)))#
#cos((13pi)/8)#
#=sqrt(1/2(1+cos(2*(13pi)/8))) #

#=sqrt(1/2(1+cos((13pi)/4))) #

#=sqrt(1/2(1+cos(3pi+pi/4)) #

#=sqrt(1/2(1-cos(pi/4)))#

#=sqrt(1/2(1-1/sqrt2))#

#=sqrt((sqrt2-1)/(2sqrt2)#

#=sqrt((2-sqrt2)/(2xx2)#

#=1/2sqrt((2-sqrt2)#

Nghi N. · Nghi N
May 14, 2016

#sqrt(2 - sqrt2)/2#

Explanation:

#cos ((13pi)/8) = cos ((3pi)/8 + cos ((16pi)/8)) = cos ((3pi)/8 + 2pi) = #
#= cos ((3pi)/8).#
Evaluate #cos ((3pi)/8)# by applying the trig identity:
#cos 2a = 2cos^2 a - 1#.
In this case, we get -->
#2cos^2 ((3pi)/8) - 1 = cos ((6pi)/8) = cos ((3pi)/4) = -sqrt2/2#
#2cos^2 ((3pi)/8) = 1 - sqrt2/2 = (2 - sqrt2)/2#
#cos^2 ((3pi)/8) = (2 - sqrt2)/4#
#cos ((3pi)/8) = +- sqrt(2 - sqrt2)/2#
#cos ((13pi)/8) = cos ((3pi)/8) = sqrt(2 - sqrt2)/2# , because
#cos ((3pi)/8)# is positive.

Related questions
See all questions in Trigonometric Functions of Any Angle
Impact of this question
11523 views around the world
You can reuse this answer
Creative Commons License