What is #cos(2arcsin(-1/2))#?

1 Answer
Karthik G
Jan 2, 2016

#1/2#

Explanation:

#cos(2arcsin(-1/2))#

The range of #arcsin(x)# is #[-pi/2, pi/2]#

Ask the question which angle in #[-pi/2, pi/2]# gives #sin(x)=-1/2#

#sin(x)=-1/2# then #x# = -pi/6#

Our problem now becomes

#cos(2(-pi/6))#
#=cos(-pi/3)#
#=cos(pi/3) = 1/2# Answer

Related questions
Impact of this question
2831 views around the world
You can reuse this answer
Creative Commons License