Question #f55f5
1 Answer
Nov 7, 2017
See the answer below...
Explanation:
LHS
#(1-cos8x)/8#
#=(2sin^2 4x)/8# [by IDENTITY]
#=(sin^2 4x)/4#
#=(sin4x)^2 /4#
#=(2sin2xcos2x)^2/4# [#color(red)(sin2x=2sinxcosx# [IDENTITY]
#=(4sin^2 2xcos^2 2x)/4#
#=(color(red)cancel 4xxsin^2 2xcos^2 2x)/color(red)(cancel4)#
#=sin^2 2xcos^2 2x# [RHS]Hope it helps...
Thank you...
