Question #6b0f2
1 Answer
Jan 16, 2018
See below...
Explanation:
We know,
#sin^2x+cos^2x=1#
#=>(sinx-cosx)^2+2sinxcosx=1#
#=>(1/2)^2+2sinxcosx=1#
#=>2sinxcosx=3/4#
#=>sinxcosx=3/8# Now,
#sin^4x+cos^4x#
#=(sin^2x+cos^2x)^2-2sin^2xcos^2x#
#=1-2xx(3/8)^2#
#=1-9/32#
#=23/32# Hope it helps...
Thank you...
