#cos 4x=8cos^4(x)-8cos^2x + 1 # and #cos 4x=8sin^4(x)-8sin^2x + 1#,can u prove it?

Question

21452 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-08T08:07:57

#LHS=cos4x#

#=2cos^2 (2x)-1#

#=2(cos (2x))^2-1#

#=2(2cos ^2x-1)^2-1#

#=2(4cos ^4x-4cos^2x+1)-1#

#=8cos ^4x-8cos^2x+2-1#

#=8cos ^4x-8cos^2x+1=RHS#

Again

#LHS=cos4x#

#=2cos^2 (2x)-1#

#=2(1-2sin^2x))^2-1#

#=2(1-4sin^2x+4sin^4x)-1#

#=2-8sin^2x+8sin^4x-1#

#=8sin^4x-8sin^2x+1=RHS#

#sin^2x+cos^2x=1#

#cos^2x = 1-sin^2x#

substitute in the equation as follows

#8cos^4x-8cos^2x+1 = 8cos^2x(cos^2x-1)+1#

#=8(1-sin^2x)(1-sin^2x-1)+1#

#=8(1-sin^2x)(-sin^2x)+1#

#=8sin^4x-8sin^2x+1#