How do I prove this? cot(x)(1-cos(2x))=sin(2x)

4 Answers
Apr 13, 2018

#LHS=cotx(1-cos2x)#

#=cosx/sinx*2sin^2x#

#=2sinx*cosx=sin2x=RHS#

c#color(purple)(ot(x)(1-cos(2x))=sin(2x)#

Explanation:

#color(green)(N.B: cos(2x) = cos^2x - sin^2x#

#color(green)(sin(2x) = 2sinxcosx#

#cot(x) = 1/tan(x) = 1/ (sinx/cosx) = cos(x)/sin(x)#

#cot(x)(1-cos(2x))#

#=> [cos(x)/sin(x)] [1- ( cos^2x - sin^2x#)]

#=> [cos(x)/sin(x)] [1- cos^2x + sin^2x#]

#=> [cos(x)/sin(x)] [( sin^2x + cos^2x) - cos^2x + sin^2x#]

#=> [cos(x)/sin(x)] [2sin^2x#]

#=>2sinxcosx#

Since

#sin(2x) = 2sinxcosx#

Hence,

#color(crimson)(cot(x)(1-cos(2x)) = sin(2x)#

# Q. E. D #

Apr 13, 2018

#cotx(1-cos2x)=sin2x#

Explanation:

convert #cotx# into sins and cosines with the identity

#cotx=cosx/sinx#

#cosx/sinx(1-cos2x)=sin2x#

turn #sin2x# in terms of a single multiple of #x# using the double angle formula

#sin2x=2cosxsinx#

#cosx/sinx(1-cos2x)=2cosxsinx#

expand the brackets

#cosx/sinx+(-cosx*cos2x)/sinx=2cosxsinx#

using one of the double angle formula for cosine

#cos2x=1-2sinx#

substitute

#cosx/sinx+(-cosx(1-2sin^2x))/sinx=2cosxsinx#

expand the brackets

#cosx/sinx+(-cosx+2cosxsin^2x)/sinx=2cosxsinx#

add the fractions

#(cosx-cosx+2cosxsin^2x)/sinx=2cosxsinx#

cancel #cosx#

#(cancel(cosx-cosx)+2cosxsin^2x)/sinx=2cosxsinx#

#(2cosxsin^cancel(2)x)/cancelsinx=2cosxsinx#

#2cosxsinx=2cosxsinx#

Apr 13, 2018

#"see explanation"#

Explanation:

#"using the "color(blue)"trigonometric identities"#

#•color(white)(x)cotx=cosx/sinx#

#•color(white)(x)cos2x=2cos^2x-1" and "sin2x=2sinxcosx#

#•color(white)(x)sin^2x+cos^2x=1#

#"consider the left side"#

#rArrcosx/sinx(1-(2cos^2x-1))#

#=cosx/sinx(2-2cos^2x)#

#=cosx/sinx(2(1-cos^2x))#

#=cosx/sinx(2sin^2x)#

#=2sinxcosx#

#=sin2x="right side "rArr"verified"#