#[cos(alpha)+cos(beta)]²+[sin(alpha)-sin(beta)]²=4cos²((alpha+beta)/2)#?

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Answer 1 · 2017-11-30T10:03:45

#LHS=[cos(alpha)+cos(beta)]^2+[sin(alpha)-sin(beta)]^2#

#=[2cos((alpha+beta)/2)cos((alpha-beta)/2)]^2+[2cos((alpha+beta)/2)sin((alpha-beta)/2)]^2#

#=4cos²((alpha+beta)/2)(cos^2((alpha-beta)/2)+sin^2((alpha-beta)/2))#

#=4cos²((alpha+beta)/2)#

Answer 2 · 2017-11-30T10:16:56

See the proof below

We need

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

#cos(a+b)=cosacosb-sinasinb#

#cos2a=2cos^2a-1#

#cos(a+b)=2cos^2((a+b)/2)-1#

Therefore,

#LHS=(cosalpha+cosbeta)^2+(sinalpha-sinbeta)^2#

#=cos^2alpha+cos^2beta+2cosalphacosbeta+sin^2alpha+sin^2beta-2sinalphasinbeta#

#=2+2cosalphacosbeta-2sinalphasinbeta#

#=2+2cos(alpha+beta)#

#=2+2(2cos^2((alpha+beta)/2)-1)#

#=2+4cos^2((alpha+beta)/2)-2#

#=4cos^2((alpha+beta)/2)#

#=RHS#

#QED#