Prove that? cos(π/10-A)cos(π/10+A)+cos(2π/5-A)cos(2π/5+A)=cos2A

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Answer 1 · 2017-12-11T06:15:05

#cos(x+y)cos(x-y)#

#=cos^2xcos^2y-sin^2xsin^2y#

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

#=cos^2x-cos^2xsin^2y-sin^2y+cos^2xsin^2y#

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

Applying this relation

#LHS=cos(π/10-A)cos(π/10+A)+cos(2π/5-A)cos(2π/5+A)#

#=cos^2(π/10)-sin^2(A)+cos^2(2π/5)-sin^2 (A)#

#=1/2(1+cos(2*π/10))-1+cos^2(A)+1/2(1+cos(4π/5)-sin^2 (A)#

#=1/2+1/2-1+1/2(cos(pi/5)+cos(pi-pi/5))+cos^2A-sin^2A#

#=1/2(cos(pi/5)-cos(pi/5))+cos2A#

#=cos2A=RHS#