show that cos²x+cos²(x+60)+cos²(x-60)=3/2
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Step-by-step explanation:
we know cos2x+1=2cos²x
cos²x=cos2x+1/2.......(1)
replacing x by (x+60)
cos²(x+60)=cos2(x+60)+1/2
=cos(2x+120)+1/2........(2)
again, replace x by (x-60)
cos²(x-60)=cos(2x-120)+1/2.......(3)cos2
LHS:
cos²x+cos²(x+60)+cos²(x-60)
=1+cos2x/2+1+cos(2x+120)/2 +cos2x(2x-120)/2
=1/2[1+cos2x+1+cos(x+60)+1+ cos(x-60)]
=1/2[3+cos2x+2cos(2x+120+2x-120)/2 +2cos(2x+120-2x+120)/2]
=1/2[3+cos2x+2cos2xcos120]
=1/2[3+cos2x+2cos2xcos(180-60)]
=1/2[3+cos2x-2cos2xcos60]
=1/2[3+cos2x-2cos2x.1/2]
=1/2[3+cos2x-cos2x]
=1/2[3]
=3/2
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