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Question
Prove the following:
cos2x + cos2(x + 120°) + cos2(x – 120°) = `3/2`
Solution
L.H.S. = cos2x + cos2(x + 120°) + cos2(x – 120°)
= `(1 + cos 2x)/2 + (1 + cos2(x + 120^circ))/2 + (1 + cos2(x - 120^circ))/2` ......`[∵ cos^2θ = (1+cos2θ)/2]`
`=3/2 + 1/2[cos2x + cos(2x + 240^circ)+cos(2x-240°)]`
= `3/2 + 1/2(cos2x + cos2x cos 240^circ-sin2x sin240°+cos2x cos240°+sin2x sin240°)`
= `3/2+1/2(cos2x+2cos2x cos240°)`
= `3/2 + 1/2[cos 2x +2 cos2x cos(180° + 60°)]`
= `3/2 + 1/2[cos2x + 2cos2x(- cos 60^circ)]`
= `3/2 + 1/2[cos2x - 2cos2x (1/2)]`
= `3/2+1/2(cos2x-cos2x)`
= `3/2+1/2(0)`
= `3/2`
= R.H.S.
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