Q. 29

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Answer :

Formula: -


(i) sin2θ + cos2θ = 1


(ii) (a + b)3 = a3 + b3 + 3a2b + 3b2a


cos12θ + 3cos10θ + 3cos8θ + cos6θ + 2cos4θ + 2cos2θ – 2


sinθ + sin2θ = 1


sinθ = 1 – sin2θ = cos2θ …… (1)


cos12θ + 3cos10θ + 3cos8θ + cos6θ + 2cos4θ + 2cos2θ – 2


taking 2 as common


= cos12θ + 3cos10θ + 3cos8θ + cos6θ + 2(cos4θ + cos2θ – 1)


using formula (ii)


= (cos4θ + cos2θ)3 + 2(cos4θ + cos2θ – 1)


using equation(1)


= (sin2θ + cos2θ)3 + 2(sin2θ + cos2θ – 1)


using formula (i)


= 1 + 2(1 – 1)


= 1


cos12θ + 3cos10θ + 3cos8θ + cos6θ + 2cos4θ + 2cos2θ – 2 = 1


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