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प्रश्न
Use De Moivres theorem and simplify the following:
`(cos2theta + "i"sin2theta)^7/(cos4theta + "i"sin4theta)^3`
उत्तर
`(cos2theta + "i"sin2theta)^7/(cos4theta + "i"sin4theta)^3`
= `(costheta + "i"sintheta)^(2 xx 7)/(costheta + "i"sintheta)^(4 xx 3)` ...[∵ cos n θ + i sin n θ = (cos θ + i sin θ)n]
= (cos θ + i sin θ)14 (cos θ + i sin θ)–12
= (cos θ + i sin θ)14−12
= (cos θ + i sin θ)2
= cos 2θ + i sin 2θ ...[∵ (cos θ + i sin θ)n = (cos n θ + i sin n θ)]
Alternate method:
`(cos2theta + "i"sin2theta)^7/(cos4theta + "i"sin4theta)^3`
= `(cos7(2theta) + "i"sin 7(2theta))/(cos3(4theta) + "i"sin3(4theta))` ...[∵ (cos θ + i sin θ)n = (cos n θ + i sin n θ)]
= `(cos 14theta + "i" sin14theta)/(cos 12theta + "i" sin 12theta)`
= `((cos 14theta + "i" sin14theta)(cos 12theta - "i" sin 12theta))/((cos12theta + "i" sin 12theta)(cos12theta - "i" sin 12theta))`
= `(cos 14theta cos 12theta - "i" cos 14theta sin 12theta + "i" sin14theta cos 12theta - "i"^2 sin 14thetasin12theta)/(cos^2 12theta - "i"^2 sin^2 12theta)`
= `((cos 14theta cos 12theta + sin 14theta sin 12theta) + "i"(sin14theta cos 12theta - cos 14thetasin12theta))/(cos^2 12theta + sin^2 12theta)`
= `(cos(14theta - 12theta) + "i" sin(14theta - 12theta))/(1)^2`
= cos 2θ + i sin 2θ
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