Question

expansion of 7 cos theta ​

Answer 1

Answer:

Step-by-step explanation:

x=cos⁡θ+i sin⁡θ \; \; \; \therefore

\dfrac{1}{x}=cos⁡θ-i sin⁡θ

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\therefore

x^m=(cos⁡θ+i sin⁡θ )^m \;=\; cos⁡mθ+i sin⁡mθ \; \; \; \; \ldots By \; De \; Moivre's \; Theorem

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Also,\dfrac{1}{x^m} =(cos⁡θ-i sin⁡θ )^m \;=\;cos⁡mθ-i sin⁡mθ\; \; \; \; \ldots

By \; De \; Moivre's \; Theorem

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\therefore

x+\dfrac{1}{x}=2 cos⁡θ \; \; and \; \; x^m+

\dfrac{1}{x^m}

\;=\; 2 cos⁡mθ

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\therefore

x+\dfrac{1}{x} \;=\; 2 cos⁡θ

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and \; \; x^m+ \dfrac{1}{x^m} =2 cos⁡mθ

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\therefore

\bigg( x+\dfrac{1}{x} \bigg)^7 \;=\; 2^7 cos^7 θ

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\therefore

x^7+7x^6 \dfrac{1}{x}+21x^5 \dfrac{1}{x^2} +35x^4 \dfrac{1}{x^3} +35x^3 \dfrac{1}{x^4} +21x^2 \dfrac{1}{x^5} +7x \dfrac{1}{x^6} +\dfrac{1}{x^7} =2^7 cos^7 θ

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\therefore

(x^7+\dfrac{1}{x^7 })+7(x^5+\dfrac{1}{x^5 } )+21(x^3+\dfrac{1}{x^3 } )+35(x+\dfrac{1}{x })=128cos^7 θ

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\therefore (2 cos⁡7θ )+7(2 cos⁡5θ )+21(2 cos⁡3θ )+35(2 cos⁡θ )=128cos^7 θ

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\therefore

cos^7 θ=\dfrac{1}{64}[cos⁡7θ+7 cos⁡5θ+21 cos⁡3θ+cos⁡θ]