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(1-costheta+isintheta)^(6)...

`(1-costheta+isintheta)^(6)`

A

`2^(6)sin^(6)""(theta)/(2)(cos3theta+isin3theta)`

B

`2^(6)sin^(6)""(theta)/(2)(-cos3theta+isin3theta)`

C

`2^(6)sin^(6)""(theta)/(2)(cot3theta+isin3theta)`

D

`2^(6)sin^(6)""(theta)/(2)(-cot3theta+isin3theta)`

Text Solution

Verified by Experts

The correct Answer is:
B
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(1+costheta-isintheta)^(n)

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(1+costheta+isintheta)^(n)+(1+costheta-isintheta)^(n)

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If theta=pi//6 then the 10th term of the series 1+(costheta+isintheta)+(costheta+isintheta)^(2)+...... is

(costheta+isintheta)^(10)+(costheta-isintheta)^(10)

Assertion (A) : (Costheta+iSintheta)^(10)+(Costheta-iSintheta)^(10)=2cos10theta Reason (R) : (1+costheta+isintheta)^n+(1+costheta-isintheta)^n= 2^(n+1)cos^n(theta/2)cos((ntheta)/(2))

((costheta+isintheta)^(5)(cos3theta-isin3theta)^(6))/((cos2theta+isin2theta)^(3)(cos4theta-isin4theta)^(5))