Give a general formula for the expansion of tan nx in terms of tan x using combinatorics notation.

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pramodpandey | College Teacher | (Level 3) Valedictorian

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We know  `(cos(theta)+isin(theta))^n=cos(ntheta)+isin(ntheta)`

 `(cos(theta)+isin(theta))^n=`

`c(n,0)cos^n(theta)+ic(n,1)cos^(n-1)(theta)sin(theta)-`

`c(n,2)cos^(n-2)(theta)sin^2(theta)-ic(n,3)cos^(n-3)(theta)sin^3(theta)+`

`c(n,4)cos^(n-4)(theta)sin^4(theta)+......... `

`cos(ntheta )=c(n,0) cos^n(theta)-c(n,2)cos^(n-2)(theta)sin^2(theta)+c(n,4)cos^(n-4)(theta)sin^4(theta)-.........`

`sin(ntheta )=c(n,1) cos^(n-1)(theta)sin(theta)-c(n,3)cos^(n-3)(theta)sin^3(theta)+c(n,5)cos^(n-5)(theta)sin^5(theta)-.........`

`tan(ntheta)=sin(ntheta)/cos(ntheta)=(c(n,1) cos^(n-1)(theta)sin(theta)-c(n,3)cos^(n-3)(theta)sin^3(theta)+c(n,5)cos^(n-5)(theta)sin^5(theta)-.........)/(c(n,0) cos^n(theta)-c(n,2)cos^(n-2)(theta)sin^2(theta)+c(n,4)cos^(n-4)(theta)sin^4(theta)-.........)`

`=(c(n,1) tan(theta)-c(n,3)tan^3(theta)+c(n,5)tan^5(theta)-.........)/(c(n,0)-c(n,2)tan^2(theta)+c(n,4)tan^4(theta)-.........)`

``

`tan (nx)={sum_(r=1)^n C (n,2r-1)(-1)^(r-1) tan ^(2r-1) (x)}/{sum_(r=0)^n C (n,2r)(-1)^r tan ^(2r) (x) }` and

`2r<=n` .

This is summation notation answer.

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