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tan nA= n C 1 tan A- n C 3 tan 3 A + n C 5 tan 5 A …/1- n C 2 tan 2 A+ n C 4 tan 4 A-…


tan nA= nC1 tan A-nC3 tan3A + nC5 tan5A …/1- nC2 tan2A+ nC4 tan4A-…


 


Grade:11

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear Sonam

we know 

cosnA + iSin nA =(cosA +isinA)n

open bracket with the help of binomial

cosnA + iSin nA =(cosA +isinA)n = cosnA + nC1 cosn-1A(isinA) + nC2cosn-2A(isinA)2 +.......

                             = (cosnA  - nC2cosn-2A(sinA)2 +.... ) + i(nC1 cosn-1A(sinA) - nC3 cosn-3A(sinA)3 .......)

 

compair Real  part

 Cos nA = (cosnA  - nC2cosn-2A(sinA)2 +.... )

              =cosnA(1-nC2 tan2A-nC4 tan3A -…)............................1

 

compair imaginery part

 Sin nA = (nC1 cosn-1A(sinA) - nC3 cosn-3A(sinA)3 .......)

              =cosnA(nC1 tanA - nC3(tanA)3 .......  ...........)  .............2

devide equatin 2 from 1

tan nA = nC1 tan A-nC3 tan3A + nC5 tan5A …/1- nC2 tan2A+ nC4 tan4A-…


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Badiuddin

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