derive the formula for finding the sum of the series of the following sequences :
 
1) Sinx + Sin2x + Sin3x + ....+ Sinnx
 
2) Cosx + Cos2x + Cos3x +.....+ Cosnx 

Asked by nitishkrnehu09 | 6th Nov, 2016, 09:07: PM

Expert Answer:

begin mathsize 16px style To space prove space that colon space sin space straight x space plus space sin 2 straight x space plus space sin 3 straight x space plus space..... plus space sin space nx space equals space fraction numerator sin open parentheses begin display style fraction numerator straight n plus 1 over denominator 2 end fraction end style close parentheses straight x. sin nx over 2 over denominator sin straight x over 2 end fraction
For space straight n space equals space 1 comma
LHS space equals space sin space straight x
RHS space equals space fraction numerator sin open parentheses begin display style fraction numerator 1 plus 1 over denominator 2 end fraction end style close parentheses straight x. sin straight x over 2 over denominator sin straight x over 2 end fraction equals fraction numerator sinx. space sin straight x over 2 over denominator sin straight x over 2 end fraction equals sinx
So space the space result space is space true space for space straight n space equals space 1.
Let space the space result space be space true space for space straight n space equals space straight k comma
rightwards double arrow sin space straight x space plus space sin 2 straight x space plus space sin 3 straight x space plus space..... plus space sin space kx space equals space fraction numerator sin open parentheses begin display style fraction numerator straight k plus 1 over denominator 2 end fraction end style close parentheses straight x. sin kx over 2 over denominator sin straight x over 2 end fraction.... left parenthesis 1 right parenthesis
To space prove space the space result space for space straight n space equals space straight k plus 1

sin space straight x space plus space sin 2 straight x space plus space sin 3 straight x space plus space..... plus space sinkx plus sin space left parenthesis straight k plus 1 right parenthesis straight x
equals fraction numerator sin open parentheses begin display style fraction numerator straight k plus 1 over denominator 2 end fraction end style close parentheses straight x. sin kx over 2 over denominator sin straight x over 2 end fraction plus sin space left parenthesis straight k plus 1 right parenthesis straight x
equals fraction numerator sin open parentheses begin display style fraction numerator straight k plus 1 over denominator 2 end fraction end style close parentheses straight x. sin kx over 2 over denominator sin straight x over 2 end fraction plus 2 sin space open parentheses fraction numerator straight k plus 1 over denominator 2 end fraction close parentheses straight x space cos open parentheses fraction numerator straight k plus 1 over denominator 2 end fraction close parentheses straight x
equals fraction numerator sin open parentheses begin display style fraction numerator straight k plus 1 over denominator 2 end fraction end style close parentheses straight x over denominator sin straight x over 2 end fraction open square brackets sin kx over 2 plus 2 space cos open parentheses fraction numerator straight k plus 1 over denominator 2 end fraction close parentheses straight x. space sin straight x over 2 close square brackets
equals fraction numerator sin open parentheses begin display style fraction numerator straight k plus 1 over denominator 2 end fraction end style close parentheses straight x over denominator sin straight x over 2 end fraction open square brackets sin kx over 2 plus 2 space cos open parentheses fraction numerator straight k plus 1 over denominator 2 end fraction close parentheses straight x. space sin straight x over 2 close square brackets
equals fraction numerator sin open parentheses begin display style fraction numerator straight k plus 1 over denominator 2 end fraction end style close parentheses straight x over denominator sin straight x over 2 end fraction open square brackets sin kx over 2 plus sin open curly brackets open parentheses fraction numerator straight k plus 1 over denominator 2 end fraction close parentheses straight x plus straight x over 2 close curly brackets minus sin open curly brackets open parentheses fraction numerator straight k plus 1 over denominator 2 end fraction close parentheses straight x minus straight x over 2 close curly brackets close square brackets space space space space space space.... left parenthesis Since space 2 cosAsinB space equals space sin left parenthesis straight A plus straight B right parenthesis space minus space sin left parenthesis straight A space minus space straight B right parenthesis right parenthesis
equals fraction numerator sin open parentheses begin display style fraction numerator straight k plus 1 over denominator 2 end fraction end style close parentheses straight x over denominator sin straight x over 2 end fraction open square brackets sin kx over 2 plus sin open parentheses fraction numerator straight k plus 2 over denominator 2 end fraction close parentheses straight x minus sin kx over 2 close square brackets
equals fraction numerator sin open parentheses begin display style fraction numerator straight k plus 1 over denominator 2 end fraction end style close parentheses straight x over denominator sin straight x over 2 end fraction open square brackets sin fraction numerator left parenthesis straight k plus 1 right parenthesis plus 1 over denominator 2 end fraction straight x close square brackets
Thus comma space it space is space true space for space straight n space equals space straight k space plus space 1
Therefore comma space it space will space be space true space for space all space values space of space straight n space that space belongs space to space the space set space of space natural space numbers
Hence space proved.

left parenthesis ii right parenthesis space Similarly comma space the space second space part space can space be space derived.
Request space space you space to space please space post space one space question space per space query. end style

Answered by Rebecca Fernandes | 6th Nov, 2016, 10:14: PM

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