Question Number 102450 by naka3546 last updated on 09/Jul/20

Commented bynaka3546 last updated on 09/Jul/20

how  to  get  both  of  them ? help  me, please

Commented bybemath last updated on 09/Jul/20

maclaurin series

Answered by Rio Michael last updated on 09/Jul/20

let f(x)=sin x⇒f(0) = 0  f ′(x) = cos x ⇒ f ′(0) = 1  f ′′(x)= −sin x ⇒f′′(0)= 0  f ′′′(x)= − cos x ⇒ f ′′′(0) = −1  f ^(iv) (x) = sinx ⇒ f ^(iv) (0) = 0  f ^v (x) = cos x ⇒ f^v (0) = 1  sin x = x −(x^3 /(3!)) + (x^5 /(5!))−...Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/((2n + 1)!))

Answered by Dwaipayan Shikari last updated on 09/Jul/20

f(x)=f(0)+f ′(0)x+((f ′′(0))/(2!))x^2 +((f ′′′(0))/(3!))x^3 +((f ′′′′(0))/(4!))x^4 +....  f(x)=sinx  f ′(x)=cosx  f′′(x)=−sinx  f ′′′(x)=−cosx  f^(iv) (x)=sinx    so  f(x)=sinx=0+x+0−(x^3 /(3!))+0+(x^5 /(5!))....  sinx=x−(x^3 /(3!))+(x^5 /(5!))−(x^7 /(7!))+........

Answered by PRITHWISH SEN 2 last updated on 09/Jul/20

cos x+isin x=e^(ix) = 1+ix−(x^2 /(2!))−i(x^3 /(3!))+(x^4 /(4!))+i(x^5 /(5!))−(x^6 /(6!))−i(x^7 /(7!))+...  Now compairing the real and imaginary part  cos x=1−(x^2 /(2!))+(x^4 /(4!))−(x^6 /(6!))+.....  and  sin x=x−(x^3 /(3!))+(x^5 /(5!))−(x^7 /(7!))+......  please check your question.  the 1st one also has a trignometrical approach  but it is too lengthy.