Math, asked by suhani4532, 3 months ago

Expansion of sin x cos x in ascending powers of x is​

Answers

Answered by MaheswariS
6

\textbf{Given:}

\mathsf{f(x)=sinx\;cosx}

\textbf{To find:}

\textsf{Maclaurin series expansion of sinx cosx}

\textbf{Solution:}

\textsf{Consider,}

\mathsf{f(x)=sinx\;cosx=\dfrac{2sinx\;cosx}{2}=\dfrac{1}{2}sin2x}

\mathsf{f(x)=\dfrac{1}{2}sin2x\;\implies\;f(0)=\dfrac{1}{2}sin(0)=0}

\mathsf{f^1(x)=\dfrac{1}{2}(cos2x)2=cos2x\;\implies\;f^1(0)=cos0=1}

\mathsf{f^2(x)=(-sin2x)2\;\implies\;f^2(0)=2(-sin0)=0}

\mathsf{f^3(x)=(-cos2x)2^2\;\implies\;f^3(0)=2^2(-cos0)=-4}

\mathsf{Now,}

\textsf{The Maclaurin series expansion of f(x) is}

\mathsf{f(x)=f(0)+\dfrac{f^1(0)}{1!}x^1+\dfrac{f^2(0)}{2!}x^2+\;.\;.\;.\;.\;.}

\mathsf{sinx\;cosx=0+\dfrac{1}{1!}x^1+\dfrac{0}{2!}x^2+\dfrac{(-4)}{3!}x^3+\;.\;.\;.\;.\;.}

\mathsf{sinx\;cosx=x^1-\dfrac{4}{1.2.3}x^3+\;.\;.\;.\;.\;.}

\implies\boxed{\mathsf{sinx\;cosx=x^1-\dfrac{2}{3}x^3+\;.\;.\;.\;.\;.}}

\textbf{Find more:}

Expansion of sinX is ascending power of X​

https://brainly.in/question/28857947

Expansion of sinX is ascending power of X​

https://brainly.in/question/28857947

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