Question

expansion of sin x cos x is ascending powers of X is​

Answer 1

$\textbf{Given:}$

$\mathsf{sinx\,cosx}$

$\textbf{To find:}$

$\textsf{Expansion of sinx cosx}$

$\textbf{Solution:}$

$\underline{\textsf{Concept used:}}$

$\boxed{\mathsf{sinx=\dfrac{x}{1!}-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\;.\;.\;.\;.\;.\;.\;.\;.}}$

$\mathsf{Consider,}$

$\mathsf{sinx\,cosx}$

$\mathsf{=\dfrac{1}{2}(2\,sinx\,cosx)}$

$\mathsf{=\dfrac{1}{2}(sin\,2x)}$

$\mathsf{=\dfrac{1}{2}\left(\dfrac{(2x)}{1!}-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^5}{5!}-\;.\;.\;.\;.\;.\;.\;.\;.\right)}$

$\mathsf{=\dfrac{1}{2}\left(\dfrac{2x}{1!}-\dfrac{8x^3}{3!}+\dfrac{32x^5}{5!}-\;.\;.\;.\;.\;.\;.\;.\;.\right)}$

$\mathsf{=\dfrac{x}{1!}-\dfrac{4x^3}{3!}+\dfrac{16x^5}{5!}-\;.\;.\;.\;.\;.\;.\;.\;.}$

$\implies\boxed{\mathsf{sinx\,cosx=x-\dfrac{4x^3}{3!}+\dfrac{16x^5}{5!}-\;.\;.\;.\;.\;.\;.\;.\;.}}$

$\textbf{Find more:}$

Expansion of sinX is ascending power of X​

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Answer 2

Answer:

expansion of sin x cos x is ascending powers of X is