Question

How do you find the Maclaurin Series for $(\sin x)(\cos x)$?

Answer 1

Hint: Maclaurin series can be said to be a function for an infinite series of sum of the functions derivative based on a condition. We have the Maclaurin series individually for \[\sin x\] and \[\cos x\]. But the product of these functions would be a little tedious work. So will convert the functions into a single component function either of \[\sin x\] or \[\cos x\]. And we will do this by making use of the trigonometric function that we know.

Complete step by step answer:
According to the question we have to find for $(\sin x)(\cos x)$,
So we will first start by converting the question completely either in terms of \[\sin x\] or \[\cos x\].
We know that,
\[\sin 2x=2\sin x\cos x\]
On rearranging we get, \[\sin x\cos x=\dfrac{\sin 2x}{2}\]
So now our question becomes, \[(\sin x)(\cos x)=\dfrac{1}{2}\sin 2x\]
We need to find the Maclaurin series for \[\dfrac{1}{2}\sin 2x\]
We know that for \[\sin x\]function, the maclaurin series is:
 \[\sin x=\sum\limits_{k=0}^{\infty }{{{(-1)}^{k}}\dfrac{{{x}^{2k+1}}}{(2k+1)!}}\]
Therefore, we have,
\[\dfrac{1}{2}\sin 2x=\dfrac{1}{2}\left( \sum\limits_{k=0}^{\infty }{\begin{align}
  & {{(-1)}^{k}}\dfrac{{{(2x)}^{2k+1}}}{(2k+1)!} \\
 & \\
\end{align}} \right)\]
This way we only had to substitute the value of x in the sine function.
We can write this as,
\[(\sin x)(\cos x)=\dfrac{1}{2}\left( \sum\limits_{k=0}^{\infty }{\begin{align}
  & {{(-1)}^{k}}\dfrac{{{(2x)}^{2k+1}}}{(2k+1)!} \\
 & \\
\end{align}} \right)\]

Note:
The question can also be solved by multiplying the individual maclaurin series for sine and cosine functions. This method will also give rise to the same answer. Cosine function has similar maclaurin series which is:
\[\cos x=\sum\limits_{k=0}^{\infty }{{{(-1)}^{k}}\dfrac{{{x}^{2k}}}{(2k)!}}\] and we already know \[\sin x=\sum\limits_{k=0}^{\infty }{{{(-1)}^{k}}\dfrac{{{x}^{2k+1}}}{(2k+1)!}}\]
Multiplying both the functions we will ultimately get,
\[(\sin x)(\cos x)=\dfrac{1}{2}\left( \sum\limits_{k=0}^{\infty }{{{(-1)}^{k}}\dfrac{{{(2x)}^{2k+1}}}{(2k+1)!}} \right)\]
While multiplying both the functions, the common factors in the maclaurin series of both the functions should be carefully written and evaluated else there is a possibility of getting a wrong answer.
Therefore, the answer is \[(\sin x)(\cos x)=\dfrac{1}{2}\left( \sum\limits_{k=0}^{\infty }{\begin{align}
  & {{(-1)}^{k}}\dfrac{{{(2x)}^{2k+1}}}{(2k+1)!} \\
 & \\
\end{align}} \right)\]