Question

How do you find the McLaurin series expansion of \[f(x)=\sin x\cos x\] ?

Answer 1

Hint: In the above question we are supposed to find the maclaurin series for \[\sin x\cos x\] . the formula for McLaurin series is \[{{f}^{n}}(0)\dfrac{{{x}^{n}}}{n!}=f(0)+f'(0)x+\dfrac{f''(0)}{2!}\] where \[f\] is the function whose series we want to derive. First, we will calculate the derivatives of the given function. Then we will evaluate the function and the derivatives at \[x=a\] . After that we will add all the values and put \[a=0\]. The resulting series is the McLaurin series.

Complete step by step answer:
A McLaurin series is a form of power series by which we can calculate an approximate value of a function at a particular input value which is close to zero.it is useful in sine function. It is a special case of the Taylor series. In other words, we can say that the McLaurin series is a Taylor series centered about zero. Therefore, the McLaurin series for the function \[f\] is
\[{{f}^{n}}(0)\dfrac{{{x}^{n}}}{n!}=f(0)+f'(0)x+\dfrac{f''(0)}{2!}\]
 Now in the above question we have \[f(x)=\sin x\cos x\].
The McLaurin expansion for \[\sin x\] is,
\[\sin x=\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}}\dfrac{{{(x)}^{2n+1}}}{(2n+1)!}\]
Also, \[\sin 2x=2\sin x\cos x\] is a trigonometric identity thus making the use of this we can write the series for \[\sin x\cos x\] .
\[\begin{align}
  & \sin 2x=\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}}\dfrac{{{(2x)}^{2n+1}}}{(2n+1)!} \\
 & \sin 2x=\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}{{2}^{2n+1}}}\dfrac{{{(x)}^{2n+1}}}{(2n+1)!} \\
\end{align}\]
Now \[\sin 2x=2\sin x\cos x\]
Or
\[\sin x\cos x=\dfrac{1}{2}\sin 2x\]
Therefore, \[\sin x\cos =\dfrac{1}{2}\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}{{2}^{2n+1}}}\dfrac{{{(x)}^{2n+1}}}{(2n+1)!}=\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}{{2}^{2n}}}\dfrac{{{(x)}^{2n+1}}}{(2n+1)!}\]
Hence the McLaurin series for the given function is \[\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}{{2}^{2n}}}\dfrac{{{(x)}^{2n+1}}}{(2n+1)!}\].

Note: In the above question we have derived the McLaurin series for \[\sin x\cos x\] by the help of basic trigonometric formulas, so remember the basic trigonometric formulas for solving these types of questions. Do the integration step by step in order to avoid confusion. Perform the calculations carefully. Keep in mind the procedure of solving this type of question.