Question
Question: How do you find the McLaurin series expansion of \[f(x)=\sin x\cos x\] ?...
How do you find the McLaurin series expansion of f(x)=sinxcosx ?
Solution
In the above question we are supposed to find the maclaurin series for sinxcosx . the formula for McLaurin series is fn(0)n!xn=f(0)+f′(0)x+2!f′′(0) 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
fn(0)n!xn=f(0)+f′(0)x+2!f′′(0)
Now in the above question we have f(x)=sinxcosx.
The McLaurin expansion for sinx is,
sinx=n=0∑∞(−1)n(2n+1)!(x)2n+1
Also, sin2x=2sinxcosx is a trigonometric identity thus making the use of this we can write the series for sinxcosx .