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Use a Maclaurin series in this table to obtain the Maclaurin series for the given function. :

f(x) =9sin ^2 (x) [Hint: use sin^2(x) 1/2(1-cos(2x)]

User Nicko Po
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The Maclaurin series for
f(x)=9 \sin ^2(x) is given by
\sum_(n-0)^(\infty) (9(-1)^n(2 x)^(2 n))/((2 n) !) . This series represents the expansion of
9 \sin ^2(x) around the point x=0 using the derivatives of the function at that point. The series includes terms for all non-negative integer values of n. The general term in the series is
(9(-1)^n(2 x)^(2 n))/((2 n) !)

The Maclaurin series expansion for
f(x)=9 \sin ^2(x) can be obtained by substituting the given hint into the function. The Maclaurin series for
\sin ^2(x) can be expressed using the hint as follows:


\sin ^2(x)=(1)/(2)(1-\cos (2 x))

Now, substitute this into the original function
f(x)=9 \sin ^2(x) :


f(x)=9 \cdot (1)/(2)(1-\cos (2 x))=(9)/(2)(1-\cos (2 x))

Now, let's express cos(2x) as a Maclaurin series. The Maclaurin series for cos(x) is well-known:


\cos (x)=\sum_(n=0)^(\infty) ((-1)^n)/((2 n) !) x^(2 n)

So, for cos(2x):


\cos (2 x)=\sum_(n-0)^(\infty) ((-1)^n)/((2 n) !)(2 x)^(2 n)

Now, substitute this into the expression for f(x):


f(x)=(9)/(2)\left(1-\sum_(n-0)^(\infty) ((-1)^n)/((2 n) !)(2 x)^(2 n)\right)

To obtain the Maclaurin series for f(x), you can distribute the 9/2 and simplify the expression:


f(x)=(9)/(2)-(9)/(2) \sum_(n-0)^(\infty) ((-1)^n)/((2 n) !)(2 x)^(2 n)

So, the Maclaurin series for
f(x)=9 \sin ^2(x) is
\sum_(n-0)^(\infty) (9(-1)^n(2 x)^(2 n))/((2 n) !)

User Abuder
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