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## Answers (1)

hellooooo

first by definition, the correct mclaurin series expansion for the function f(x) = sin(x) is

∞

∑ ( -1 )^n * [ x^(2n+1) / (2n+1)! ] ..............but how can we get this? as the following:

n=0

first, i am going to define some terms such as

n = the number of derivatives & number of factorials

f^(n)(x) = is first original and the derived equations.

f^(n)(0) = plugging 0 in the original equation and derived equations.

an = is coefficient numbers of the variables.

IT WILL BE AS TABLE:

n ---- f^(n)(x) ------f^n(0)-------an

0------sin(x)---------0-----------(0/0!) = 0/1 = 0 (0! is equal to one )

1------cos(x)--------1-----------1/1! = 1

2-----(-sin(x)--------0-----------0/2! = 0

3----(-cos(x))-----(-1)--------(-1/3!) = -1/(1*2*3) = -1/6

4------sin(x) --------0------------ 0/4! = 0

5-----cos(x)---------1--------(1/5!) = 1/(1*2*3*4*5) = 1/120

now sin(3x) = 0 + 1*x^1 + 0 - [ 1*x^3/3! ] + 0 + [ 1*x^5/5! ] then it will be

since its alternating, and its increment for both: the exponent of the variable and to the denominator (with factorials) are odd number, it will be as 2n+1

but if its increment even numbers, it will be 2n

so the series is

∞

∑ ( -1 )^n * [ x^(2n+1) / (2n+1)! ] = x^1 - [ x^3/3! ] + [ x^5/5! ] -....

n = 0