# How do you find the taylor series of #f(x)=sinx# at #a=pi/6#?

##### 2 Answers

Using Calculator:

#### Explanation:

Solution Strategy:

Use the definition of Taylor series for a function,

1)

Expansion

a special case of Taylor series

2) Find the

We see that all even derivatives are zero and all odd derivative toggle between -1 and 1. Thus we write

Now let's evaluate

Using Calculator:

It is likely that your calculator is using Taylor or likely MAcLaurin series to compute,

#### Explanation:

I believe the question is asking for a Taylor series centered around

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

First, calculate the first few derivatives of

#f(x) = sin(x)#

#f'(x) = cos(x)#

#f''(x) = -sin(x)#

#f'''(x) = -cos(x)#

#f^(4)(x) = sin(x)#

Since

Now, we plug in

#f(pi/6) = sin(pi/6) = 1/2#

#f'(pi/6) = cos(pi/6) = sqrt3/2#

#f''(pi/6) = -sin(pi/6)= -1/2#

#f'''(pi/6) = -cos(pi/6)= -sqrt3/2#

#f^((4))(pi/6) = sin(pi/6) = 1/2#

The formula for the

#f^((n))(a)/(n!)*(x-a)^n#

So, we can write out the first few terms of our Taylor polynomial for

#sin(x) = (1/2)/(0!) * (x-pi/6)^0+(sqrt3/2)/(1!) * (x-pi/6)^1 + (-1/2)/(2!) * (x-pi/6)^2 + (-sqrt3/2)/(3!) * (x-pi/6)^3 + (1/2)/(4!) * (x-pi/6)^4+...#

Simplifying this a bit, we get:

#sin(x) = 1/2 + (sqrt3(x-pi/6))/2 - (x-pi/6)^2/(2*2!) - (sqrt3(x-pi/6)^3)/(2*3!) + (x-pi/6)^4/(2*4!)+...#

This is technically the final answer but the formal way to write it is with summation notation. If we group every two terms together, we can write this polynomial as:

#sin(x) = sum_(n=0)^oo (-1)^n(((x-pi/6)^(2n))/(2*(2n)!) + (sqrt3(x-pi/6)^(2n+1))/(2*(2n+1)!))#

And remember: don't be bogged down by how big or complex this summation is; it's just a set of instructions for writing out the polynomial we just made.

*Final Answer*